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

basic properties.

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

same grounds

log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

What if there is a degree in the base or argument of the logarithm? 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 logarithm is observed: a > 0, a ≠ 1, x >

A task. Find the value 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:

A task. Find the value 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 2.7 and twice the year of birth of Leo 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

Take the logarithm of expressions

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

By 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 number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

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

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

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

These formulas will help calculate the 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:

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

A 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.

A 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 considered separately. But after transformations quite normal numbers turn out. Based on this fact, many test papers. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

It is easy to see that the last rule follows their 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 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 sign of the logarithm into the logarithm itself. This is what is most often required.

A task. Find the value of the expression: log7 496.

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

A task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:

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

Formulas of logarithms. Logarithms are examples of solutions.

They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have 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 was done. The result is 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 bases are different? What if they are not exact powers of the same number?

Formulas for transition to a new base come to the rescue. We 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 put c = x, we get:

It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is in the denominator.

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

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

A task. Find the value of the expression: log5 16 log2 25.

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

Now let's flip the second logarithm:

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.

A 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 it 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 process of solving it is required to represent a number as a logarithm to a given base. In this case, the 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's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

A task. Find the value of the expression:

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

If someone is not in the know, this was a real task from the Unified State Examination 🙂

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.

logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument is one - the logarithm 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.

Basic properties of the logarithm

The above properties need to be known, since, on their basis, almost all problems and examples are solved based on logarithms. The remaining exotic properties can be derived by mathematical manipulations with these formulas

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

When calculating the formulas for the sum and difference of logarithms (3.4) are encountered 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 deuce.
The base ten logarithm is called decimal logarithm and simply denote lg(x).

It can be seen from the record that the basics are not written in the record. For example

The natural logarithm is the logarithm whose basis is the exponent (denoted ln(x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo 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 base two logarithm is

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

The integral or antiderivative logarithm is determined by the dependence

The above material is enough for you to solve a wide class of problems related to logarithms and logarithms. For the sake of understanding the material, I will give only a few common examples from school curriculum and universities.

Examples for logarithms

Take the logarithm of expressions

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

By properties 3,5 we calculate

2.
By the difference property of logarithms, we have

3.
Using properties 3.5 we find

4. where .

A seemingly complex expression using a series of rules is simplified to the form

Finding Logarithm Values

Example 2 Find x if

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

Substitute in the record and mourn

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Calculate log(x) if

Solution: Take the logarithm of the variable to write the logarithm through the sum of the terms

This is just the beginning of acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the acquired knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another no less important topic- logarithmic inequalities ...

Basic properties of logarithms

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

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

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

These formulas will help calculate the 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:

A task. Find the value of the expression: log6 4 + log6 9.

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

A 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.

A 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 considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? 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 their 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.

A task. Find the value of the expression: log7 496.

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

A task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:

I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Transition to a new foundation

Formulas for transition to a new base come to the rescue. We 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 put 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 tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

A task. Find the value of the expression: log5 16 log2 25.

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

Now let's flip the second logarithm:

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.

A 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 it 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 process of solving it is required to represent a number as a logarithm to a given base. In this case, the 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's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

A task. Find the value of the expression:

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

If someone is not in the know, this was a real task from the Unified State Examination 🙂

Logarithmic unit and logarithmic zero

logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is 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.

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

Let's explain it easier. For example, \(\log_(2)(8)\) is equal to the power \(2\) must be raised to 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 the logarithm

Any logarithm has the following "anatomy":

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

How to calculate the logarithm?

To calculate the logarithm, you need to answer the question: to what degree 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. That's why:

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

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

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

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

d) To what power must \(\sqrt(7)\) be raised to get \(\sqrt(7)\)? In the first - any number in the first degree is equal to itself.

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

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

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

Example : Calculate the 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 the logarithm: \(\log_(a)(c)=b\) \(\Leftrightarrow\) \(a^(b)=c\)
\((4\sqrt(2))^(x)=8\)		What links \(4\sqrt(2)\) 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 degree properties: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\)
\(2^(\frac(5)(2)x)=2^(3)\)		The bases are equal, we proceed to the 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 equality work. Of course, \(x=2\).

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

The most ingenious will say: "X is a little less than two." How exactly is this number to be written? To answer this question, they came up with the logarithm. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).

I want to emphasize that \(\log_(3)(8)\), as well as any logarithm is just a number. Yes, it looks unusual, but it is short. Because if we wanted to write it in the form decimal fraction, then 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 reduced to the same base. So here you can not do without the logarithm. Let's use the definition of the logarithm: \(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
\(\log_(4)(10)=5x-4\)		Flip the equation so x is on the left
\(5x-4=\log_(4)(10)\)		Before us. Move \(4\) to the right. And don't be afraid of the logarithm, treat it like a regular number.
\(5x=\log_(4)(10)+4\)		Divide the equation by 5
\(x=\)\(\frac(\log_(4)(10)+4)(5)\)		Here is our root. Yes, it looks unusual, but the answer is not chosen.

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

Decimal and natural logarithms

As stated in the definition of the logarithm, its base can be any positive number, except for the unit \((a>0, a\neq1)\). 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 the Euler number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).

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

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

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

Basic logarithmic identity

Logarithms have many properties. One of them is called "Basic logarithmic identity" and looks like this:

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

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

Recall the short definition of the 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 the rest of the 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 you can write \(\log_(2)(4)\) instead of two.

But \(\log_(3)(9)\) is also equal to \(2\), so you can also write \(2=\log_(3)(9)\) . Similarly 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 the two as a logarithm with any base anywhere (even in an equation, even in an expression, even in an inequality) - we just write the squared base as an argument.

It's the same with a 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 value of an expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)

Solution :

Answer : \(1\)

The main properties of the logarithm, the graph of the logarithm, the domain of definition, the set of values, the basic formulas, the increase and decrease are given. Finding the derivative of the logarithm is considered. And also the integral, expansion in power series and representation by means of complex numbers.

Definition of logarithm

Logarithm with base a is the y function (x) = log x, inverse to the exponential function with base a: x (y) = a y.

Decimal logarithm is the logarithm to the base of the number 10 : log x ≡ log 10 x.

natural logarithm is the logarithm to the base of e: ln x ≡ log e x.

2,718281828459045... ;
.

The graph of the logarithm is obtained from the graph of the exponential function by mirror reflection about the straight line y \u003d x. On the left are graphs of the function y (x) = log x for four values bases of the logarithm:a= 2 , a = 8 , a = 1/2 and a = 1/8 . The graph shows that for a > 1 the logarithm is monotonically increasing. As x increases, the growth slows down significantly. At 0 < a < 1 the logarithm is monotonically decreasing.

Properties of the logarithm

Domain, set of values, ascending, descending

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


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

Private values

The base 10 logarithm is called decimal logarithm and is marked like this:

base logarithm e called natural logarithm:

Basic logarithm formulas

Properties of the logarithm following 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 the logarithm. When taking a logarithm, the products of factors are converted to sums of terms.

Potentiation is the mathematical operation inverse to logarithm. When potentiating, the given base is raised to the power of the expression on which the potentiation is performed. In this case, the sums of terms are converted into products of factors.

Proof of the 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
.
Apply the property of the exponential function
:
.

Let us prove the base change formula.
;
.
Setting c = b , we have:

Inverse function

The reciprocal of the base a logarithm is the exponential function with exponent a.

If , then

Derivative of the logarithm

Derivative of logarithm modulo x :
.
Derivative of the nth order:
.
Derivation of 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 in terms of 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 clearly defined. If we 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

For , the expansion takes place:

References:
I.N. Bronstein, K.A. Semendyaev, Handbook of Mathematics for Engineers and Students of Higher Educational Institutions, Lan, 2009.

With the development of society, the complexity of production, mathematics also developed. Movement from simple to complex. From the usual accounting method of addition and subtraction, with their repeated repetition, they came to the concept of multiplication and division. The reduction of the multiply repeated operation became the concept of exponentiation. The first tables of the dependence of numbers on the base and the number of exponentiation were compiled back in the 8th century by the Indian mathematician Varasena. From them, you can count the time of occurrence of logarithms.

Historical outline

The revival of Europe in the 16th century also stimulated the development of mechanics. T required a large amount of computation associated with multiplication and division of multi-digit numbers. The ancient tables did a great service. They made it possible to replace complex operations with simpler ones - addition and subtraction. A big step forward was the work of the mathematician Michael Stiefel, published in 1544, in which he realized the idea of many mathematicians. This made it possible to use tables not only for degrees in the form of prime numbers, but also for arbitrary rational ones.

In 1614, the Scotsman John Napier, developing these ideas, first introduced the new term "logarithm of a number." New complex tables were compiled for calculating the logarithms of sines and cosines, as well as tangents. This greatly reduced the work of astronomers.

New tables began to appear, which were successfully used by scientists for three centuries. A lot of time passed before the new operation in algebra acquired its finished form. The logarithm was defined and its properties were studied.

Only in the 20th century, with the advent of the calculator and the computer, mankind abandoned the ancient tables that had been successfully operating throughout the 13th centuries.

Today we call the logarithm of b to base a the number x, which is the power of a, to get the number b. This is written as a formula: x = log a(b).

For example, log 3(9) will be equal to 2. This is obvious if you follow the definition. If we raise 3 to the power of 2, we get 9.

Thus, the formulated definition puts only one restriction, the numbers a and b must be real.

Varieties of logarithms

The classical definition is called the real logarithm and is actually a solution to the equation a x = b. The option a = 1 is borderline and is of no interest. Note: 1 to any power is 1.

Real value of the logarithm defined only if the base and the argument is greater than 0, and the base must not be equal to 1.

Special place in the field of mathematics play logarithms, which will be named depending on the value of their base:

Rules and restrictions

The fundamental property of logarithms is the rule: the logarithm of a product is equal to the logarithmic sum. log abp = log a(b) + log a(p).

As a variant of this statement, it will be: log c (b / p) \u003d log c (b) - log c (p), the quotient function is equal to the difference of the functions.

It is easy to see from the previous two rules that: log a(b p) = p * log a(b).

Other properties include:

Comment. Do not make a common mistake - the logarithm of the sum is not equal to the sum of the logarithms.

For many centuries, the operation of finding the logarithm was a rather time-consuming task. Mathematicians used famous formula logarithmic theory of polynomial expansion:

ln (1 + x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ... + ((-1)^(n + 1))*(( x^n)/n), where n is natural number greater than 1, which determines the accuracy of the calculation.

Logarithms with other bases were calculated using the theorem on the transition from one base to another and the property of the logarithm of the product.

Since this method is very laborious and when solving practical problems difficult to implement, they used pre-compiled tables of logarithms, which greatly accelerated the entire work.

In some cases, specially compiled graphs of logarithms were used, which gave less accuracy, but significantly speeded up the search. desired value. The curve of the function y = log a(x), built on several points, allows using the usual ruler to find the values of the function at any other point. For a long time, engineers used the so-called graph paper for these purposes.

In the 17th century, the first auxiliary analog computing conditions appeared, which to XIX century acquired a finished look. The most successful device was named logarithmic ruler. Despite the simplicity of the device, its appearance significantly accelerated the process of all engineering calculations, and this is difficult to overestimate. Currently, few people are familiar with this device.

The advent of calculators and computers made it pointless to use any other devices.

Equations and inequalities

For solutions various equations and inequalities using logarithms, the following formulas apply:

Transition from one base to another: log a(b) = log c(b) / log c(a);
Consequently previous version: log a(b) = 1 / log b(a).

To solve inequalities, it is useful to know:

The value of the logarithm will only be positive if both the base and the argument are both greater than or less than one; if at least one condition is violated, the value of the logarithm will be negative.
If the logarithm function is applied to the right and left sides of the inequality, and the base of the logarithm is greater than one, then the sign of the inequality is preserved; otherwise, it changes.

Task examples

Consider several options for using logarithms and their properties. Examples with solving equations:

Consider the option of placing the logarithm in the degree:

Task 3. Calculate 25^log 5(3). Solution: in the conditions of the problem, the notation is similar to the following (5^2)^log5(3) or 5^(2 * log 5(3)). Let's write it differently: 5^log 5(3*2), or the square of a number as a function argument can be written as the square of the function itself (5^log 5(3))^2. Using the properties of logarithms, this expression is 3^2. Answer: as a result of the calculation we get 9.

Practical use

Being a purely mathematical tool, it seems far from real life that the logarithm suddenly took on great importance in describing objects real world. It is difficult to find a science where it is not used. This fully applies not only to the natural, but also to the humanities fields of knowledge.

Logarithmic dependencies

Here are some examples of numerical dependencies:

Mechanics and physics

Historically, mechanics and physics have always developed using mathematical methods research and at the same time served as a stimulus for the development of mathematics, including logarithms. The theory of most laws of physics is written in the language of mathematics. We give only two examples of describing physical laws using the logarithm.

It is possible to solve the problem of calculating such a complex quantity as the speed of a rocket using the Tsiolkovsky formula, which laid the foundation for the theory of space exploration:

V = I * ln(M1/M2), where

V is the final speed of the aircraft.
I is the specific impulse of the engine.
M 1 is the initial mass of the rocket.
M 2 - final mass.

Another important example- this is the use in the formula of another great scientist, Max Planck, which serves to evaluate the equilibrium state in thermodynamics.

S = k * ln (Ω), where

S is a thermodynamic property.
k is the Boltzmann constant.
Ω is the statistical weight of different states.

Chemistry

Less obvious would be the use of formulas in chemistry containing the ratio of logarithms. Here are just two examples:

The Nernst equation, the condition of the redox potential of the medium in relation to the activity of substances and the equilibrium constant.
The calculation of such constants as the autoprolysis index and the acidity of the solution is also not complete without our function.

Psychology and biology

And it’s completely incomprehensible what the psychology has to do with it. It turns out that the strength of sensation is well described by this function as the inverse ratio of the intensity value of the stimulus to the lower intensity value.

After the above examples, it is no longer surprising that the theme of logarithms is also widely used in biology. Whole volumes can be written about biological forms corresponding to logarithmic spirals.

Other areas

It seems that the existence of the world is impossible without connection with this function, and it governs all laws. Especially when the laws of nature are connected with a geometric progression. It is worth referring to the MatProfi website, and there are many such examples in the following areas of activity:

The list could be endless. Having mastered the basic laws of this function, you can plunge into the world of infinite wisdom.

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

Definition

natural logarithm is the function y = ln x, inverse to the exponent, x \u003d e y , and which 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 definitions, the base of the natural logarithm is the number e:
e ≅ 2.718281828459045...;
.

Graph of the function y = ln x.

Graph of the natural logarithm (functions y = ln x) is obtained from the graph of the exponent by mirror reflection about the straight line y = x .

The natural logarithm is defined for positive values of x . It monotonically increases on its domain of definition.

As 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 rather 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

log 1 = 0

Basic formulas for natural logarithms

Formulas arising 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 change formula:

The proofs of these formulas are presented in the "Logarithm" section.

Inverse function

The reciprocal 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 the modulo x:
.
Derivative of the nth order:
.
Derivation of formulas > > >

Integral

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

Expressions in terms of complex numbers

Consider a function of a 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 we put
, where n is an integer,
then it will be the same number for different n.

That's why natural logarithm, as a function of a complex variable, is not a single-valued function.

Power series expansion

For , the expansion takes place:

References:
I.N. Bronstein, K.A. Semendyaev, Handbook of Mathematics for Engineers and Students of Higher Educational Institutions, Lan, 2009.