Solving logarithmic equations is the final lesson. Formulas of logarithms. Logarithms solution examples
The listed equalities when converting expressions with logarithms are used both from right to left and from left to right.
It is worth noting that it is not necessary to memorize the consequences of the properties: when carrying out transformations, you can get by with the basic properties of logarithms and other facts (for example, those for b≥0), from which the corresponding consequences follow. " By-effect This approach only manifests itself in the fact that the solution will be a little longer. For example, in order to do without the consequence, which is expressed by the formula 
, and starting only from the basic properties of logarithms, you will have to carry out a chain of transformations of the following form: 
.
The same can be said about the last property from the above list, which corresponds to the formula 
, since it also follows from the basic properties of logarithms. The main thing to understand is that it is always possible for the degree of a positive number with a logarithm in the exponent to swap the base of the degree and the number under the sign of the logarithm. In fairness, we note that examples involving the implementation of transformations of this kind are rare in practice. We will give a few examples below.
Converting numeric expressions with logarithms
We remembered the properties of logarithms, now it's time to learn how to put them into practice to transform expressions. It is natural to start with the transformation of numeric expressions, and not expressions with variables, since it is more convenient and easier to learn the basics on them. So we will do, and we will start with a very simple examples to learn how to choose the desired property of the logarithm, but we will gradually complicate the examples, up to the moment when several properties in a row will need to be applied to obtain the final result.
Selecting the desired property of logarithms
There are not so few properties of logarithms, and it is clear that you need to be able to choose the appropriate one from them, which in this particular case will lead to the desired result. Usually this is not difficult to do by comparing the form of the logarithm or expression being converted with the types of the left and right parts of the formulas expressing the properties of logarithms. If the left or right side of one of the formulas matches the given logarithm or expression, then most likely it is this property that should be used during the transformation. The following examples this is clearly demonstrated.
Let's start with examples of transforming expressions using the definition of a logarithm, which corresponds to the formula a log a b =b , a>0 , a≠1 , b>0 .
Example.
Calculate, if possible: a) 5 log 5 4 , b) 10 log(1+2 π) , c) 
, d) 2 log 2 (−7) , e) .
Solution.
In the example, letter a) clearly shows the structure a log a b , where a=5 , b=4 . These numbers satisfy the conditions a>0 , a≠1 , b>0 , so you can safely use the equality a log a b =b . We have 5 log 5 4=4 .
b) Here a=10 , b=1+2 π , conditions a>0 , a≠1 , b>0 are fulfilled. In this case, the equality 10 lg(1+2 π) =1+2 π takes place.
c) And in this example we are dealing with a degree of the form a log a b , where and b=ln15 . So 
.
Despite belonging to the same form a log a b (here a=2 , b=−7 ), the expression under the letter d) cannot be converted by the formula a log a b =b . The reason is that it doesn't make sense because it contains a negative number under the logarithm sign. Moreover, the number b=−7 does not satisfy the condition b>0 , which makes it impossible to resort to the formula a log a b =b , since it requires the conditions a>0 , a≠1 , b>0 . So, we can't talk about computing the value 2 log 2 (−7) . In this case, writing 2 log 2 (−7) = −7 would be an error.
Similarly, in the example under the letter e) it is impossible to give a solution of the form 
, since the original expression does not make sense.
Answer:
a) 5 log 5 4 =4 , b) 10 log(1+2 π) =1+2 π , c) , d), e) expressions do not make sense.
It is often useful to convert a positive number as a power of some positive non-one number with a logarithm in the exponent. It is based on the same definition of the logarithm a log a b =b , a>0 , a≠1 , b>0 , but the formula is applied from right to left, that is, in the form b=a log a b . For example, 3=e ln3 or 5=5 log 5 5 .
Let's move on to using the properties of logarithms to transform expressions.
Example.
Find the value of the expression: a) log −2 1, b) log 1 1, c) log 0 1, d) log 7 1, e) ln1, f) lg1, g) log 3.75 1, h) log 5 π 7 1 .
Solution.
In the examples under the letters a), b) and c), the expressions log −2 1 , log 1 1 , log 0 1 are given, which do not make sense, since the base of the logarithm should not contain a negative number, zero or one, because we have defined logarithm only for a positive and non-unit base. Therefore, in examples a) - c) there can be no question of finding the value of the expression.
In all other tasks, obviously, in the bases of the logarithms there are positive and non-unit numbers 7, e, 10, 3.75 and 5 π 7, respectively, and units are everywhere under the signs of the logarithms. And we know the property of the logarithm of unity: log a 1=0 for any a>0 , a≠1 . Thus, the values of expressions b) - f) are equal to zero.
Answer:
a), b), c) the expressions do not make sense, d) log 7 1=0, e) ln1=0, f) log1=0, g) log 3.75 1=0, h) log 5 e 7 1=0 .
Example.
Calculate: a) , b) lne , c) lg10 , d) log 5 π 3 −2 (5 π 3 −2), e) log −3 (−3) , f) log 1 1 .
Solution.
It is clear that we have to use the property of the logarithm of the base, which corresponds to the formula log a a=1 for a>0 , a≠1 . Indeed, in tasks under all letters, the number under the sign of the logarithm coincides with its base. Thus, I want to say right away that the value of each of the given expressions is 1 . However, do not rush to conclusions: in tasks under the letters a) - d) the values of the expressions are really equal to one, and in tasks e) and f) the original expressions do not make sense, so it cannot be said that the values of these expressions are equal to 1.
Answer:
a) , b) lne=1 , c) lg10=1 , d) log 5 π 3 −2 (5 π 3 −2)=1, e), f) expressions do not make sense.
Example.
Find the value: a) log 3 3 11 , b) 
, c) , d) log −10 (−10) 6 .
Solution.
Obviously, under the signs of the logarithms are some degrees of base. Based on this, we understand that the property of the degree of the base is useful here: log a a p =p, where a>0, a≠1 and p is any real number. Considering this, we have the following results: a) log 3 3 11 =11 , b) 
, V) 
. Is it possible to write a similar equality for the example under the letter d) of the form log −10 (−10) 6 =6? No, you can't, because log −10 (−10) 6 doesn't make sense.
Answer:
a) log 3 3 11 =11, b) , V) d) the expression does not make sense.
Example.
Express the expression as the sum or difference of logarithms in the same base: a) 
, b) , c) log((−5) (−12)) .
Solution.
a) The product is under the sign of the logarithm, and we know the property of the logarithm of the product log a (x y)=log a x+log a y , a>0 , a≠1 , x>0 , y>0 . In our case, the number in the base of the logarithm and the numbers in the product are positive, that is, they satisfy the conditions of the selected property, therefore, we can safely apply it: 
.
b) Here we use the property of the logarithm of the quotient , where a>0 , a≠1 , x>0 , y>0 . In our case, the base of the logarithm is a positive number e, the numerator and denominator π are positive, which means they satisfy the conditions of the property, so we have the right to use the chosen formula: 
.
c) First, note that the expression lg((−5) (−12)) makes sense. But at the same time, we do not have the right to apply the formula for the logarithm of the product log a (x y)=log a x+log a y , a>0 , a≠1 , x>0 , y>0 , since the numbers −5 and −12 are negative and do not satisfy the conditions x>0 , y>0 . That is, it is impossible to carry out such a transformation: log((−5)(−12))=log(−5)+log(−12). But what to do? In such cases, the original expression needs to be pre-transformed to avoid negative numbers. We will talk in detail about similar cases of converting expressions with negative numbers under the sign of the logarithm in one of, but for now we will give a solution to this example, which is clear in advance and without explanation: lg((−5)(−12))=lg(5 12)=lg5+lg12.
Answer:
A) , b) , c) lg((−5) (−12))=lg5+lg12 .
Example.
Simplify the expression: a) log 3 0.25 + log 3 16 + log 3 0.5, b) .
Solution.
Here, all the same properties of the logarithm of the product and the logarithm of the quotient that we used in the previous examples will help us, only now we will apply them from right to left. That is, we convert the sum of logarithms to the logarithm of the product, and the difference of the logarithms to the logarithm of the quotient. We have
A) log 3 0.25+log 3 16+log 3 0.5=log 3 (0.25 16 0.5)=log 3 2.
b) 
.
Answer:
A) log 3 0.25+log 3 16+log 3 0.5=log 3 2, b) 
.
Example.
Get rid of the degree under the sign of the logarithm: a) log 0.7 5 11, b) 
, c) log 3 (−5) 6 .
Solution.
It is easy to see that we are dealing with expressions like log a b p . The corresponding property of the logarithm is log a b p =p log a b , where a>0 , a≠1 , b>0 , p is any real number. That is, under the conditions a>0 , a≠1 , b>0 from the logarithm of the degree log a b p we can go to the product p·log a b . Let's carry out this transformation with the given expressions.
a) In this case a=0.7 , b=5 and p=11 . So log 0.7 5 11 =11 log 0.7 5 .
b) Here , the conditions a>0 , a≠1 , b>0 are fulfilled. That's why 
c) The expression log 3 (−5) 6 has the same structure log a b p , a=3 , b=−5 , p=6 . But for b, the condition b>0 is not satisfied, which makes it impossible to apply the formula log a b p =p log a b . So why can't you get the job done? It is possible, but a preliminary transformation of the expression is required, which we will discuss in detail below in the paragraph under the heading . The solution will be like this: log 3 (−5) 6 =log 3 5 6 =6 log 3 5.
Answer:
a) log 0.7 5 11 =11 log 0.7 5 ,
b)
c) log 3 (−5) 6 =6 log 3 5 .
Quite often, the formula for the logarithm of the degree when carrying out transformations has to be applied from right to left in the form p log a b \u003d log a b p (this requires the same conditions for a, b and p). For example, 3 ln5=ln5 3 and lg2 log 2 3=log 2 3 lg2 .
Example.
a) Calculate the value of log 2 5 if it is known that lg2≈0.3010 and lg5≈0.6990. b) Write the fraction as a logarithm to base 3.
Solution.
a) The formula for the transition to a new base of the logarithm allows us to represent this logarithm as a ratio of decimal logarithms, the values of which are known to us: . It remains only to carry out the calculations, we have 
.
b) Here it is enough to use the formula for the transition to a new base, and apply it from right to left, that is, in the form 
. We get 
.
Answer:
a) log 2 5≈2.3223, b) .
At this stage, we have rather carefully considered the transformation of the most simple expressions using the basic properties of logarithms and the definition of a logarithm. In these examples, we had to use one property and nothing else. Now, with a clear conscience, you can move on to examples whose transformation requires the use of several properties of logarithms and other additional transformations. We will deal with them in the next paragraph. But before that, let us briefly dwell on examples of the application of consequences from the basic properties of logarithms.
Example.
a) Get rid of the root under the sign of the logarithm. b) Convert the fraction to a base 5 logarithm. c) Get rid of the powers under the sign of the logarithm and at its base. d) Calculate the value of the expression 
. e) Replace the expression with a power with base 3.
Solution.
a) If we recall the corollary from the property of the logarithm of the degree 
, then you can immediately answer: 
.
b) Here we use the formula 
from right to left, we have 
.
c) B this case formula leads to the result . We get 
.
d) And here it suffices to apply the corollary to which the formula corresponds 
. So 
.
e) The property of the logarithm allows us to achieve the desired result: 
.
Answer:
A) . b) . V) 
. G) 
. e) .
Consistently Applying Multiple Properties
Real tasks for transforming expressions using the properties of logarithms are usually more complicated than those that we dealt with in the previous paragraph. In them, as a rule, the result is not obtained in one step, but the solution already consists in the sequential application of one property after another, together with additional identical transformations, such as opening brackets, reducing like terms, reducing fractions, etc. So let's get closer to such examples. There is nothing complicated about this, the main thing is to act carefully and consistently, observing the order in which the actions are performed.
Example.
Calculate the value of an expression (log 3 15−log 3 5) 7 log 7 5.
Solution.
The difference of logarithms in brackets by the property of the logarithm of the quotient can be replaced by the logarithm log 3 (15:5) , and then calculate its value log 3 (15:5)=log 3 3=1 . And the value of the expression 7 log 7 5 by the definition of the logarithm is 5 . Substituting these results into the original expression, we get (log 3 15−log 3 5) 7 log 7 5 =1 5=5.
Here is a solution without explanation:
(log 3 15−log 3 5) 7 log 7 5 =log 3 (15:5) 5=
= log 3 3 5=1 5=5 .
Answer:
(log 3 15−log 3 5) 7 log 7 5 =5.
Example.
What is the value of the numerical expression log 3 log 2 2 3 −1 ?
Solution.
Let's first transform the logarithm, which is under the sign of the logarithm, according to the formula of the logarithm of the degree: log 2 2 3 =3. So log 3 log 2 2 3 =log 3 3 and then log 3 3=1 . So log 3 log 2 2 3 −1=1−1=0 .
Answer:
log 3 log 2 2 3 −1=0 .
Example.
Simplify the expression.
Solution.
The formula for the transition to a new base of the logarithm allows the ratio of logarithms to one base to be represented as log 3 5 . In this case, the original expression will take the form . By definition of the logarithm 3 log 3 5 =5 , that is 
, and the value of the resulting expression, by virtue of the same definition of the logarithm, is equal to two.
Here is a short version of the solution, which is usually given: 
.
Answer:
.
For a smooth transition to the information of the next paragraph, let's take a look at the expressions 5 2+log 5 3 , and lg0.01 . Their structure does not fit any of the properties of logarithms. So what happens if they cannot be converted using the properties of logarithms? It is possible if you carry out preliminary transformations that prepare these expressions for applying the properties of logarithms. So 5 2+log 5 3 =5 2 5 log 5 3 =25 3=75, 
and lg0,01=lg10 −2 = −2 . Further we will understand in detail how such preparation of expressions is carried out.
Preparing expressions to apply the properties of logarithms
Logarithms in the converted expression very often differ in the structure of the notation from the left and right parts of formulas that correspond to the properties of logarithms. But just as often, transforming these expressions involves using the properties of logarithms: to use them, you only need preliminary preparation. And this preparation consists in carrying out certain identical transformations that bring logarithms to a form convenient for applying properties.
In fairness, we note that almost any transformation of expressions can act as preliminary transformations, from the banal reduction of similar terms to the application trigonometric formulas. This is understandable, since the converted expressions can contain any mathematical objects: brackets, modules, fractions, roots, degrees, etc. Thus, one must be prepared to perform any required transformation in order to further benefit from the properties of logarithms.
Let's say right away that in this paragraph we do not set ourselves the task of classifying and analyzing all conceivable preliminary transformations that allow us to apply the properties of logarithms or the definition of a logarithm in the future. Here we will focus on only four of them, which are the most characteristic and most often encountered in practice.
And now in detail about each of them, after which, within the framework of our topic, it remains only to deal with the transformation of expressions with variables under the signs of logarithms.
Selection of powers under the sign of the logarithm and in its base
Let's start right away with an example. Let us have a logarithm. Obviously, in this form, its structure is not conducive to the use of the properties of logarithms. Is it possible to somehow transform this expression in order to simplify it, or even better calculate its value? To answer this question, let's take a closer look at the numbers 81 and 1/9 in the context of our example. It is easy to see here that these numbers can be represented as a power of 3 , indeed, 81=3 4 and 1/9=3 −2 . In this case, the original logarithm is presented in the form and it becomes possible to apply the formula . So, 
.
Analysis of the analyzed example gives rise to the following idea: if possible, you can try to highlight the degree under the sign of the logarithm and at its base in order to apply the property of the logarithm of the degree or its consequence. It remains only to figure out how to single out these degrees. We will give some recommendations on this issue.
Sometimes it is quite obvious that the number under the sign of the logarithm and / or in its base represents some integer power, as in the example discussed above. Almost constantly you have to deal with powers of two, which are well familiar: 4=2 2 , 8=2 3 , 16=2 4 , 32=2 5 , 64=2 6 , 128=2 7 , 256=2 8 , 512= 2 9 , 1024=2 10 . The same can be said about the degrees of the triple: 9=3 2 , 27=3 3 , 81=3 4 , 243=3 5 , ... In general, it does not hurt if there is table of powers of natural numbers within ten. It is also not difficult to work with integer powers of ten, hundred, thousand, etc.
Example.
Calculate the value or simplify the expression: a) log 6 216 , b) , c) log 0.000001 0.001 .
Solution.
a) Obviously, 216=6 3 , so log 6 216=log 6 6 3 =3 .
b) The table of powers of natural numbers allows us to represent the numbers 343 and 1/243 as powers of 7 3 and 3 −4, respectively. Therefore, the following transformation of the given logarithm is possible: 
c) Since 0.000001=10 −6 and 0.001=10 −3, then log 0.000001 0.001=log 10 −6 10 −3 =(−3)/(−6)=1/2.
Answer:
a) log 6 216=3, b) 
, c) log 0.000001 0.001=1/2 .
In more complex cases, to highlight the powers of numbers, you have to resort to.
Example.
Convert expression to more plain sight log 3 648 log 2 3 .
Solution.
Let's see what is the decomposition of the number 648 into prime factors:
That is, 648=2 3 3 4 . Thus, log 3 648 log 2 3=log 3 (2 3 3 4) log 2 3.
Now we convert the logarithm of the product to the sum of logarithms, after which we apply the properties of the logarithm of the degree:
log 3 (2 3 3 4) log 2 3=(log 3 2 3 + log 3 3 4) log 2 3=
=(3 log 3 2+4) log 2 3 .
By virtue of the corollary of the property of the logarithm of the degree, which corresponds to the formula , the product log32 log23 is the product , and it is known to be equal to one. Considering this, we get 3 log 3 2 log 2 3+4 log 2 3=3 1+4 log 2 3=3+4 log 2 3.
Answer:
log 3 648 log 2 3=3+4 log 2 3.
Quite often, expressions under the sign of the logarithm and in its base are products or ratios of the roots and / or powers of some numbers, for example, , . Similar expressions can be represented as a degree. To do this, the transition from roots to degrees is carried out, and and are applied. These transformations allow you to select the degrees under the sign of the logarithm and in its base, and then apply the properties of logarithms.
Example.
Calculate: a) 
, b).
Solution.
a) The expression in the base of the logarithm is the product of powers with the same bases, by the corresponding property of powers we have 5 2 5 −0.5 5 −1 =5 2−0.5−1 =5 0.5.
Now let's convert the fraction under the sign of the logarithm: let's move from the root to the degree, after which we will use the property of the ratio of degrees with the same bases: 
.
It remains to substitute the results obtained in the original expression, use the formula and finish the transformation: 
b) Since 729=3 6 , and 1/9=3 −2 , the original expression can be rewritten as .
Next, apply the property of the root of the exponent, move from the root to the exponent, and use the ratio property of the powers to convert the base of the logarithm to a power: 
.
Taking into account the last result, we have 
.
Answer:
A) 
, b).
It is clear that in the general case, to obtain powers under the sign of the logarithm and in its base, various transformations of various expressions may be required. Let's give a couple of examples.
Example.
What is the value of the expression: a) 
, b) 
.
Solution.
Further, we note that the given expression has the form log A B p , where A=2 , B=x+1 and p=4 . We transformed numerical expressions of this kind according to the property of the logarithm of the degree log a b p \u003d p log a b, therefore, with a given expression, I want to do the same, and go from log 2 (x + 1) 4 to 4 log 2 (x + 1) . And now let's calculate the value of the original expression and the expression obtained after the transformation, for example, with x=−2 . We have log 2 (−2+1) 4 =log 2 1=0 , and 4 log 2 (−2+1)=4 log 2 (−1)- meaningless expression. This raises a legitimate question: “What did we do wrong”?
And the reason is as follows: we performed the transformation log 2 (x+1) 4 =4 log 2 (x+1) , based on the formula log a b p =p log a b , but we have the right to apply this formula only if the conditions a >0 , a≠1 , b>0 , p - any real number. That is, the transformation we have done takes place if x+1>0 , which is the same x>−1 (for A and p, the conditions are met). However, in our case, the ODZ of the variable x for the original expression consists not only of the interval x> −1 , but also of the interval x<−1 . Но для x<−1 мы не имели права осуществлять преобразование по выбранной формуле.
The need to take into account ODZ
Let's continue to analyze the transformation of the expression log 2 (x+1) 4 we have chosen, and now let's see what happens to the ODZ when passing to the expression 4 log 2 (x+1) . In the previous paragraph, we found the ODZ of the original expression - this is the set (−∞, −1)∪(−1, +∞) . Now let's find the area allowed values variable x for the expression 4 log 2 (x+1) . It is determined by the condition x+1>0 , which corresponds to the set (−1, +∞) . It is obvious that when going from log 2 (x+1) 4 to 4·log 2 (x+1), the range of admissible values narrows. And we agreed to avoid reforms that lead to a narrowing of the ODZ, as this can lead to various negative consequences.
Here it is worth noting for yourself that it is useful to control the ODZ at each step of the transformation and not allow it to narrow. And if suddenly at some stage of the transformation there was a narrowing of the ODZ, then it is worth looking very carefully at whether this transformation is permissible and whether we had the right to carry it out.
In fairness, we say that in practice we usually have to work with expressions in which the ODZ of variables is such that it allows us to use the properties of logarithms without restrictions in the form already known to us, both from left to right and from right to left, when carrying out transformations. You quickly get used to this, and you begin to carry out the transformations mechanically, without thinking about whether it was possible to carry them out. And at such moments, as luck would have it, more complex examples slip through, in which the inaccurate application of the properties of logarithms leads to errors. So you need to always be on the alert, and make sure that there is no narrowing of the ODZ.
It does not hurt to separately highlight the main transformations based on the properties of logarithms, which must be carried out very carefully, which can lead to a narrowing of the DPV, and as a result, to errors:
Some transformations of expressions according to the properties of logarithms can also lead to the opposite - the expansion of the ODZ. For example, going from 4 log 2 (x+1) to log 2 (x+1) 4 extends the ODZ from the set (−1, +∞) to (−∞, −1)∪(−1, +∞) . Such transformations take place if you remain within the ODZ for the original expression. So the transformation just mentioned 4 log 2 (x+1)=log 2 (x+1) 4 takes place on the ODZ variable x for the original expression 4 log 2 (x+1) , that is, when x+1> 0 , which is the same as (−1, +∞) .
Now that we have discussed the nuances that you need to pay attention to when converting expressions with variables using the properties of logarithms, it remains to figure out how these conversions should be carried out correctly.
X+2>0 . Does it work in our case? To answer this question, let's take a look at the DPV of the x variable. It is determined by the system of inequalities 
, which is equivalent to the condition x+2>0 (if necessary, see the article solution of systems of inequalities). Thus, we can safely apply the property of the logarithm of the degree.
We have
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =
=3 7 log(x+2)−log(x+2)−5 4 log(x+2)=
=21 log(x+2)−log(x+2)−20 log(x+2)=
=(21−1−20)lg(x+2)=0 .
You can act differently, since the ODZ allows you to do this, for example like this: 
Answer:
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =0.
And what to do when the conditions associated with the properties of logarithms are not met on the ODZ? We will deal with this with examples.
Let us be required to simplify the expression lg(x+2) 4 −lg(x+2) 2 . The transformation of this expression, unlike the expression from the previous example, does not allow the free use of the property of the logarithm of the degree. Why? The ODZ of the variable x in this case is the union of two intervals x>−2 and x<−2 . При x>−2 we can safely apply the property of the logarithm of the degree and proceed as in the example above: log(x+2) 4 −log(x+2) 2 =4 log(x+2)−2 log(x+2)=2 log(x+2). But the ODZ contains another interval x+2<0 , для которого последнее преобразование будет некорректно. Что же делать при x+2<0 ? В подобных случаях на помощь приходит . Определение модуля позволяет выражение x+2 при x+2<0 представить как −|x+2| . Тогда при x+2<0 от lg(x+2) 4 −lg(x+2) 2 переходим к log(−|x+2|) 4 −log(−|x+2|) 2 and further, due to the power properties of lg|x+2| 4−lg|x+2| 2. The resulting expression can be transformed according to the property of the logarithm of the degree, since |x+2|>0 for any values of the variable. We have log|x+2| 4−lg|x+2| 2 =4 log|x+2|−2 log|x+2|=2 log|x+2|. Now you can get rid of the module, since it has done its job. Since we are transforming at x+2<0 , то 2·lg|x+2|=2·lg(−(x+2)) . Итак, можно считать, что мы справились с поставленной задачей. Ответ: . Полученный результат можно записать компактно с использованием модуля как .
Let's consider one more example to make working with modules familiar. Let us conceive from the expression 
pass to the sum and difference of the logarithms of the linear binomials x−1 , x−2 and x−3 . First we find the ODZ: 
On the interval (3, +∞), the values of the expressions x−1 , x−2 and x−3 are positive, so we can safely apply the properties of the logarithm of the sum and difference: 
And on the interval (1, 2), the values of the expression x−1 are positive, and the values of the expressions x−2 and x−3 are negative. Therefore, on the interval under consideration, we represent x−2 and x−3 using the modulo as −|x−2| and −|x−3| respectively. Wherein 
Now we can apply the properties of the logarithm of the product and the quotient, since on the considered interval (1, 2) the values of the expressions x−1 , |x−2| and |x−3| - positive.
We have 
The results obtained can be combined:
In general, similar reasoning allows, based on the formulas for the logarithm of the product, ratio and degree, to obtain three practically useful results that are quite convenient to use:
- The logarithm of the product of two arbitrary expressions X and Y of the form log a (X·Y) can be replaced by the sum of the logarithms log a |X|+log a |Y| , a>0 , a≠1 .
- The special logarithm log a (X:Y) can be replaced by the difference of the logarithms log a |X|−log a |Y| , a>0 , a≠1 , X and Y are arbitrary expressions.
- From the logarithm of some expression B to an even power p of the form log a B p, one can pass to the expression p log a |B| , where a>0 , a≠1 , p is an even number and B is an arbitrary expression.
Similar results are given, for example, in instructions for solving exemplary and logarithmic equations in the collection of problems in mathematics for applicants to universities, edited by M. I. Skanavi.
Example.
Simplify the expression 
.
Solution.
It would be good to apply the properties of the logarithm of the degree, sum and difference. But can we do it here? To answer this question, we need to know the ODZ.
Let's define it: 
It is quite obvious that the expressions x+4 , x−2 and (x+4) 13 on the range of possible values of the variable x can take both positive and negative values. Therefore, we will have to work through modules. 
Module properties allow you to rewrite as , so 
Also, nothing prevents you from using the property of the logarithm of the degree, and then bring like terms: 
Another sequence of transformations leads to the same result: 
and since the expression x−2 can take both positive and negative values on the ODZ, when taking an even exponent 14
The final videos from a long series of lessons about solving logarithmic equations. This time we will work primarily with the logarithm ODZ - it is precisely because of the incorrect accounting (or even ignoring) of the domain of definition that most errors occur when solving such problems.
In this short video tutorial, we will analyze the application of the addition and subtraction formulas for logarithms, as well as deal with fractional rational equations, which many students also have problems with.
What will be discussed? The main formula I would like to deal with looks like this:
log a (f g ) = log a f + log a g
This is the standard transition from the product to the sum of logarithms and vice versa. You probably know this formula from the very beginning of the study of logarithms. However, there is one hitch here.
As long as the variables a , f and g are ordinary numbers, there are no problems. This formula works great.
However, as soon as functions appear instead of f and g, the problem of expanding or narrowing the domain of definition arises, depending on which way to convert. Judge for yourself: in the logarithm written on the left, the domain of definition is as follows:
fg > 0
But in the sum written on the right, the domain of definition is already somewhat different:
f > 0
g > 0
This set of requirements is more stringent than the original one. In the first case, we will be satisfied with the option f< 0, g < 0 (ведь их произведение положительное, поэтому неравенство fg >0 is being executed).
Thus, when passing from the left construction to the right one, the domain of definition becomes narrower. If at first we had a sum, and we rewrite it as a product, then the domain of definition is expanded.
In other words, in the first case, we could lose roots, and in the second, we could get extra ones. This must be taken into account when solving real logarithmic equations.
So the first task is:
[Figure caption] On the left we see the sum of the logarithms in the same base. Therefore, these logarithms can be added:
[Figure caption] As you can see, on the right we have replaced zero by the formula:
a = log b b a
Let's rearrange our equation a little more:
log 4 (x − 5) 2 = log 4 1
Before us is the canonical form of the logarithmic equation, we can cross out the log sign and equate the arguments:
(x − 5) 2 = 1
|x−5| = 1
Pay attention: where did the module come from? Let me remind you that the root of the exact square is exactly equal to the modulus:
[Figure caption] Then we solve the classical equation with the modulus:
|f| = g (g > 0) ⇒f = ±g
x − 5 = ±1 ⇒ x 1 = 5 − 1 = 4; x 2 = 5 + 1 = 6
Here are two candidates for the answer. Are they solutions to the original logarithmic equation? No way!
We have no right to leave everything just like that and write down the answer. Take a look at the step where we replace the sum of the logarithms with one logarithm of the product of the arguments. The problem is that in the original expressions we have functions. Therefore, it should be required:
x(x − 5) > 0; (x − 5)/x > 0.
When we transformed the product, getting an exact square, the requirements changed:
(x − 5) 2 > 0
When is this requirement met? Yes, almost always! Except for the case when x − 5 = 0. That is, the inequality will be reduced to one punctured point:
x − 5 ≠ 0 ⇒ x ≠ 5
As you can see, there has been an expansion of the domain of definition, which we talked about at the very beginning of the lesson. Therefore, extra roots may also appear.
How to prevent the emergence of these extra roots? It's very simple: we look at our obtained roots and compare them with the domain of the original equation. Let's count:
x (x − 5) > 0
We will solve using the interval method:
x (x − 5) = 0 ⇒ x = 0; x = 5
We mark the received numbers on a straight line. All points are punctured because the inequality is strict. We take any number greater than 5 and substitute:
[Figure caption] We are interested in the intervals (−∞; 0) ∪ (5; ∞). If we mark our roots on the segment, we will see that x = 4 does not suit us, because this root lies outside the domain of the original logarithmic equation.
We return to the population, cross out the root x \u003d 4 and write down the answer: x \u003d 6. This is the final answer to the original logarithmic equation. Everything, the task is solved.
We pass to the second logarithmic equation:
[Figure caption] We solve it. Note that the first term is a fraction, and the second is the same fraction, but inverted. Don't be intimidated by the lgx expression - it's just a base 10 logarithm, we can write:
lgx = log 10 x
Since we have two inverted fractions, I propose to introduce a new variable:
[Figure caption] Therefore, our equation can be rewritten as follows:
t + 1/t = 2;
t + 1/t − 2 = 0;
(t 2 − 2t + 1)/t = 0;
(t − 1) 2 /t = 0.
As you can see, the numerator of the fraction is an exact square. A fraction is zero when its numerator zero, and the denominator is different from zero:
(t − 1) 2 = 0; t ≠ 0
We solve the first equation:
t − 1 = 0;
t = 1.
This value satisfies the second requirement. Therefore, it can be argued that we have completely solved our equation, but only with respect to the variable t . Now let's remember what t is:
[Figure caption]We got the ratio:
lgx = 2 lgx + 1
2 lgx − lgx = −1
logx = −1
We bring this equation to the canonical form:
lgx = lg 10 −1
x = 10 −1 = 0.1
As a result, we got the only root, which, in theory, is the solution to the original equation. However, let's still play it safe and write out the domain of the original equation:
[Figure caption] Therefore, our root satisfies all the requirements. We have found a solution to the original logarithmic equation. Answer: x = 0.1. Problem solved.
There is only one key point in today's lesson: when using the formula for the transition from product to sum and vice versa, be sure to keep in mind that the domain of definition can narrow or expand depending on which direction the transition is made.
How to understand what is happening: contraction or expansion? Very simple. If earlier the functions were together, and now they have become separate, then the scope of definition has narrowed (because there are more requirements). If at first the functions were separate, and now they are together, then the domain of definition is expanded (less requirements are imposed on the product than on individual factors).
In view of this remark, I would like to note that the second logarithmic equation does not require these transformations at all, i.e. we do not add or multiply the arguments anywhere. However, here I would like to draw your attention to another wonderful trick that allows you to significantly simplify the solution. It's about changing a variable.
However, remember that no substitution does not free us from the scope. That is why after all the roots were found, we were not too lazy and returned to the original equation to find its ODZ.
Often when changing a variable, an annoying mistake occurs when students find the value of t and think that the solution is over. No way!
When you have found the value of t , you need to return to the original equation and see what exactly we denoted by this letter. As a result, we have to solve one more equation, which, however, will be much simpler than the original one.
This is precisely the point of introducing a new variable. We split the original equation into two intermediate ones, each of which is solved much easier.
How to solve "nested" logarithmic equations
Today we continue to study logarithmic equations and analyze constructions when one logarithm is under the sign of another logarithm. We will solve both equations using the canonical form.
Today we continue to study logarithmic equations and analyze constructions when one logarithm is under the sign of another. We will solve both equations using the canonical form. Let me remind you that if we have the simplest logarithmic equation of the form log a f (x) \u003d b, then we perform the following steps to solve such an equation. First of all, we need to replace the number b :
b = log a a b
Note that a b is an argument. Similarly, in the original equation, the argument is the function f(x). Then we rewrite the equation and get this construction:
log a f(x) = log a a b
After that, we can perform the third step - get rid of the sign of the logarithm and simply write:
f(x) = a b
As a result, we get a new equation. In this case, no restrictions are imposed on the function f(x). For example, in its place can also be a logarithmic function. And then we again get a logarithmic equation, which we again reduce to the simplest and solve through the canonical form.
But enough of the lyrics. Let's solve the real problem. So task number 1:
log 2 (1 + 3 log 2 x ) = 2
As you can see, we have a simple logarithmic equation. The role of f (x) is the construction 1 + 3 log 2 x, and the number b is the number 2 (the role of a is also two). Let's rewrite this two as follows:
It is important to understand that the first two deuces came to us from the base of the logarithm, that is, if there were 5 in the original equation, then we would get that 2 = log 5 5 2. In general, the base depends solely on the logarithm, which is initially given in the problem. And in our case this number is 2.
So, we rewrite our logarithmic equation, taking into account the fact that the two, which is on the right, is actually also a logarithm. We get:
log 2 (1 + 3 log 2 x ) = log 2 4
We pass to the last step of our scheme - we get rid of the canonical form. We can say, just cross out the signs of log. However, from the point of view of mathematics, it is impossible to “cross out log” - it is more correct to say that we simply simply equate the arguments:
1 + 3 log 2 x = 4
From here it is easy to find 3 log 2 x :
3 log 2 x = 3
log 2 x = 1
We again got the simplest logarithmic equation, let's bring it back to the canonical form. To do this, we need to make the following changes:
1 = log 2 2 1 = log 2 2
Why is there a deuce at the base? Because in our canonical equation on the left is the logarithm precisely in base 2. We rewrite the problem taking into account this fact:
log 2 x = log 2 2
Again, we get rid of the sign of the logarithm, i.e., we simply equate the arguments. We have the right to do this, because the bases are the same, and no more additional actions were performed either on the right or on the left:
That's all! Problem solved. We have found a solution to the logarithmic equation.
Note! Although the variable x is in the argument (that is, there are requirements for the domain of definition), we will not make any additional requirements.
As I said above, this check is redundant if the variable occurs in only one argument of only one logarithm. In our case, x really is only in the argument and only under one log sign. Therefore, no additional checks are required.
However, if you don't trust this method, you can easily verify that x = 2 is indeed a root. It is enough to substitute this number into the original equation.
Let's move on to the second equation, it's a little more interesting:
log 2 (log 1/2 (2x − 1) + log 2 4) = 1
If we denote the expression inside the large logarithm by the function f (x), we get the simplest logarithmic equation with which we started today's video lesson. Therefore, it is possible to apply the canonical form, for which it is necessary to represent the unit in the form log 2 2 1 = log 2 2.
Rewriting our big equation:
log 2 (log 1/2 (2x − 1) + log 2 4) = log 2 2
We get rid of the sign of the logarithm, equating the arguments. We have the right to do this, because the bases are the same on the left and on the right. Also, note that log 2 4 = 2:
log 1/2 (2x − 1) + 2 = 2
log 1/2 (2x − 1) = 0
Before us again is the simplest logarithmic equation of the form log a f (x) \u003d b. We pass to the canonical form, i.e. we represent zero in the form log 1/2 (1/2)0 = log 1/2 1.
We rewrite our equation and get rid of the log sign by equating the arguments:
log 1/2 (2x − 1) = log 1/2 1
2x − 1 = 1
Again, we received an immediate response. No additional checks are required, because in the original equation, only one logarithm contains the function in the argument.
Therefore, no additional checks are required. We can safely say that x = 1 is the only root of this equation.
But if in the second logarithm instead of four there would be some function of x (or 2x would not be in the argument, but in the base) - then it would be necessary to check the domain of definition. Otherwise, there is a great chance of running into extra roots.
Where do these extra roots come from? This point needs to be understood very clearly. Look at the original equations: everywhere the function x is under the sign of the logarithm. Therefore, since we have written log 2 x , we automatically set the requirement x > 0. Otherwise, this record simply does not make sense.
However, as we solve the logarithmic equation, we get rid of all the signs of log and get simple constructions. There are no more restrictions here, because linear function defined for any value of x.
It is this problem, when the final function is defined everywhere and always, and the initial one is by no means everywhere and not always, that is the reason why extra roots very often appear in the solution of logarithmic equations.
But I repeat once again: this happens only in a situation where the function is either in several logarithms, or at the base of one of them. In the problems that we are considering today, there are in principle no problems with expanding the domain of definition.
Cases of different grounds
This lesson is dedicated to complex structures. The logarithms in today's equations will no longer be solved "blank" - first you need to perform some transformations.
We start solving logarithmic equations with completely different bases, which are not exact powers of each other. Don't be intimidated by such tasks - they are no more difficult to solve than the most simple designs which we have discussed above.
But before proceeding directly to the problems, let me remind you of the formula for solving the simplest logarithmic equations using the canonical form. Consider a problem like this:
log a f(x) = b
It is important that the function f (x) is just a function, and the numbers a and b should be exactly the numbers (without any variables x). Of course, literally in a minute we will also consider such cases when instead of variables a and b there are functions, but this is not about that now.
As we remember, the number b must be replaced by a logarithm in the same base a, which is on the left. This is done very simply:
b = log a a b
Of course, the words "any number b" and "any number a" mean such values that satisfy the domain of definition. In particular, in this equation we are talking only the base a > 0 and a ≠ 1.
However, this requirement is met automatically, because the original problem already contains a logarithm to the base a - it will certainly be greater than 0 and not equal to 1. Therefore, we continue the solution of the logarithmic equation:
log a f(x) = log a a b
Such a notation is called the canonical form. Its convenience is that we can immediately get rid of the log sign by equating the arguments:
f(x) = a b
It is this technique that we will now use to solve logarithmic equations with a variable base. So let's go!
log 2 (x 2 + 4x + 11) = log 0.5 0.125
What's next? Someone will now say that you need to calculate the right logarithm, or reduce them to one base, or something else. And indeed, now you need to bring both bases to the same form - either 2 or 0.5. But let's learn the following rule once and for all:
If there are decimal fractions in the logarithmic equation, be sure to convert these fractions from decimal notation into the usual. Such a transformation can significantly simplify the solution.
Such a transition must be performed immediately, even before any actions and transformations are performed. Let's get a look:
log 2 (x 2 + 4x + 11) = log 1/2 1/8
What does such a record give us? We can represent 1/2 and 1/8 as a negative exponent:
[Figure caption] We have the canonical form. Equate the arguments and get the classical quadratic equation:
x 2 + 4x + 11 = 8
x 2 + 4x + 3 = 0
Before us is the given quadratic equation, which is easily solved using the Vieta formulas. You should see similar calculations in high school literally orally:
(x + 3)(x + 1) = 0
x 1 = -3
x 2 = -1
That's all! The original logarithmic equation is solved. We have two roots.
Let me remind you that in this case it is not required to define the scope, since the function with the variable x is present in only one argument. Therefore, the scope is performed automatically.
So the first equation is solved. Let's move on to the second one:
log 0.5 (5x 2 + 9x + 2) = log 3 1/9
log 1/2 (5x 2 + 9x + 2) = log 3 9 −1
And now note that the argument of the first logarithm can also be written as a power with a negative exponent: 1/2 = 2 −1. Then you can take out the powers on both sides of the equation and divide everything by −1:
[Figure caption] And now we have done very important step in solving a logarithmic equation. Perhaps someone did not notice something, so let me explain.
Take a look at our equation: log is on the left and right, but the base 2 logarithm is on the left, and the base 3 logarithm is on the right. degree.
Therefore, these are logarithms with different bases, which are not reduced to each other by simple exponentiation. The only way to solve such problems is to get rid of one of these logarithms. In this case, since we are still considering quite simple tasks, the logarithm on the right was simply calculated, and we got the simplest equation - exactly the one we talked about at the very beginning of today's lesson.
Let's represent the number 2, which is on the right, as log 2 2 2 = log 2 4. And then get rid of the sign of the logarithm, after which we are left with just a quadratic equation:
log 2 (5x 2 + 9x + 2) = log 2 4
5x2 + 9x + 2 = 4
5x2 + 9x − 2 = 0
Before us is the usual quadratic equation, but it is not reduced, because the coefficient at x 2 is different from unity. Therefore, we will solve it using the discriminant:
D = 81 − 4 5 (−2) = 81 + 40 = 121
x 1 = (−9 + 11)/10 = 2/10 = 1/5
x 2 \u003d (−9 - 11) / 10 \u003d -2
That's all! We found both roots, which means we got the solution to the original logarithmic equation. Indeed, in the original problem, the function with the variable x is present in only one argument. Consequently, no additional checks on the domain of definition are required - both roots that we have found certainly meet all possible restrictions.
This could be the end of today's video tutorial, but in conclusion I would like to say again: be sure to convert all decimal fractions to ordinary ones when solving logarithmic equations. In most cases, this greatly simplifies their solution.
Rarely, very rarely, there are problems in which getting rid of decimal fractions only complicates the calculations. However, in such equations, as a rule, it is initially clear that it is not necessary to get rid of decimal fractions.
In most other cases (especially if you are just starting to train in solving logarithmic equations), feel free to get rid of decimal fractions and translate them into ordinary ones. Because practice shows that in this way you will greatly simplify the subsequent solution and calculations.
Subtleties and tricks of the solution
Today we move on to more complex tasks and we will solve a logarithmic equation, which is based not on a number, but on a function.
And even if this function is linear, small changes will have to be made to the solution scheme, the meaning of which boils down to additional requirements imposed on the domain of definition of the logarithm.
Difficult tasks
This lesson will be quite long. In it, we will analyze two rather serious logarithmic equations, in the solution of which many students make mistakes. During my practice as a tutor in mathematics, I constantly encountered two types of errors:
- The appearance of extra roots due to the expansion of the domain of definition of logarithms. To avoid making such offensive mistakes, just keep a close eye on each transformation;
- Loss of roots due to the fact that the student forgot to consider some "subtle" cases - it is on such situations that we will focus today.
This is the last lesson on logarithmic equations. It will be long, we will analyze complex logarithmic equations. Make yourself comfortable, make yourself some tea, and we'll begin.
The first equation looks quite standard:
log x + 1 (x - 0.5) = log x - 0.5 (x + 1)
Immediately, we note that both logarithms are inverted copies of each other. Let's remember the wonderful formula:
log a b = 1/log b a
However, this formula has a number of limitations that arise if instead of the numbers a and b there are functions of the variable x:
b > 0
1 ≠ a > 0
These requirements are imposed on the base of the logarithm. On the other hand, in a fraction, we are required to have 1 ≠ a > 0, since not only is the variable a in the argument of the logarithm (hence, a > 0), but the logarithm itself is in the denominator of the fraction. But log b 1 = 0, and the denominator must be non-zero, so a ≠ 1.
So, the restrictions on the variable a are preserved. But what happens to the variable b ? On the one hand, b > 0 follows from the base, on the other hand, the variable b ≠ 1, because the base of the logarithm must be different from 1. In total, it follows from the right side of the formula that 1 ≠ b > 0.
But here's the problem: the second requirement (b ≠ 1) is missing from the first inequality on the left logarithm. In other words, when performing this transformation, we must check separately that the argument b is different from one!
Here, let's check it out. Let's apply our formula:
[Figure caption] 1 ≠ x - 0.5 > 0; 1 ≠ x + 1 > 0
So we got that already from the original logarithmic equation it follows that both a and b must be greater than 0 and not equal to 1. So, we can easily flip the logarithmic equation:
I propose to introduce a new variable:
log x + 1 (x − 0.5) = t
In this case, our construction will be rewritten as follows:
(t 2 − 1)/t = 0
Note that in the numerator we have the difference of squares. We reveal the difference of squares using the abbreviated multiplication formula:
(t − 1)(t + 1)/t = 0
A fraction is zero when its numerator is zero and its denominator is non-zero. But the numerator contains the product, so we equate each factor to zero:
t1 = 1;
t2 = −1;
t ≠ 0.
As you can see, both values of the variable t suit us. However, the solution does not end there, because we need to find not t , but the value of x . We return to the logarithm and get:
log x + 1 (x − 0.5) = 1;
log x + 1 (x − 0.5) = −1.
Let's bring each of these equations to canonical form:
log x + 1 (x − 0.5) = log x + 1 (x + 1) 1
log x + 1 (x − 0.5) = log x + 1 (x + 1) −1
We get rid of the sign of the logarithm in the first case and equate the arguments:
x − 0.5 = x + 1;
x - x \u003d 1 + 0.5;
Such an equation has no roots, therefore, the first logarithmic equation also has no roots. But with the second equation, everything is much more interesting:
(x − 0.5)/1 = 1/(x + 1)
We solve the proportion - we get:
(x − 0.5)(x + 1) = 1
I remind you that when solving logarithmic equations, it is much more convenient to give all common decimal fractions, so let's rewrite our equation as follows:
(x − 1/2)(x + 1) = 1;
x 2 + x - 1/2x - 1/2 - 1 = 0;
x 2 + 1/2x - 3/2 = 0.
Before us is the given quadratic equation, it is easily solved using the Vieta formulas:
(x + 3/2) (x − 1) = 0;
x 1 \u003d -1.5;
x2 = 1.
We got two roots - they are candidates for solving the original logarithmic equation. In order to understand what roots will really go into the answer, let's go back to the original problem. Now we'll check each of our roots to see if they match the scope:
1.5 ≠ x > 0.5; 0 ≠ x > −1.
These requirements are tantamount to a double inequality:
1 ≠ x > 0.5
From here we immediately see that the root x = −1.5 does not suit us, but x = 1 is quite satisfied. Therefore x = 1 is the final solution of the logarithmic equation.
Let's move on to the second task:
log x 25 + log 125 x 5 = log 25 x 625
At first glance, it may seem that all logarithms different grounds and various arguments. What to do with such structures? First of all, note that the numbers 25, 5, and 625 are powers of 5:
25 = 5 2 ; 625 = 5 4
And now we will use the remarkable property of the logarithm. The fact is that you can take out the degrees from the argument in the form of factors:
log a b n = n ∙ log a b
Restrictions are also imposed on this transformation when there is a function in place of b. But with us b is just a number, and no additional restrictions arise. Let's rewrite our equation:
2 ∙ log x 5 + log 125 x 5 = 4 ∙ log 25 x 5
We got an equation with three terms containing the log sign. Moreover, the arguments of all three logarithms are equal.
It's time to flip the logarithms to bring them to the same base - 5. Since the variable b is a constant, there is no change in the scope. We just rewrite:
[Figure caption] As expected, the same logarithms “crawled out” in the denominator. I suggest changing the variable:
log 5 x = t
In this case, our equation will be rewritten as follows:
Let's write out the numerator and open the brackets:
2 (t + 3) (t + 2) + t (t + 2) - 4t (t + 3) = 2 (t 2 + 5t + 6) + t 2 + 2t - 4t 2 - 12t = 2t 2 + 10t + 12 + t 2 + 2t − 4t 2 − 12t = −t 2 + 12
We return to our fraction. The numerator must be zero:
[Figure caption] And the denominator is different from zero:
t ≠ 0; t ≠ −3; t ≠ −2
The last requirements are fulfilled automatically, since they are all "tied" to integers, and all answers are irrational.
So, fractional rational equation solved, the values of the variable t are found. We return to the solution of the logarithmic equation and remember what t is:
[Figure caption] We bring this equation to the canonical form, we get a number with an irrational degree. Don't let this confuse you - even such arguments can be equated:
[Figure caption] We have two roots. More precisely, two candidates for answers - let's check them for compliance with the scope. Since the base of the logarithm is the variable x, we require the following:
1 ≠ x > 0;
With the same success, we assert that x ≠ 1/125, otherwise the base of the second logarithm will turn into one. Finally, x ≠ 1/25 for the third logarithm.
In total, we got four restrictions:
1 ≠ x > 0; x ≠ 1/125; x ≠ 1/25
Now the question is: do our roots meet these requirements? Certainly satisfied! Because 5 to any power will be Above zero, and the requirement x > 0 is fulfilled automatically.
On the other hand, 1 \u003d 5 0, 1/25 \u003d 5 −2, 1/125 \u003d 5 −3, which means that these restrictions for our roots (which, let me remind you, have an irrational number in the indicator) are also met, and both answers are solutions to the problem.
So we've got the final answer. Key Points There are two tasks in this one:
- Be careful when reversing the logarithm when the argument and base are reversed. Such transformations impose unnecessary restrictions on the domain of definition.
- Do not be afraid to convert logarithms: you can not only flip them, but also open them according to the sum formula and generally change them according to any formulas that you studied when solving logarithmic expressions. However, always remember that some transformations expand the scope, and some narrow it down.
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We continue to study logarithms. In this article we will talk about calculation of logarithms, this process is called logarithm. First, we will deal with the calculation of logarithms by definition. Next, consider how the values of logarithms are found using their properties. After that, we will dwell on the calculation of logarithms through the initially given values of other logarithms. Finally, let's learn how to use tables of logarithms. The whole theory is provided with examples with detailed solutions.
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Computing logarithms by definition
In the simplest cases, it is possible to quickly and easily perform finding the logarithm by definition. Let's take a closer look at how this process takes place.
Its essence is to represent the number b in the form a c , whence, by the definition of the logarithm, the number c is the value of the logarithm. That is, by definition, finding the logarithm corresponds to the following chain of equalities: log a b=log a a c =c .
So, the calculation of the logarithm, by definition, comes down to finding such a number c that a c \u003d b, and the number c itself is the desired value of the logarithm.
Given the information of the previous paragraphs, when the number under the sign of the logarithm is given by some degree of the base of the logarithm, then you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show examples.
Example.
Find log 2 2 −3 , and also calculate the natural logarithm of e 5.3 .
Solution.
The definition of the logarithm allows us to say right away that log 2 2 −3 = −3 . Indeed, the number under the sign of the logarithm is equal to the base 2 to the −3 power.
Similarly, we find the second logarithm: lne 5.3 =5.3.
Answer:
log 2 2 −3 = −3 and lne 5.3 =5.3 .
If the number b under the sign of the logarithm is not given as the degree of the base of the logarithm, then you need to carefully consider whether it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the sign of the logarithm is equal to the base to the power of 1, or 2, or 3, ...
Example.
Compute the logarithms log 5 25 , and .
Solution.
It is easy to see that 25=5 2 , this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2 .
We proceed to the calculation of the second logarithm. A number can be represented as a power of 7: 
(see if necessary). Hence, 
.
Let's rewrite the third logarithm in the following form. Now you can see that 
, whence we conclude that 
. Therefore, by the definition of the logarithm 
.
Briefly, the solution could be written as follows:
Answer:
log 5 25=2 , 
And .
When there is a sufficiently large value under the sign of the logarithm natural number, then it does not hurt to decompose it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore, to calculate this logarithm by definition.
Example.
Find the value of the logarithm.
Solution.
Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of one and the property of the logarithm of a number equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1 . That is, when the number 1 or the number a is under the sign of the logarithm, equal to the base of the logarithm, then in these cases the logarithms are 0 and 1, respectively.
Example.
What are the logarithms and lg10 ?
Solution.
Since , it follows from the definition of the logarithm 
.
In the second example, the number 10 under the sign of the logarithm is the same as its base, so the decimal logarithm of ten equal to one, that is, lg10=lg10 1 =1 .
Answer:
AND lg10=1 .
Note that computing logarithms by definition (which we discussed in the previous paragraph) implies the use of the equality log a a p =p , which is one of the properties of logarithms.
In practice, when the number under the sign of the logarithm and the base of the logarithm are easily represented as a power of some number, it is very convenient to use the formula 
, which corresponds to one of the properties of logarithms. Consider an example of finding the logarithm, illustrating the use of this formula.
Example.
Calculate the logarithm of .
Solution.
Answer:
.
The properties of logarithms not mentioned above are also used in the calculation, but we will talk about this in the following paragraphs.
Finding logarithms in terms of other known logarithms
The information in this paragraph continues the topic of using the properties of logarithms in their calculation. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's take an example for clarification. Let's say we know that log 2 3≈1.584963 , then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .
In the above example, it was enough for us to use the property of the logarithm of the product. However, much more often you have to use a wider arsenal of properties of logarithms in order to calculate the original logarithm in terms of the given ones.
Example.
Calculate the logarithm of 27 to base 60 if it is known that log 60 2=a and log 60 5=b .
Solution.
So we need to find log 60 27 . It is easy to see that 27=3 3 , and the original logarithm, due to the property of the logarithm of the degree, can be rewritten as 3·log 60 3 .
Now let's see how log 60 3 can be expressed in terms of known logarithms. The property of the logarithm of a number equal to the base allows you to write the equality log 60 60=1 . On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2 log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2 log 60 2−log 60 5=1−2 a−b.
Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3 (1−2 a−b)=3−6 a−3 b.
Answer:
log 60 27=3 (1−2 a−b)=3−6 a−3 b.
Separately, it is worth mentioning the meaning of the formula for the transition to a new base of the logarithm of the form 
. It allows you to move from logarithms with any base to logarithms with a specific base, the values of which are known or it is possible to find them. Usually, from the original logarithm, according to the transition formula, they switch to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow them to be calculated with a certain degree of accuracy. In the next section, we will show how this is done.
Tables of logarithms, their use
For an approximate calculation of the values of the logarithms, one can use logarithm tables. The most commonly used table of base 2 logarithms, the table natural logarithms and a table of decimal logarithms. When working in the decimal number system, it is convenient to use a table of logarithms to base ten. With its help, we will learn to find the values of logarithms.
The presented table allows, with an accuracy of one ten-thousandth, to find the values of the decimal logarithms of numbers from 1.000 to 9.999 (with three decimal places). The principle of finding the value of the logarithm using the table of decimal logarithms will be analyzed in specific example- so much clearer. Let's find lg1,256 .
In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). The third digit of the number 1.256 (number 5) is found in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (number 6) is found in the first or last line to the right of the double line (this number is circled in green). Now we find the numbers in the cells of the table of logarithms at the intersection of the marked row and marked columns (these numbers are highlighted orange). The sum of the marked numbers gives the desired value of the decimal logarithm up to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.
Is it possible, using the above table, to find the values of the decimal logarithms of numbers that have more than three digits after the decimal point, and also go beyond the limits from 1 to 9.999? Yes, you can. Let's show how this is done with an example.
Let's calculate lg102.76332 . First you need to write number in standard form : 102.76332=1.0276332 10 2 . After that, the mantissa should be rounded up to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm of the resulting number, that is, we take lg102.76332≈lg1.028·10 2 . Now apply the properties of the logarithm: lg1.028 10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 according to the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the whole process of calculating the logarithm looks like this: lg102.76332=lg1.0276332 10 2 ≈lg1.028 10 2 = lg1.028+lg10 2 =lg1.028+2≈0.012+2=2.012.
In conclusion, it is worth noting that using the table of decimal logarithms, you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.
For example, let's calculate log 2 3 . According to the formula for the transition to a new base of the logarithm, we have . From the table of decimal logarithms we find lg3≈0.4771 and lg2≈0.3010. Thus, 
.
Bibliography.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the Beginnings of Analysis: A Textbook for Grades 10-11 of General Educational Institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).
The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)
Note that the logarithm of a non-positive number is not defined. Also, the base of the logarithm must be a positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the base -2 logarithm of 4 is 2.
Basic logarithmic identity
a log a b = b (a > 0, a ≠ 1) (2)It is important that the domains of definition of the right and left parts of this formula are different. Left side is defined only for b>0, a>0 and a ≠ 1. The right side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic "identity" in solving equations and inequalities can lead to a change in the DPV.
Two obvious consequences of the definition of the logarithm
log a a = 1 (a > 0, a ≠ 1) (3)log a 1 = 0 (a > 0, a ≠ 1) (4)
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the zero power, we get one.
The logarithm of the product and the logarithm of the quotient
log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)
I would like to warn schoolchildren against the thoughtless use of these formulas when solving logarithmic equations and inequalities. When they are used "from left to right", the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f(x) and g(x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x) , we are forced to restrict ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of admissible values, and this is categorically unacceptable, since it can lead to the loss of solutions. A similar problem exists for formula (6).
The degree can be taken out of the sign of the logarithm
log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)And again I would like to call for accuracy. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! Taking the power out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of admissible values. All these remarks apply not only to the power of 2, but also to any even power.
Formula for moving to a new base
log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)That rare case when the ODZ does not change during the conversion. If you have chosen the base c wisely (positive and not equal to 1), the formula for moving to a new base is perfectly safe.
If we choose the number b as a new base c, we get an important special case formulas (8):
Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)
Some simple examples with logarithms
Example 1 Calculate: lg2 + lg50.
Solution. lg2 + lg50 = lg100 = 2. We used the formula for the sum of logarithms (5) and the definition of the decimal logarithm.
Example 2 Calculate: lg125/lg5.
Solution. lg125/lg5 = log 5 125 = 3. We used the new base transition formula (8).
Table of formulas related to logarithms
| a log a b = b (a > 0, a ≠ 1) |
| log a a = 1 (a > 0, a ≠ 1) |
| log a 1 = 0 (a > 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b p = p log a b (a > 0, a ≠ 1, b > 0) |
| log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) |
| log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) |
