Logarithms: examples and solutions. Logarithm - properties, formulas, graph Basic properties of logarithms

\(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)

Let's explain it more simply. For example, \(\log_(2)(8)\) is equal to the power to which \(2\) must be raised to get \(8\). From this it is clear that \(\log_(2)(8)=3\).

Examples:	\(\log_(5)(25)=2\)	because \(5^(2)=25\)
	\(\log_(3)(81)=4\)	because \(3^(4)=81\)
	\(\log_(2)\)\(\frac(1)(32)\) \(=-5\)	because \(2^(-5)=\)\(\frac(1)(32)\)

Argument and base of logarithm

Any logarithm has the following “anatomy”:

The argument of a logarithm is usually written at its level, and the base is written in subscript closer to the logarithm sign. And this entry reads like this: “logarithm of twenty-five to base five.”

How to calculate logarithm?

To calculate the logarithm, you need to answer the question: to what power should the base be raised to get the argument?

For example, calculate the logarithm: a) \(\log_(4)(16)\) b) \(\log_(3)\)\(\frac(1)(3)\) c) \(\log_(\sqrt (5))(1)\) d) \(\log_(\sqrt(7))(\sqrt(7))\) e) \(\log_(3)(\sqrt(3))\)

a) To what power must \(4\) be raised to get \(16\)? Obviously the second one. That's why:

\(\log_(4)(16)=2\)

\(\log_(3)\)\(\frac(1)(3)\) \(=-1\)

c) To what power must \(\sqrt(5)\) be raised to get \(1\)? What power makes any number one? Zero, of course!

\(\log_(\sqrt(5))(1)=0\)

d) To what power must \(\sqrt(7)\) be raised to obtain \(\sqrt(7)\)? Firstly, any number to the first power is equal to itself.

\(\log_(\sqrt(7))(\sqrt(7))=1\)

e) To what power must \(3\) be raised to obtain \(\sqrt(3)\)? From we know that is a fractional power, which means the square root is the power of \(\frac(1)(2)\) .

\(\log_(3)(\sqrt(3))=\)\(\frac(1)(2)\)

Example : Calculate logarithm \(\log_(4\sqrt(2))(8)\)

Solution :

\(\log_(4\sqrt(2))(8)=x\)		We need to find the value of the logarithm, let's denote it as x. Now let's use the definition of a logarithm: \(\log_(a)(c)=b\) \(\Leftrightarrow\) \(a^(b)=c\)
\((4\sqrt(2))^(x)=8\)		What connects \(4\sqrt(2)\) and \(8\)? Two, because both numbers can be represented by twos: \(4=2^(2)\) \(\sqrt(2)=2^(\frac(1)(2))\) \(8=2^(3)\)
\(((2^(2)\cdot2^(\frac(1)(2))))^(x)=2^(3)\)		On the left we use the properties of the degree: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\)
\(2^(\frac(5)(2)x)=2^(3)\)		The bases are equal, we move on to equality of indicators
\(\frac(5x)(2)\) \(=3\)		Multiply both sides of the equation by \(\frac(2)(5)\)
		The resulting root is the value of the logarithm

Answer : \(\log_(4\sqrt(2))(8)=1,2\)

Why was the logarithm invented?

To understand this, let's solve the equation: \(3^(x)=9\). Just match \(x\) to make the equation work. Of course, \(x=2\).

Now solve the equation: \(3^(x)=8\).What is x equal to? That's the point.

The smartest ones will say: “X is a little less than two.” How exactly to write this number? To answer this question, the logarithm was invented. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).

I want to emphasize that \(\log_(3)(8)\), like any logarithm is just a number. Yes, it looks unusual, but it’s short. Because if we wanted to write it as a decimal, it would look like this: \(1.892789260714.....\)

Example : Solve the equation \(4^(5x-4)=10\)

Solution :

\(4^(5x-4)=10\)		\(4^(5x-4)\) and \(10\) cannot be brought to the same base. This means you can’t do without a logarithm. Let's use the definition of logarithm: \(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
\(\log_(4)(10)=5x-4\)		Let's flip the equation so that X is on the left
\(5x-4=\log_(4)(10)\)		Before us. Let's move \(4\) to the right. And don't be afraid of the logarithm, treat it like an ordinary number.
\(5x=\log_(4)(10)+4\)		Divide the equation by 5
\(x=\)\(\frac(\log_(4)(10)+4)(5)\)		This is our root. Yes, it looks unusual, but they don’t choose the answer.

Answer : \(\frac(\log_(4)(10)+4)(5)\)

Decimal and natural logarithms

As stated in the definition of a logarithm, its base can be any positive number except one \((a>0, a\neq1)\). And among all the possible bases, there are two that occur so often that a special short notation was invented for logarithms with them:

Natural logarithm: a logarithm whose base is Euler's number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).

That is, \(\ln(a)\) is the same as \(\log_(e)(a)\)

Decimal Logarithm: A logarithm whose base is 10 is written \(\lg(a)\).

That is, \(\lg(a)\) is the same as \(\log_(10)(a)\), where \(a\) is some number.

Basic logarithmic identity

Logarithms have many properties. One of them is called the “Basic Logarithmic Identity” and looks like this:

\(a^(\log_(a)(c))=c\)

This property follows directly from the definition. Let's see exactly how this formula came about.

Let us recall a short notation of the definition of logarithm:

if \(a^(b)=c\), then \(\log_(a)(c)=b\)

That is, \(b\) is the same as \(\log_(a)(c)\). Then we can write \(\log_(a)(c)\) instead of \(b\) in the formula \(a^(b)=c\). It turned out \(a^(\log_(a)(c))=c\) - the main logarithmic identity.

You can find other properties of logarithms. With their help, you can simplify and calculate the values of expressions with logarithms, which are difficult to calculate directly.

Example : Find the value of the expression \(36^(\log_(6)(5))\)

Solution :

Answer : \(25\)

How to write a number as a logarithm?

As mentioned above, any logarithm is just a number. The converse is also true: any number can be written as a logarithm. For example, we know that \(\log_(2)(4)\) is equal to two. Then instead of two you can write \(\log_(2)(4)\).

But \(\log_(3)(9)\) is also equal to \(2\), which means we can also write \(2=\log_(3)(9)\) . Likewise with \(\log_(5)(25)\), and with \(\log_(9)(81)\), etc. That is, it turns out

\(2=\log_(2)(4)=\log_(3)(9)=\log_(4)(16)=\log_(5)(25)=\log_(6)(36)=\ log_(7)(49)...\)

Thus, if we need, we can write two as a logarithm with any base anywhere (be it in an equation, in an expression, or in an inequality) - we simply write the base squared as an argument.

It’s the same with the triple – it can be written as \(\log_(2)(8)\), or as \(\log_(3)(27)\), or as \(\log_(4)(64) \)... Here we write the base in the cube as an argument:

\(3=\log_(2)(8)=\log_(3)(27)=\log_(4)(64)=\log_(5)(125)=\log_(6)(216)=\ log_(7)(343)...\)

And with four:

\(4=\log_(2)(16)=\log_(3)(81)=\log_(4)(256)=\log_(5)(625)=\log_(6)(1296)=\ log_(7)(2401)...\)

And with minus one:

\(-1=\) \(\log_(2)\)\(\frac(1)(2)\) \(=\) \(\log_(3)\)\(\frac(1)( 3)\) \(=\) \(\log_(4)\)\(\frac(1)(4)\) \(=\) \(\log_(5)\)\(\frac(1 )(5)\) \(=\) \(\log_(6)\)\(\frac(1)(6)\) \(=\) \(\log_(7)\)\(\frac (1)(7)\) \(...\)

And with one third:

\(\frac(1)(3)\) \(=\log_(2)(\sqrt(2))=\log_(3)(\sqrt(3))=\log_(4)(\sqrt( 4))=\log_(5)(\sqrt(5))=\log_(6)(\sqrt(6))=\log_(7)(\sqrt(7))...\)

Any number \(a\) can be represented as a logarithm with base \(b\): \(a=\log_(b)(b^(a))\)

Example : Find the meaning of the expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)

Solution :

Answer : \(1\)

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 - without them, not a single serious logarithmic problem can be solved. 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: log a x and log a y. Then they can be added and subtracted, and:

log a x+ log a y= log a (x · y);
log a x− log a y= log a (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 lesson “What is a logarithm”). Take a look at the examples and see:

Log 6 4 + log 6 9.

Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.

Task. Find the value of the expression: log 2 48 − log 2 3.

The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.

Task. Find the value of the expression: log 3 135 − log 3 5.

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

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: log 7 49 6 .

Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the meaning of the expression:
[Caption for the picture]

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:

[Caption for the picture]

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.

Now let's look at the main fraction. The numerator and denominator contain the same number: log 2 7. Since log 2 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 log be given a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:
[Caption for the picture]
In particular, if we put c = x, we get:
[Caption for the picture]

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 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: log 5 16 log 2 25.

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

Now let’s “reverse” the second logarithm:

[Caption for the picture]

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: log 9 100 lg 3.

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

[Caption for the picture]

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

[Caption for the picture]

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 an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.

The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.

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

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:
[Caption for the picture]

Note that log 25 64 = log 5 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:

[Caption for the picture]

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.

log a a= 1 is a logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a 0 = 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.

Follows from its definition. And so the logarithm of the number b based on A is defined as the exponent to which a number must be raised a to get the number b(logarithm exists only for positive numbers).

From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation a x =b. For example, log 2 8 = 3 because 8 = 2 3 . The formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b based on a equals With. It is also clear that the topic of logarithms is closely related to the topic of powers of a number.

With logarithms, as with any numbers, you can do operations of addition, subtraction and transform in every possible way. But due to the fact that logarithms are not entirely ordinary numbers, their own special rules apply here, which are called main properties.

Adding and subtracting logarithms.

Let's take two logarithms with the same bases: log a x And log a y. Then it is possible to perform addition and subtraction operations:

log a x+ log a y= log a (x·y);

log a x - log a y = log a (x:y).

log a(x 1 . x 2 . x 3 ... x k) = log a x 1 + log a x 2 + log a x 3 + ... + log a x k.

From logarithm quotient theorem One more property of the logarithm can be obtained. It is common knowledge that log a 1= 0, therefore

log a 1 /b= log a 1 - log a b= -log a b.

This means there is an equality:

log a 1 / b = - log a b.

Logarithms of two reciprocal numbers for the same reason will differ from each other solely by sign. So:

Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.

We continue to study logarithms. In this article we will talk about calculating logarithms, this process is called logarithm. First we will understand the calculation of logarithms by definition. Next, let's look at how the values of logarithms are found using their properties. After this, we will focus on calculating logarithms through the initially specified values of other logarithms. Finally, let's learn how to use logarithm tables. The entire theory is provided with examples with detailed solutions.

Page navigation.

Calculating logarithms by definition

In the simplest cases it is possible to perform quite quickly and easily finding the logarithm by definition. Let's take a closer look at how this process happens.

Its essence is to represent the number b in the form a c, from which, by the definition of a logarithm, the number c is the value of the logarithm. That is, by definition, the following chain of equalities corresponds to finding the logarithm: log a b=log a a c =c.

So, calculating a logarithm by definition comes down to finding a number c such that a c = b, and the number c itself is the desired value of the logarithm.

Taking into account the information in the previous paragraphs, when the number under the logarithm sign is given by a certain power of the logarithm base, you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show solutions to examples.

Example.

Find log 2 2 −3, and also calculate the natural logarithm of the number e 5,3.

Solution.

The definition of the logarithm allows us to immediately say that log 2 2 −3 =−3. Indeed, the number under the logarithm sign is equal to 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 logarithm sign is not specified as a power of the base of the logarithm, then you need to carefully look to see if 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 logarithm sign is equal to the base to the power of 1, or 2, or 3, ...

Example.

Calculate 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.

Let's move on to calculating the second logarithm. The 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 , from which we conclude that . Therefore, by the definition of logarithm .

Briefly, the solution could be written as follows: .

Answer:

log 5 25=2 , And .

When there is a sufficiently large natural number under the logarithm sign, it does not hurt to factor it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore 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 under the sign of the logarithm there is a number 1 or a number a equal to the base of the logarithm, then in these cases the logarithms are equal to 0 and 1, respectively.

Example.

What are logarithms and log10 equal to?

Solution.

Since , then from the definition of logarithm it follows .

In the second example, the number 10 under the logarithm sign coincides with its base, so the decimal logarithm of ten is equal to one, that is, lg10=lg10 1 =1.

Answer:

AND lg10=1 .

Note that the calculation of 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 a number under the logarithm sign and the base of the logarithm are easily represented as a power of a certain number, it is very convenient to use the formula , which corresponds to one of the properties of logarithms. Let's look at an example of finding a logarithm that illustrates the use of this formula.

Example.

Calculate the logarithm.

Solution.

Answer:

Properties of logarithms not mentioned above are also used in calculations, but we will talk about this in the following paragraphs.

Finding logarithms through other known logarithms

The information in this paragraph continues the topic of using the properties of logarithms when calculating them. 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 give 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 a product. However, much more often it is necessary to use a wider arsenal of properties of logarithms in order to calculate the original logarithm through the given ones.

Example.

Calculate the logarithm of 27 to base 60 if you know 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 power, can be rewritten as 3·log 60 3 .

Now let's see how to express log 60 3 in terms of known logarithms. The property of the logarithm of a number equal to the base allows us 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 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, using the transition formula, they move to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow their values to be calculated with a certain degree of accuracy. In the next paragraph we will show how this is done.

Logarithm tables and their uses

For approximate calculation of logarithm values can be used logarithm tables. The most commonly used base 2 logarithm table, natural logarithm table, and decimal logarithm table. When working in the decimal number system, it is convenient to use a table of logarithms based on base ten. With its help we will learn to find the values of logarithms.

The presented table allows you to find the values of the decimal logarithms of numbers from 1,000 to 9,999 (with three decimal places) with an accuracy of one ten-thousandth. We will analyze the principle of finding the value of a logarithm using a table of decimal logarithms using a specific example - it’s clearer this way. Let's find log1.256.

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 (digit 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 (digit 6) is found in the first or last line to the right of the double line (this number is circled with a green line). Now we find the numbers in the cells of the logarithm table at the intersection of the marked row and marked columns (these numbers are highlighted in orange). The sum of the marked numbers gives the desired value of the decimal logarithm accurate to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.

Is it possible, using the table above, to find the values of decimal logarithms of numbers that have more than three digits after the decimal point, as well as those that go beyond the range 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 down number in standard form: 102.76332=1.0276332·10 2. After this, the mantissa should be rounded 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 log102.76332≈lg1.028·10 2. Now we 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 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.

In conclusion, it is worth noting that using a 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 transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, .

Bibliography.

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).

The basic properties of the logarithm, logarithm graph, domain of definition, set of values, basic formulas, increasing and decreasing are given. Finding the derivative of a logarithm is considered. As well as integral, power series expansion and representation using complex numbers.

Content

Domain, set of values, increasing, decreasing

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


Domain	0 < x < + ∞	0 < x < + ∞
Range of values	- ∞ < y < + ∞	- ∞ < y < + ∞
Monotone	monotonically increases	monotonically decreases
Zeros, y = 0	x = 1	x = 1
Intercept points with the ordinate axis, x = 0	No	No
	+ ∞	- ∞
	- ∞	+ ∞

Private values

The logarithm to base 10 is called decimal logarithm and is denoted as follows:

Logarithm to base e called natural logarithm:

Basic formulas for logarithms

Properties of the logarithm arising from the definition of the inverse function:

The main property of logarithms and its consequences

Base replacement formula

Logarithm is the mathematical operation of taking a logarithm. When taking logarithms, products of factors are converted into sums of terms.
Potentiation is the mathematical operation inverse to logarithm. During potentiation, a given base is raised to the degree of expression over which potentiation is performed. In this case, the sums of terms are transformed into products of factors.

Proof of basic formulas for logarithms

Formulas related to logarithms follow from formulas for exponential functions and from the definition of an inverse function.

Consider the property of the exponential function
.
Then
.
Let's apply the property of the exponential function
:
.

Let us prove the base replacement formula.
;
.
Assuming c = b, we have:

Inverse function

The inverse of a logarithm to base a is an exponential function with exponent a.

If , then

Derivative of logarithm

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

To find the derivative of a logarithm, it must be reduced to the base e.
;
.

Integral

The integral of the logarithm is calculated by integrating by parts: .
So,

Expressions using complex numbers

Consider the complex number function z:
.
Let's express a complex number z via module r and argument φ :
.
Then, using the properties of the logarithm, we have:
.
Or

However, the argument φ not uniquely defined. If you put
, where n is an integer,
then it will be the same number for different n.

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

Power series expansion

When the expansion takes place:

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