Addition of logarithms examples. Converting Logarithmic Expressions

The basic properties of the natural logarithm, graph, domain of definition, set of values, basic formulas, derivative, integral, expansion in power series and representation of the function ln x using complex numbers.

Definition

Natural logarithm is the function y = ln x, the inverse of the exponential, x = e y, and is the logarithm to the base of the number e: ln x = log e x.

The natural logarithm is widely used in mathematics because its derivative has the simplest form: (ln x)′ = 1/ x.

Based on definitions, the base of the natural logarithm is the number e:
e ≅ 2.718281828459045...;
.

Graph of the function y = ln x.

Graph of natural logarithm (functions y = ln x) is obtained from the exponential graph mirror image relative to the straight line y = x.

The natural logarithm is defined at positive values variable x. It increases monotonically in its domain of definition.

At x → 0 the limit of the natural logarithm is minus infinity (-∞).

As x → + ∞, the limit of the natural logarithm is plus infinity (+ ∞). For large x, the logarithm increases quite slowly. Any power function x a with a positive exponent a grows faster than the logarithm.

Properties of the natural logarithm

Domain of definition, set of values, extrema, increase, decrease

The natural logarithm is a monotonically increasing function, so it has no extrema. The main properties of the natural logarithm are presented in the table.

ln x values

ln 1 = 0

Basic formulas for natural logarithms

Formulas following from the definition of the inverse function:

The main property of logarithms and its consequences

Base replacement formula

Any logarithm can be expressed in terms of natural logarithms using the base substitution formula:

Proofs of these formulas are presented in the section "Logarithm".

Inverse function

The inverse of the natural logarithm is the exponent.

If , then

If, then.

Derivative ln x

Derivative of the natural logarithm:
.
Derivative of the natural logarithm of modulus x:
.
Derivative of nth order:
.
Deriving formulas > > >

Integral

The integral is calculated by integration by parts:
.
So,

Expressions using complex numbers

Consider the function of the complex variable z:
.
Let's express the complex variable z via module r and argument φ :
.
Using the properties of the logarithm, we have:
.
Or
.
The argument φ is not uniquely defined. If you put
, where n is an integer,
it will be the same number for different n.

Therefore, the natural logarithm, as a function of a complex variable, is not a single-valued function.

Power series expansion

When the expansion takes place:

Used literature:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.

When converting expressions with logarithms, the listed equalities 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, the fact that for b≥0), from which the corresponding consequences follow. " Side 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 list above, which is answered by 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 power of a positive number with a logarithm in the exponent to swap the base of the power and the number under the logarithm sign. To be fair, we note that examples implying the implementation of transformations of this kind are rare in practice. We will give a few examples below in the text.

Converting numeric expressions with logarithms

We have remembered the properties of logarithms, now it’s time to learn how to apply them in practice to transform expressions. It is natural to start with converting numerical expressions rather than expressions with variables, since they are more convenient and easier to learn the basics. That's what we'll do, and we'll 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 point when, to obtain final result you will need to apply several properties in a row.

Selecting the desired property of logarithms

There are many 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 required result. Usually this is not difficult to do by comparing the type of converted logarithm or expression with the types of left and right parts of formulas expressing the properties of logarithms. If left or right side one of the formulas coincides with a 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 under the letter a) the structure a log a b is clearly visible, 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, the conditions a>0, a≠1, b>0 are met. In this case, the equality 10 log(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 type a log a b (here a=2, b=−7), the expression under the letter g) cannot be converted using the formula a log a b =b. The reason is that it is meaningless 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 fulfillment of the conditions a>0, a≠1, b>0. So, we can't talk about calculating the value of 2 log 2 (−7) . In this case, writing 2 log 2 (−7) =−7 would be an error.

Similarly, in the example under 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.

A transformation is often useful in which a positive number is represented as a power of some positive and non-unity 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) log1, 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 base that is positive and different from unity. Therefore, in examples a) - c) there can be no question of finding the meaning of the expression.

In all other tasks, obviously, the bases of the logarithms contain positive and non-unity numbers 7, e, 10, 3.75 and 5·π 7, respectively, and under the signs of the logarithms there are units everywhere. 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) – e) are equal to zero.

Answer:

a), b), c) 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 the tasks under all the letters, the number under the logarithm sign coincides with its base. Thus, I would like to immediately say that the value of each of the given expressions is 1. However, you should 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 logarithms there are some powers of the base. Based on this, we understand that here we will need the property of the degree of the base: log a a p =p, where a>0, a≠1 and p is any real number. Taking this into account, 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 the expression log −10 (−10) 6 makes no sense.

Answer:

a) log 3 3 11 =11, b) , V) , d) the expression does not make sense.

Example.

Present the expression as a sum or difference of logarithms using the same base: a) , b) , c) log((−5)·(−12)) .

Solution.

a) Under the sign of the logarithm there is a product, 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 quotient logarithm, 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 log((−5)·(−12)) makes sense. But at the same time, for it 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 are −5 and −12 – negative and do not satisfy the conditions x>0, y>0. That is, you cannot carry out such a transformation: log((−5)·(−12))=log(−5)+log(−12). So what should we do? In such cases, the original expression needs a preliminary transformation to avoid negative numbers. About similar cases We will discuss the transformation of expressions with negative numbers under the logarithm sign in detail in one of the pages, but for now we will give a solution to this example, which is clear in advance and without explanation: log((−5)·(−12))=log(5·12)=log5+lg12.

Answer:

A) , b) , c) log((−5)·(−12))=log5+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 previous examples will help us, only now we will apply them from right to left. That is, we transform the sum of logarithms into the logarithm of the product, and the difference of logarithms into 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 logarithm sign: a) log 0.7 5 11, b) , c) log 3 (−5) 6 .

Solution.

It is easy to see that we are dealing with expressions of the form log a b p . The corresponding property of the logarithm has the form log a b p =p·log a b, where a>0, a≠1, b>0, p - any real number. That is, if the conditions a>0, a≠1, b>0 are met, from the logarithm of the power log a b p we can proceed 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 satisfied. 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 what, you can’t cope with the task? 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, when carrying out transformations, the formula for the logarithm of a power has to be applied from right to left in the form p·log a b=log a b p (the same conditions must be met for a, b and p). For example, 3·ln5=ln5 3 and log2·log 2 3=log 2 3 lg2.

Example.

a) Calculate the value of log 2 5 if it is known that log2≈0.3010 and log5≈0.6990. b) Express the fraction as a logarithm to base 3.

Solution.

a) The formula for transition to a new logarithm base allows us to present this logarithm as a ratio of decimal logarithms, the values of which are known to us: . All that remains is to carry out the calculations, we have .

b) Here it is enough to use the formula for moving 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 quite 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 apply one property and nothing more. Now with clear conscience you can proceed to examples, the transformation of which 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 look at examples of the application of consequences from the basic properties of logarithms.

Example.

a) Get rid of the root under the logarithm sign. b) Convert the fraction to base 5 logarithm. c) Free yourself from powers under the sign of the logarithm and in 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 give the answer: .

b) Here we use the formula from right to left, we have .

c) B in this case the formula gives the result . We get .

d) And here it is enough to apply the corollary to which the formula corresponds . So .

e) Property of the logarithm allows us to achieve the desired result: .

Answer:

A) . b) . V) . G) . d) .

Consecutive application of several properties

Real tasks on transforming expressions using the properties of logarithms are usually more complicated than those 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 parentheses, casting similar 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 of actions.

Example.

Calculate the value of an expression (log 3 15−log 3 5) 7 log 7 5.

Solution.

The difference between the logarithms in parentheses, according to the property of the quotient logarithm, 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 definition of a logarithm is equal to 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.

We first transform the logarithm under the logarithm sign using the formula for the logarithm of the power: log 2 2 3 =3. Thus, 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 moving to a new logarithm base 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 short version solutions, which are usually given: .

Answer:

To smoothly transition to the information in the next paragraph, let's take a look at the expressions 5 2+log 5 3, and log0.01. Their structure does not fit any of the properties of logarithms. So what happens, they cannot be converted using the properties of logarithms? It is possible if you carry out preliminary transformations that prepare these expressions for the application of the properties of logarithms. So 5 2+log 5 3 =5 2 5 log 5 3 =25 3=75, and log0.01=log10 −2 =−2. Next we will look in detail at how such expression preparation is carried out.

Preparing Expressions to Use the Properties of Logarithms

Logarithms in the expression being converted 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 no less often, the transformation of these expressions involves the use of the properties of logarithms: to use them you only need preliminary preparation. And this preparation consists of carrying out certain identity transformations, bringing logarithms to a form convenient for applying the properties.

To be fair, we note that almost any transformation of expressions can act as preliminary transformations, from the banal reduction of similar terms to the use of trigonometric formulas. This is understandable, since the expressions being converted can contain any mathematical objects: brackets, modules, fractions, roots, powers, etc. Thus, you need to be prepared to perform any necessary transformation in order to further be able to take advantage of the properties of logarithms.

Let us say right away that at this point we do not set ourselves the task of classifying and analyzing all conceivable preliminary transformations that would allow us to subsequently apply the properties of logarithms or the definition of a logarithm. Here we will focus on only four of them, which are the most typical and most often encountered in practice.

And now about each of them in detail, after which, within the framework of our topic, all that remains is to understand the transformation of expressions with variables under the signs of logarithms.

Identification of powers under the logarithm sign and at 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 convert this expression to simplify it, or better yet, 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. Here it is easy to notice 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 thought: if possible, you can try to isolate the degree under the sign of the logarithm and in its base in order to apply the property of the logarithm of the degree or its consequences. It remains only to figure out how to distinguish these degrees. Let's give some recommendations on this issue.

Sometimes it is quite obvious that the number under the logarithm sign and/or in its base represents some integer power, as in the example discussed above. Almost constantly we 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 powers of three: 9 = 3 2, 27 = 3 3, 81 = 3 4, 243 = 3 5, ... In general, it won’t hurt if you have before your eyes table of powers of natural numbers within a dozen. It is also not difficult to work with integer powers of ten, one 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 you to represent the numbers 343 and 1/243 as powers 7 3 and 3 −4, respectively. Therefore, the following transformation of a 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 difficult cases to distinguish powers of numbers one has to resort to .

Example.

Convert the expression to more simple view log 3 648 log 2 3 .

Solution.

Let's see what 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 transform the logarithm of the product into the sum of logarithms, after which we apply the properties of the logarithm of the power:
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 a corollary from the property of the logarithm of the power, which corresponds to the formula , the product log32·log23 is the product of , and, as is known, it is equal to one. Taking this into account, 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 represent products or ratios of the roots and/or powers of some numbers, for example, , . Such expressions can be expressed as powers. To do this, a transition is made from roots to powers, and and are used. These transformations make it possible to isolate the powers 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, according to corresponding property we have degrees 5 2 ·5 −0.5 ·5 −1 =5 2−0.5−1 =5 0.5.

Now let’s transform the fraction under the logarithm sign: we’ll move from the root to the power, after which we’ll use the property of the ratio of powers with the same bases: .

It remains to substitute the obtained results into 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, we apply the property of the root of a power, move from the root to the power, and use the property of the ratio of powers to convert the base of the logarithm to a power: .

Considering last result, we have .

Answer:

A) , b) .

It is clear that in general case to obtain powers under the sign of the logarithm and in its base, various transformations may be required various expressions. Let's give a couple of examples.

Example.

What is the meaning of the expression: a) , b) .

Solution.

We further 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 type according to the property of the logarithm of the power log a b p =p·log a b , therefore, with a given expression I want to do the same, and move from log 2 (x+1) 4 to 4·log 2 (x+1) . Now let's calculate the value of the original expression and the expression obtained after the transformation, for example, when x=−2. We have log 2 (−2+1) 4 =log 2 1=0 , and 4 log 2 (−2+1)=4 log 2 (−1)- a meaningless expression. This raises a logical question: “What did we do wrong?”

And the reason is this: 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 this formula we have the right to apply only if the conditions are met: 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 as x>−1 (for A and p, the conditions are met). However, in our case, the ODZ of 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 DL

Let's continue to analyze the transformation of the expression we have chosen log 2 (x+1) 4 , and now let's see what happens to the ODZ when moving 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 range of acceptable values of the 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 moving from log 2 (x+1) 4 to 4·log 2 (x+1), the range of permissible values narrows. And we agreed to avoid transformations that lead to a narrowing of the DL, as this can lead to various negative consequences.

Here it is worth noting for yourself that it is useful to control the OA at each step of the transformation and prevent its narrowing. And if suddenly at some stage of the transformation there was a narrowing of the DL, then it is worth looking very carefully at whether this transformation is permissible and whether we had the right to carry out it.

To be fair, let’s say that in practice we usually have to work with expressions in which the ODZ of variables is such that, when carrying out transformations, we can use the properties of logarithms without restrictions in the form already known to us, both from left to right and from right to left. You quickly get used to this, and you begin to carry out transformations mechanically, without thinking whether it was possible to carry them out. And at such moments, as luck would have it, more complex examples slip through in which careless application of the properties of logarithms leads to errors. So you need to always be on the lookout and make sure that there is no narrowing of the ODZ.

It would 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 OD, and as a result, to errors:

Some transformations of expressions based on the properties of logarithms can also lead to the opposite - expansion of the ODZ. For example, the transition from 4·log 2 (x+1) to log 2 (x+1) 4 expands the ODZ from the set (−1, +∞) to (−∞, −1)∪(−1, +∞) . Such transformations take place if we remain within the framework of the original expression. So the just mentioned transformation 4·log 2 (x+1)=log 2 (x+1) 4 takes place on the ODZ of the variable x for the original expression 4·log 2 (x+1), that is, for x+1> 0, which is the same as (−1, +∞).

Now that we have discussed the nuances that you need to pay attention to when transforming expressions with variables using the properties of logarithms, it remains to figure out how to correctly carry out these transformations.

X+2>0 . Does it work in our case? To answer this question, let's look at the ODZ of the variable x. It is determined by the system of inequalities , which is equivalent to the condition x+2>0 (if necessary, see the article solving systems of inequalities). Thus, we can safely apply the property of the logarithm of the power.

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)·log(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.

But what to do when the conditions accompanying the properties of logarithms are not met in the ODZ? We will understand this with examples.

Let us be required to simplify the expression log(x+2) 4 − log(x+2) 2 . The transformation of this expression, unlike the expression from the previous example, does not allow free use of the property of the logarithm of the power. Why? The ODZ of variable x in this case is the union of two intervals x>−2 and x<−2 . При x>−2 we can easily apply the property of the logarithm of a power and act 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 one more 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 properties of the degree k lg|x+2| 4 −lg|x+2| 2. The resulting expression can be transformed using the property of the logarithm of a power, since |x+2|>0 for any value of the variable. We have log|x+2| 4 −lg|x+2| 2 =4·lg|x+2|−2·lg|x+2|=2·lg|x+2|. Now you can free yourself from the module, since it has done its job. Since we carry out the transformation 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 look at one more example so that working with modules becomes familiar. Let us conceive from the expression go to the sum and difference of logarithms of 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 easily 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 considered interval we represent x−2 and x−3 using the modulus as −|x−2| and −|x−3| respectively. At the same time

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, which 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 logarithms log a |X|+log a |Y| , a>0 , a≠1 .
The logarithm of a particular form log a (X:Y) can be replaced by the difference of 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 we can go 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 exponential and logarithmic equations in a collection of problems in mathematics for those entering universities, edited by M. I. Skanavi.

Example.

Simplify the expression .

Solution.

It would be good to apply the properties of the logarithm of the power, sum and difference. But can we do this here? To answer this question we need to know the DZ.

Let's define it:

It is quite obvious that the expressions x+4, x−2 and (x+4) 13 in the range of permissible values of the variable x can take on both positive and negative values. Therefore, we will have to operate through modules.

The module properties allow you to rewrite it as , so

Also, nothing prevents you from using the property of the logarithm of a power, and then bringing similar terms:

Another sequence of transformations leads to the same result:

and since on the ODZ the expression x−2 can take both positive and negative values, then when taking an even exponent 14

main properties.

logax + logay = loga(x y);
logax − logay = loga (x: y).

identical grounds

Log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x >

Task. Find the meaning of the expression:

Transition to a new foundation

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

Task. Find the meaning of the expression:

Basic properties of the logarithm

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy.

Basic properties of logarithms

Knowing this rule, you will know and exact value exhibitors, and the date of birth of Leo Tolstoy.

Examples for logarithms

Logarithm expressions

Example 1.
A). x=10ac^2 (a>0,c>0).

Using properties 3.5 we calculate

4. Where .

Example 2. Find x if

Example 3. Let the value of logarithms be given

Calculate log(x) if

Basic properties of logarithms

Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly regular numbers, there are rules here, which are called main properties.

You definitely need to know these rules - without them not a single serious problem can be solved. logarithmic problem. In addition, there are very few of them - you can learn everything in one day. So let's get started.

Adding and subtracting logarithms

Consider two logarithms with the same bases: logax and logay. Then they can be added and subtracted, and:

logax + logay = loga(x y);
logax − logay = loga (x: y).

So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Please note: key point Here - identical grounds. If the reasons are different, these rules do not work!

These formulas will help you calculate logarithmic expression even when its individual parts are not counted (see the lesson “What is a logarithm”). Take a look at the examples and see:

Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact tests. What about the controls? similar expressions in all seriousness (sometimes with virtually no changes) are offered on the Unified State Examination.

Extracting the exponent from the logarithm

It's easy to notice that last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the meaning of the expression:

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:

I think to last example clarification required. Where have logarithms gone? Until the very last moment we work only with the denominator.

Logarithm formulas. Logarithms examples solutions.

We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator contain the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.

Transition to a new foundation

Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?

Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

In particular, if we set c = x, we get:

From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.

These formulas are rarely found in conventional numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let’s “reverse” the second logarithm:

Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.

Task. Find the value of the expression: log9 100 lg 3.

The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:

Now let's get rid of the decimal logarithm by moving to a new base:

Basic logarithmic identity

Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.

The second formula is actually a paraphrased definition. That's what it's called: .

In fact, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: the result is the same number a. Read this paragraph carefully again - many people get stuck on it.

Like the formulas for transition to a new base, the main logarithmic identity sometimes it is the only possible solution.

Task. Find the meaning of the expression:

Note that log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Considering the rules for multiplying powers with the same basis, we get:

If anyone doesn’t know, this was a real task from the Unified State Exam :)

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.

logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument contains one - logarithm equal to zero! Because a0 = 1 is direct consequence from the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.

Basic properties of the logarithm

It is necessary to know the above properties, since almost all problems and examples related to logarithms are solved on their basis. The rest of the exotic properties can be derived through mathematical manipulations with these formulas

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

When calculating the formula for the sum and difference of logarithms (3.4) you come across quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.

Common cases of logarithms

Some of the common logarithms are those in which the base is even ten, exponential or two.
The logarithm to base ten is usually called the decimal logarithm and is simply denoted by lg(x).

It is clear from the recording that the basics are not written in the recording. For example

A natural logarithm is a logarithm whose base is an exponent (denoted by ln(x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

And another important logarithm to base two is denoted by

The derivative of the logarithm of a function is equal to one divided by the variable

The integral or antiderivative logarithm is determined by the relationship

The given material is enough for you to solve a wide class of problems related to logarithms and logarithms. To help you understand the material, I will give just a few common examples from school curriculum and universities.

Examples for logarithms

Logarithm expressions

Example 1.
A). x=10ac^2 (a>0,c>0).

Using properties 3.5 we calculate

2.
By the property of difference of logarithms we have

3.
Using properties 3.5 we find

4. Where .

In appearance complex expression using a number of rules is simplified to form

Finding logarithm values

Example 2. Find x if

Solution. For calculation, we apply to the last term 5 and 13 properties

We put it on record and mourn

Since the bases are equal, we equate the expressions

Logarithms. Entry level.

Let the value of logarithms be given

Calculate log(x) if

Solution: Let's take a logarithm of the variable to write the logarithm through the sum of its terms

This is just the beginning of our acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the knowledge you gain to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge to another equally important topic - logarithmic inequalities...

Basic properties of logarithms

Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.

You definitely need to know these rules - not a single serious logarithmic problem can be solved without them. In addition, there are very few of them - you can learn everything in one day. So let's get started.

Adding and subtracting logarithms

Consider two logarithms with the same bases: logax and logay. Then they can be added and subtracted, and:

logax + logay = loga(x y);
logax − logay = loga (x: y).

So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Please note: the key point here is identical grounds. If the reasons are different, these rules do not work!

These formulas will help you calculate a logarithmic expression even when its individual parts are not considered (see the lesson “What is a logarithm”). Take a look at the examples and see:

Task. Find the value of the expression: log6 4 + log6 9.

Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.

Extracting the exponent from the logarithm

Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

How to solve logarithms

This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the meaning of the expression:

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:

I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.

Transition to a new foundation

Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

In particular, if we set c = x, we get:

These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let’s “reverse” the second logarithm:

Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.

Task. Find the value of the expression: log9 100 lg 3.

The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:

Now let's get rid of the decimal logarithm by moving to a new base:

Basic logarithmic identity

Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.

The second formula is actually a paraphrased definition. That's what it's called: .

Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the meaning of the expression:

Note that log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:

If anyone doesn’t know, this was a real task from the Unified State Exam :)

Logarithmic unit and logarithmic zero

logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.

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Let's start with properties of the logarithm of one. Its formulation is as follows: the logarithm of unity is equal to zero, that is, log a 1=0 for any a>0, a≠1. The proof is not difficult: since a 0 =1 for any a satisfying the above conditions a>0 and a≠1, then the equality log a 1=0 to be proved follows immediately from the definition of the logarithm.

Let us give examples of the application of the considered property: log 3 1=0, log1=0 and .

Let's move on to to the following property: logarithm of the number, equal to the base, equal to one, that is, log a a=1 for a>0, a≠1. Indeed, since a 1 =a for any a, then by definition of the logarithm log a a=1.

Examples of the use of this property of logarithms are the equalities log 5 5=1, log 5.6 5.6 and lne=1.

For example, log 2 2 7 =7, log10 -4 =-4 and .

Logarithm of the product of two positive numbers x and y equal to the product logarithms of these numbers: log a (x y)=log a x+log a y, a>0 , a≠1 . Let us prove the property of the logarithm of a product. Due to the properties of the degree a log a x+log a y =a log a x ·a log a y, and since by the main logarithmic identity a log a x =x and a log a y =y, then a log a x ·a log a y =x·y. Thus, a log a x+log a y =x·y, from which, by the definition of a logarithm, the equality being proved follows.

Let's show examples of using the property of the logarithm of a product: log 5 (2 3)=log 5 2+log 5 3 and .

The property of the logarithm of a product can be generalized to the product finite number n positive numbers x 1 , x 2 , …, x n as log a (x 1 ·x 2 ·…·x n)= log a x 1 +log a x 2 +…+log a x n . This equality can be proven without problems.

For example, the natural logarithm of a product can be replaced by the sum of three natural logarithms numbers 4 , e , and .

Logarithm of the quotient of two positive numbers x and y equal to the difference logarithms of these numbers. The property of the logarithm of a quotient corresponds to a formula of the form , where a>0, a≠1, x and y are some positive numbers. The validity of this formula is proven as well as the formula for the logarithm of a product: since , then by definition of a logarithm.

Here is an example of using this property of the logarithm: .

Let's move on to property of the logarithm of the power. The logarithm of a degree is equal to the product of the exponent and the logarithm of the modulus of the base of this degree. Let us write this property of the logarithm of a power as a formula: log a b p =p·log a |b|, where a>0, a≠1, b and p are numbers such that the degree b p makes sense and b p >0.

First we prove this property for positive b. The basic logarithmic identity allows us to represent the number b as a log a b , then b p =(a log a b) p , and the resulting expression, due to the property of power, is equal to a p·log a b . So we come to the equality b p =a p·log a b, from which, by the definition of a logarithm, we conclude that log a b p =p·log a b.

It remains to prove this property for negative b. Here we note that the expression log a b p for negative b makes sense only for even exponents p (since the value of the degree b p must be greater than zero, otherwise the logarithm will not make sense), and in this case b p =|b| p. Then b p =|b| p =(a log a |b|) p =a p·log a |b|, from where log a b p =p·log a |b| .

For example, and ln(-3) 4 =4·ln|-3|=4·ln3 .

It follows from the previous property property of the logarithm from the root: the logarithm of the nth root is equal to the product of the fraction 1/n by the logarithm of the radical expression, that is, , where a>0, a≠1, n – natural number, greater than one, b>0.

The proof is based on the equality (see), which is valid for any positive b, and the property of the logarithm of the power: .

Here is an example of using this property: .

Now let's prove formula for moving to a new logarithm base kind . To do this, it is enough to prove the validity of the equality log c b=log a b·log c a. The basic logarithmic identity allows us to represent the number b as a log a b , then log c b=log c a log a b . It remains to use the property of the logarithm of the degree: log c a log a b =log a b log c a. This proves the equality log c b=log a b·log c a, which means the formula for moving to a new logarithm base has also been proven.

Let's show a couple of examples of using this property of logarithms: and .

The formula for moving to a new base allows you to move on to working with logarithms that have a “convenient” base. For example, with its help you can switch to natural or decimal logarithms so that you can calculate the value of the logarithm from the table of logarithms. The formula for moving to a new logarithm base also allows, in some cases, to find the value of a given logarithm when the values of some logarithms with other bases are known.

Often used special case formulas for transition to a new base of the logarithm with c=b of the form . This shows that log a b and log b a – . For example, .

The formula is also often used , which is convenient for finding logarithm values. To confirm our words, we will show how it can be used to calculate the value of a logarithm of the form . We have . To prove the formula it is enough to use the formula for transition to a new base of the logarithm a: .

It remains to prove the properties of comparison of logarithms.

Let us prove that for any positive numbers b 1 and b 2, b 1 log a b 2 , and for a>1 – the inequality log a b 1

Finally, it remains to prove the last of the listed properties of logarithms. Let us limit ourselves to the proof of its first part, that is, we will prove that if a 1 >1, a 2 >1 and a 1 1 is true log a 1 b>log a 2 b . The remaining statements of this property of logarithms are proved according to a similar principle.

Let's use the opposite method. Suppose that for a 1 >1, a 2 >1 and a 1 1 is true log a 1 b≤log a 2 b . Based on the properties of logarithms, these inequalities can be rewritten as And respectively, and from them it follows that log b a 1 ≤log b a 2 and log b a 1 ≥log b a 2, respectively. Then, according to the properties of powers with the same bases, the equalities b log b a 1 ≥b log b a 2 and b log b a 1 ≥b log b a 2 must hold, that is, a 1 ≥a 2 . So we came to a contradiction to the condition a 1

References.

Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.

Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).