Learning logarithms from scratch. What is the logarithm

So, before us are powers of two. If you take a number from the bottom line, you can easily find the degree to which you have to raise a deuce to get this number. For example, to get 16, you need to raise two to the fourth degree. And to get 64, you need to raise two to the sixth degree. This can be seen from the table.

And now - in fact, the definition of the logarithm:

The base-a logarithm of argument x is the degree to which the number a must be raised to obtain the number x.

Designation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is equal to.

For example, 2 3 \u003d 8 ⇒ log 2 8 \u003d 3 (the base 2 logarithm of 8 is three, since 2 3 \u003d 8). With the same success, log 2 64 \u003d 6, since 2 6 \u003d 64.

The operation of finding the logarithm of a number on a given basis is called the logarithm. So, we supplement our table with a new line:

2 1	2 2	2 3	2 4	2 5	2 6
2	4	8	16	32	64
log 2 2 \u003d 1	log 2 4 \u003d 2	log 2 8 \u003d 3	log 2 16 \u003d 4	log 2 32 \u003d 5	log 2 64 \u003d 6

Unfortunately, not all logarithms are considered so easy. For example, try to find log 2 5. The number 5 is not in the table, but the logic suggests that the logarithm will lie somewhere on the segment. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.

Such numbers are called irrational: decimal digits can be written indefinitely, and they never repeat. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.

It is important to understand that the logarithm is an expression with two variables (base and argument). Many at first confused where the foundation is and where the argument is. To avoid annoying misunderstandings, just look at the picture:

Before us is nothing more than the definition of a logarithm. Remember: the logarithm is a degree, in which you need to raise the foundation to get an argument. It is the base that is raised to the power - in the picture it is highlighted in red. It turns out that the base is always below! I tell this wonderful rule to my students in the first lesson - and there is no confusion.

We figured out the definition - it remains to learn how to count logarithms, i.e. get rid of the log sign. To begin with, we note that two important facts follow from the definition:

The argument and the base must always be greater than zero. This follows from the definition of degree by a rational indicator, to which the definition of a logarithm is reduced.
The base must be different from one, because the unit remains to any extent one. Because of this, the question "to what degree must a unit be raised in order to get a deuce" is meaningless. There is no such degree!

Such restrictions are called valid range (DLD). It turns out that the ODZ of the logarithm looks like this: log a x \u003d b ⇒ x\u003e 0, a\u003e 0, a ≠ 1.

Note that there are no restrictions on the number b (the logarithm value). For example, the logarithm may well be negative: log 2 0.5 \u003d −1, because 0.5 \u003d 2 −1.

However, now we consider only numerical expressions, where it is not required to know the logistic linear differential equation. All restrictions are already taken into account by the drafters of the tasks. But when the logarithmic equations and inequalities go, the requirements of ODZ will become mandatory. Indeed, the basis and argument can be quite non-weak constructions that do not necessarily correspond to the above restrictions.

Now consider the general scheme for calculating logarithms. It consists of three steps:

Represent the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it is better to get rid of decimal fractions;
Solve the equation for variable b: x \u003d a b;
The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be seen already in the first step. The requirement that the base be more than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similarly with decimal fractions: if you immediately translate them into regular fractions, there will be many times fewer errors.

Let's see how this scheme works with specific examples:

A task. Calculate the logarithm: log 5 25

We represent the basis and argument as the degree of the five: 5 \u003d 5 1; 25 \u003d 5 2;
We compose and solve the equation:
log 5 25 \u003d b ⇒ (5 1) b \u003d 5 2 ⇒ 5 b \u003d 5 2 ⇒ b \u003d 2;
Received the answer: 2.

A task. Calculate the logarithm:

A task. Calculate the logarithm: log 4 64

We represent the basis and argument as a power of two: 4 \u003d 2 2; 64 \u003d 2 6;
We compose and solve the equation:
log 4 64 \u003d b ⇒ (2 2) b \u003d 2 6 ⇒ 2 2b \u003d 2 6 ⇒ 2b \u003d 6 ⇒ b \u003d 3;
Received the answer: 3.

A task. Calculate the logarithm: log 16 1

We represent the basis and argument as a power of two: 16 \u003d 2 4; 1 \u003d 2 0;
We compose and solve the equation:
log 16 1 \u003d b ⇒ (2 4) b \u003d 2 0 ⇒ 2 4b \u003d 2 0 ⇒ 4b \u003d 0 ⇒ b \u003d 0;
Received the answer: 0.

A task. Calculate the logarithm: log 7 14

We represent the basis and argument as the degree of seven: 7 \u003d 7 1; 14 does not appear as a power of seven, since 7 1< 14 < 7 2 ;
From the previous paragraph it follows that the logarithm is not considered;
The answer is unchanged: log 7 14.

A short note to the last example. How to make sure that a number is not an exact degree of another number? Very simple - just factor it into simple factors. And if such factors cannot be collected in degrees with the same exponents, then the initial number is not an exact degree.

A task. Find out if the exact powers of a number are: 8; 48; 81; 35; fourteen.

8 \u003d 2 · 2 · 2 \u003d 2 3 - the exact degree, because there is only one factor;
48 \u003d 6 · 8 \u003d 3 · 2 · 2 · 2 · 2 \u003d 3 · 2 4 - is not an exact degree, since there are two factors: 3 and 2;
81 \u003d 9 · 9 \u003d 3 · 3 · 3 · 3 \u003d 3 4 - the exact degree;
35 \u003d 7 · 5 - again is not an exact degree;
14 \u003d 7 · 2 - again not an exact degree;

We also note that the primes themselves are always exact degrees of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and designation.

The decimal logarithm of argument x is the base 10 logarithm, i.e. the power to raise the number 10 to get the number x. Designation: log x.

For example, log 10 \u003d 1; lg 100 \u003d 2; lg 1000 \u003d 3 - etc.

From now on, when a phrase like “Find lg 0.01” is found in a textbook, be aware that this is not a typo. This is the decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x \u003d log 10 x

Everything that is true for ordinary logarithms is also true for decimals.

Natural logarithm

There is another logarithm that has its own notation. In a sense, it is even more important than decimal. This is a natural logarithm.

The natural logarithm of the argument x is the base logarithm of e, i.e. the degree to which the number e must be raised to obtain the number x. Designation: ln x.

Many will ask: what else is the number e? This is an irrational number, its exact meaning cannot be found and written down. I will give only the first figures of it:
e \u003d 2.718281828459 ...

We will not go deep into what this number is and why it is necessary. Just remember that e is the base of the natural logarithm:
ln x \u003d log e x

Thus, ln e \u003d 1; ln e 2 \u003d 2; ln e 16 \u003d 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number is irrational. Except, of course, units: ln 1 \u003d 0.

For natural logarithms, all the rules that are true for ordinary logarithms are true.

The logarithm of the number b (b\u003e 0) at the base a (a\u003e 0, a ≠ 1) Is the exponent to which the number a must be raised to obtain b.

The base 10 logarithm of b can be written as lg (b), and the base logarithm of e (the natural logarithm) is ln (b).

Often used in solving problems with logarithms:

Logarithm Properties

There are four main logarithm properties.

Let a\u003e 0, a ≠ 1, x\u003e 0 and y\u003e 0.

Property 1. Logarithm of the product

Logarithm of the product equal to the sum of the logarithms:

log a (x ⋅ y) \u003d log a x + log a y

Property 2. Logarithm of the quotient

Logarithm of private equal to the difference of the logarithms:

log a (x / y) \u003d log a x - log a y

Property 3. Logarithm of degree

Logarithm of degree equal to the product of the degree by the logarithm:

If the base of the logarithm is in degree, then another formula applies:

Property 4. Logarithm of the root

This property can be obtained from the property of the logarithm of the degree, since the root of the nth degree is equal to the degree 1 / n:

The formula for the transition from the logarithm in one base to the logarithm in another

This formula is also often used when solving various tasks on logarithms:

Special case:

Comparison of logarithms (inequalities)

Suppose we have 2 functions f (x) and g (x) under logarithms with the same bases and between them there is an inequality sign:

To compare them, you first need to look at the base of the logarithms of a:

If a\u003e 0, then f (x)\u003e g (x)\u003e 0
If 0< a < 1, то 0 < f(x) < g(x)

How to solve problems with logarithms: examples

Logarithm Jobs included in the exam in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the relevant sections. Also, tasks with logarithms are found in the bank of tasks in mathematics. You can find all examples through site search.

What is the logarithm

Logarithms have always been considered a complex topic in a school math course. There are many different definitions of the logarithm, but most textbooks for some reason use the most complex and unsuccessful ones.

We will determine the logarithm simply and clearly. To do this, compile a table:

So, before us are powers of two.

Logarithms - properties, formulas, how to solve

If you take a number from the bottom line, you can easily find the degree to which you have to raise a deuce to get this number. For example, to get 16, you need to raise two to the fourth degree. And to get 64, you need to raise two to the sixth degree. This can be seen from the table.

And now - in fact, the definition of the logarithm:

on the basis of a from the argument x is the degree to which the number a must be raised to obtain the number x.

Designation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is equal to.

For example, 2 3 \u003d 8 ⇒log 2 8 \u003d 3 (the base 2 logarithm of 8 is three, since 2 3 \u003d 8). With the same success, log 2 64 \u003d 6, since 2 6 \u003d 64.

The operation of finding the logarithm of a number for a given base is called. So, we supplement our table with a new line:

2 1	2 2	2 3	2 4	2 5	2 6
2	4	8	16	32	64
log 2 2 \u003d 1	log 2 4 \u003d 2	log 2 8 \u003d 3	log 2 16 \u003d 4	log 2 32 \u003d 5	log 2 64 \u003d 6

How to count logarithms

We figured out the definition - it remains to learn how to count logarithms, i.e. get rid of the log sign. To begin with, we note that two important facts follow from the definition:

The argument and the base must always be greater than zero. This follows from the definition of degree by a rational indicator, to which the definition of a logarithm is reduced.
The base must be different from one, because the unit remains to any extent one. Because of this, the question "to what degree must a unit be raised in order to get a deuce" is meaningless. There is no such degree!

Such restrictions are called valid range (DLD). It turns out that the ODZ of the logarithm looks like this: log a x \u003d b ⇒x\u003e 0, a\u003e 0, a ≠ 1.

Note that there are no restrictions on the number b (the logarithm value). For example, the logarithm may well be negative: log 2 0.5 \u003d −1, because 0.5 \u003d 2 −1.

Now consider the general scheme for calculating logarithms. It consists of three steps:

Represent the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it is better to get rid of decimal fractions;
Solve the equation for variable b: x \u003d a b;
The resulting number b will be the answer.

Let's see how this scheme works with specific examples:

A task. Calculate the logarithm: log 5 25

We represent the basis and argument as the degree of the five: 5 \u003d 5 1; 25 \u003d 5 2;

We compose and solve the equation:
log 5 25 \u003d b ⇒ (5 1) b \u003d 5 2 ⇒5 b \u003d 5 2 ⇒ b \u003d 2;

Received the answer: 2.

A task. Calculate the logarithm:

A task. Calculate the logarithm: log 4 64

We represent the basis and argument as a power of two: 4 \u003d 2 2; 64 \u003d 2 6;
We compose and solve the equation:
log 4 64 \u003d b ⇒ (2 2) b \u003d 2 6 ⇒2 2b \u003d 2 6 ⇒2b \u003d 6 ⇒ b \u003d 3;
Received the answer: 3.

A task. Calculate the logarithm: log 16 1

We represent the basis and argument as a power of two: 16 \u003d 2 4; 1 \u003d 2 0;
We compose and solve the equation:
log 16 1 \u003d b ⇒ (2 4) b \u003d 2 0 ⇒2 4b \u003d 2 0 ⇒4b \u003d 0 ⇒ b \u003d 0;
Received the answer: 0.

A task. Calculate the logarithm: log 7 14

We represent the basis and argument as the degree of seven: 7 \u003d 7 1; 14 does not appear as a power of seven, since 7 1< 14 < 7 2 ;
From the previous paragraph it follows that the logarithm is not considered;
The answer is unchanged: log 7 14.

A short note to the last example. How to make sure that a number is not an exact degree of another number? Very simple - just factor it into simple factors. If there are at least two different factors in the expansion, the number is not an exact power.

A task. Find out if the exact powers of a number are: 8; 48; 81; 35; fourteen.

We also note that the primes themselves are always exact degrees of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and designation.

from argument x is the base 10 logarithm, i.e. the power to raise the number 10 to get the number x. Designation: log x.

For example, log 10 \u003d 1; lg 100 \u003d 2; lg 1000 \u003d 3 - etc.

Everything that is true for ordinary logarithms is also true for decimals.

Natural logarithm

There is another logarithm that has its own notation. In a sense, it is even more important than decimal. This is a natural logarithm.

from argument x is the base logarithm of e, i.e. the degree to which the number e must be raised to obtain the number x. Designation: ln x.

Many will ask: what is the number e? This is an irrational number, its exact meaning cannot be found and written down. I will give only the first figures of it:
e \u003d 2.718281828459 ...

We will not go deep into what this number is and why it is necessary. Just remember that e is the base of the natural logarithm:
ln x \u003d log e x

For natural logarithms, all the rules that are true for ordinary logarithms are true.

Logarithm. Properties of the logarithm (degree of the logarithm).

How to represent a number as a logarithm?

We use the definition of the logarithm.

The logarithm is an indicator of the degree to which the base must be raised in order to get the number under the sign of the logarithm.

Thus, in order to represent a certain number c as a logarithm on the basis of a, one must put a power under the sign of the logarithm with the same base as the base of the logarithm, and write down the number c in the exponent:

In the form of a logarithm, you can imagine absolutely any number - positive, negative, integer, fractional, rational, irrational:

In order not to confuse a and c under stressful conditions of the control or exam, you can use this rule to remember:

what is below goes down, what is above goes up.

For example, you need to represent the number 2 as a base 3 logarithm.

We have two numbers - 2 and 3. These numbers are the base and the exponent, which we write under the sign of the logarithm. It remains to determine which of these numbers needs to be written down to the base of the degree, and which up to the indicator.

Base 3 in the logarithm entry is at the bottom, which means that when we represent the two in the form of a logarithm on base 3, 3, we also write down to the base.

2 stands above the triple. And in the degree record, we write the deuce above the triple, that is, in the exponent:

Logarithms First level.

Logarithms

Logarithm positive number b on the basis of awhere a\u003e 0, a ≠ 1is called the exponent to which the number should be raised a, To obtain b.

Logarithm Definition can be summarized as follows:

This equality holds for b\u003e 0, a\u003e 0, a ≠ 1. It is usually called logarithmic identity.
The action of finding the logarithm of a number is called logarithm.

Logarithm Properties:

Logarithm of the product:

Logarithm of quotient from division:

Replacing the base of the logarithm:

Logarithm of degree:

Root Logarithm:

Power Logarithm:

Decimal and natural logarithms.

Decimal logarithm numbers call the base 10 logarithm of this number and write & nbsp lg b
Natural logarithm numbers are called the logarithm of this number at the base ewhere e - an irrational number approximately equal to 2.7. At the same time they write ln b.

Other notes on algebra and geometry

Basic properties of logarithms

Logarithms, like any numbers, can be added, subtracted and converted in every way. But since logarithms are not quite ordinary numbers, there are rules that are called basic properties.

You must know these rules - 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.

Logarithm Addition and Subtraction

Consider two logarithms with the same base: log a x and log a y. Then they can be added and subtracted, moreover:

log a x + log a y \u003d log a (x · y);
log a x - log a y \u003d log a (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 equal grounds. If the grounds are different, these rules do not work!

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

Log 6 4 + log 6 9.

Since the bases of the logarithms are the same, we use the sum formula:
log 6 4 + log 6 9 \u003d log 6 (4 · 9) \u003d log 6 36 \u003d 2.

A 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 \u003d log 2 (48: 3) \u003d log 2 16 \u003d 4.

A task. Find the meaning of the expression: log 3 135 - log 3 5.

Again, the bases are the same, so we have:
log 3 135 - log 3 5 \u003d log 3 (135: 5) \u003d log 3 27 \u003d 3.

As you can see, the original expressions are made up of “bad” logarithms that are not counted separately. But after the transformations, quite normal numbers are obtained. On this fact many tests are built. Yes, control - such expressions in all seriousness (sometimes - almost unchanged) are offered at the exam.

Removing exponent from the logarithm

Now let's complicate the task a bit. What if there is a degree in the base or argument of the logarithm? Then an indicator of this degree can be taken out 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 all the same - in some cases this will significantly reduce the amount of computation.

Of course, all these rules make sense when observing the ODZ logarithm: a\u003e 0, a ≠ 1, x\u003e 0. And also: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers in front of the logarithm into the logarithm itself.

How to solve logarithms

This is what is most often required.

A task. Find the value of the expression: log 7 49 6.

Let's get rid of the degree in the argument by the first formula:
log 7 49 6 \u003d 6 log 7 49 \u003d 6 2 \u003d 12

A task. Find the value of the expression:

Note that the denominator is the logarithm, the base and argument of which are exact degrees: 16 \u003d 2 4; 49 \u003d 7 2. We have:

I think the last example needs clarification. Where did the logarithms disappear? Until the very last moment, we work only with the denominator. They presented the basis and argument of the logarithm there in the form of degrees and carried out indicators - they received a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have 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 was done. The result was the answer: 2.

Transition to a new foundation

Speaking about the rules of addition and subtraction of logarithms, I specifically emphasized that they work only on the same grounds. But what if the grounds are different? What if they are not exact powers of the same number?

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

Let the logarithm of log a x be given. Then for any number c such that c\u003e 0 and c ≠ 1, the equality

In particular, if we put c \u003d x, we get:

From the second formula it follows that you can swap the base and the argument of the logarithm, but at the same time the whole expression is “flipped”, i.e. the logarithm is in the denominator.

These formulas are rarely found in ordinary numerical terms. 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 the transition to a new foundation. Consider a couple of these:

A task. Find the value of the expression: log 5 16 · log 2 25.

Note that the arguments of both logarithms contain exact degrees. We take out the indicators: log 5 16 \u003d log 5 2 4 \u003d 4log 5 2; log 2 25 \u003d log 2 5 2 \u003d 2log 2 5;

And now, “flip” the second logarithm:

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

A task. Find the value of the expression: log 9 100 · log 3.

The base and argument of the first logarithm are exact degrees. We write this and get rid of the indicators:

Now get rid of the decimal logarithm, moving to a new base:

Basic logarithmic identity

Often in the process of solving it is required to represent the number as a logarithm for a given basis.

In this case, the formulas will help us:

In the first case, the number n becomes an indicator of the degree in the argument. The number n can be absolutely anything, because it is just the value of the logarithm.

The second formula is actually a rephrased definition. It is called:.

In fact, what happens if the number b is raised to such an extent that the number b in this degree gives the number a? That's right: this is the very number a. Carefully read this paragraph again - many on it "hang."

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

A task. Find the value of the expression:

Note that log 25 64 \u003d log 5 8 - just took out the square from the base and the argument of the logarithm. Given the rules of multiplication of degrees with the same base, we get:

If someone is not in the know, this was a real challenge from the exam 🙂

Logarithmic unit and logarithmic zero

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

log a a \u003d 1 is this. Remember once and for all: the logarithm for any base a from this base itself is equal to one.
log a 1 \u003d 0 is this. The base a can be anything, but if there is one in the argument, the logarithm is zero! Because a 0 \u003d 1 is a direct consequence of the definition.

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

Valid Values \u200b\u200bArea (SDL) of the logarithm

Now let’s talk about restrictions (ODZ - the area of \u200b\u200bpermissible values \u200b\u200bof variables).

We remember that, for example, the square root cannot be extracted from negative numbers; or if we have a fraction, then the denominator cannot be equal to zero. Logarithms have similar limitations:

That is, both the argument and the base must be greater than zero, and the base also cannot be equal.

Why is that?

Let's start with a simple one: let's say that. Then, for example, the number does not exist, since no matter what degree we raise, it always turns out. Moreover, it does not exist for any. But at the same time it can equal anything (for the same reason - it is equal to any degree). Therefore, the object is of no interest, and it was simply thrown out of mathematics.

We have a similar problem in the case: this is in any positive degree, but it cannot be raised to negative at all, since we get a division by zero (let me remind you).

When we encounter the problem of raising to a fractional degree (which is represented as a root:. For example, (that is), but does not exist.

Therefore, negative grounds are easier to throw away than to mess with them.

Well, since the basis of a is only positive for us, no matter what degree we raise it, we always get a strictly positive number. So the argument must be positive. For example, it does not exist, since it will by no means be a negative number (and even zero, therefore it also does not exist).

In problems with logarithms, the first thing you need to do is write the ODZ. I will give an example:

Solve the equation.

Recall the definition: the logarithm is the degree to which the base must be raised in order to get an argument. And by condition, this degree is equal to:.

We get the usual quadratic equation:. We solve it using the Vieta theorem: the sum of the roots is equal, and the product. Easy to pick, these are numbers and.

But if you immediately take and write down both of these numbers in the answer, you can get 0 points for the task. Why? Let’s think what will happen if we substitute these roots in the initial equation?

This is clearly wrong, since the basis cannot be negative, that is, the root is “external”.

In order to avoid such unpleasant tricks, it is necessary to write the ODZ before solving the equation:

Then, having received the roots and, immediately discard the root, and write the correct answer.

Example 1 (try to solve it yourself) :

Find the root of the equation. If there are several roots, indicate the smaller one in the answer.

Decision:

First of all, we write ODZ:

Now we recall what the logarithm is: to what extent do you need to raise the base to get an argument? In the second. I.e:

It would seem that the smaller root is equal. But this is not so: according to the ODZ, the root is third-party, that is, it is not the root of this equation at all. Thus, the equation has only one root:.

Answer: .

Basic logarithmic identity

Recall the definition of the logarithm in a general form:

We substitute in the second equality instead of the logarithm:

This equality is called basic logarithmic identity. Although in essence this equality is simply differently written logarithm definition:

This is the degree to which you must raise to get.

For example:

Solve the following examples:

Example 2

Find the meaning of the expression.

Decision:

Recall the rule from the section: that is, when raising a power to a power, the indicators multiply. Apply it:

Example 3

Prove that.

Decision:

Logarithm Properties

Unfortunately, the tasks are not always so simple - often first you need to simplify the expression, bring it to its familiar form, and only then it will be possible to calculate the value. This is easiest to do, knowing logarithm properties. So let's learn the basic properties of logarithms. I will prove each of them, because it’s easier to remember any rule if you know where it comes from.

All these properties must be remembered; without them, most problems with logarithms cannot be solved.

And now about all the properties of the logarithms in more detail.

Property 1:

Evidence:

Let then.

We have:

Property 2: Sum of Logarithms

The sum of the logarithms with the same bases is equal to the logarithm of the product: .

Evidence:

Let then. Let then.

Example:Find the meaning of the expression:.

Decision: .

The formula just learned helps to simplify the sum of the logarithms, not the difference, so these logarithms cannot be combined right away. But you can do the opposite - “break” the first logarithm into two: But the promised simplification:
.
Why is this needed? Well, for example: what is equal?

Now it is obvious that.

Now simplify yourself:

Tasks:

Answers:

Property 3: Logarithm difference:

Evidence:

Everything is exactly the same as in paragraph 2:

Let then.

Let then. We have:

An example from the last paragraph is now even easier:

An example is more complicated:. Guess yourself how to solve it?

It should be noted here that we do not have a single formula about the logarithms squared. This is something akin to an expression - this is not immediately simplified.

Therefore, we digress from the formulas about the logarithms, and think, what kind of formulas do we use in mathematics most often? Starting from grade 7!

It - . You need to get used to the fact that they are everywhere! And in indicative, and in trigonometric, and in irrational problems, they are found. Therefore, they must be remembered.

If you look closely at the first two terms, it becomes clear that this square difference:

Answer to check:

Simplify yourself.

Examples

Answers

Property 4: Removing the exponent from the logarithm argument:

Evidence:And here we also use the definition of the logarithm: let, then. We have:

You can understand this rule like this:

That is, the degree of the argument is moved ahead of the logarithm as a coefficient.

Example:Find the meaning of the expression.

Decision: .

Decide for yourself:

Examples:

Answers:

Property 5: Removing the exponent from the base of the logarithm:

Evidence:Let then.

We have:
Remember: from grounds degree is taken out as the opposite number, unlike the previous case!

Property 6: Extraction of the exponent from the base and argument of the logarithm:

Or if the degrees are the same:.

Property 7: Transition to a new foundation:

Evidence:Let then.

We have:

Property 8: Replacing the base and argument of the logarithm:

Evidence:This is a special case of formula 7: if we substitute, we get:

Let's look at a few more examples.

Example 4

Find the meaning of the expression.

We use the property of logarithms No. 2 - the sum of the logarithms with the same base is equal to the logarithm of the product:

Example 5

Find the meaning of the expression.

Decision:

We use the property of logarithms No. 3 and No. 4:

Example 6

Find the meaning of the expression.

Decision:

We use property No. 7 - we will pass to the base 2:

Example 7

Find the meaning of the expression.

Decision:

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On the exam and the exam and in general in life

With the development of society, the complexity of production, mathematics also developed. The movement from simple to complex. From ordinary accounting by the method of addition and subtraction, with their repeated repetition, we came to the concept of multiplication and division. The reduction of the multiply repeated operation of multiplication has become the concept of exponentiation. The first tables of the dependence of numbers on the base and the number of exponentiation were compiled in the VIII century by the Indian mathematician Varasena. From them it is possible to count the time of occurrence of logarithms.

Historical essay

The revival of Europe in the 16th century also stimulated the development of mechanics. T demanded a large amount of computationrelated to the multiplication and division of multi-valued numbers. Ancient tables provided a great service. They allowed replacing complex operations with simpler ones - addition and subtraction. A great step forward was the work of mathematician Michael Stiefel, published in 1544, in which he realized the idea of \u200b\u200bmany mathematicians. This made it possible to use tables not only for degrees in the form of primes, but also for arbitrary rational ones.

In 1614, the Scotsman John Napier, developing these ideas, first introduced the new term "the logarithm of 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 that were successfully used by scientists for three centuries. It took a long time before a new operation in algebra acquired its finished form. A definition of the logarithm was given, and its properties were studied.

Only in the XX century, with the advent of a calculator and a computer, did mankind abandon the ancient tables that worked successfully throughout the XIII centuries.

Today, we call the logarithm of b at the base of a the number x, which is the power of a, to get the number b. In the form of a formula, this is written: x \u003d log a (b).

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

So, 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 \u003d b. Option a \u003d 1 is borderline and is not of interest. Note: 1 to any degree is 1.

The real meaning of the logarithm defined only with a base and an argument greater than 0, while the base should not be equal to 1.

Special place in mathematics play logarithms, which will be called depending on the size of their base:

Rules and restrictions

The fundamental property of logarithms is the rule: the logarithm of the product is equal to the logarithmic sum. log abp \u003d log a (b) + log a (p).

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

From the previous two rules it is easily seen that: log a (b p) \u003d p * log a (b).

Among other properties, we can distinguish:

Comment. No need to make a common mistake - the logarithm of the sum is not equal to the sum of the logarithms.

For many centuries, the logarithm search operation has been a rather time-consuming task. Mathematicians used the well-known formula of the logarithmic theory of decomposition into a polynomial:

ln (1 + x) \u003d x - (x ^ 2) / 2 + (x ^ 3) / 3 - (x ^ 4) / 4 + ... + ((-1) ^ (n + 1)) * (( x ^ n) / n), where n is a positive integer 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 time consuming and in solving practical problems difficult to implement, then used pre-compiled tables of logarithms, which greatly accelerated the whole work.

In some cases, specially designed logarithm graphs were used, which gave less accuracy, but significantly accelerated the search for the desired value. The curve of the function y \u003d log a (x), constructed over several points, allows using the usual ruler to find the values \u200b\u200bof the function at any other point. For a long time, engineers used the so-called graph paper.

In the 17th century, the first auxiliary analog computing conditions appeared, which by the 19th century had acquired a complete form. The most successful device was called the slide rule. Despite the simplicity of the device, its appearance significantly accelerated the process of all engineering calculations, and it is difficult to overestimate. Currently, few people are familiar with this device.

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

Equations and Inequalities

To solve various equations and inequalities using logarithms, the following formulas are used:

The transition from one base to another: log a (b) \u003d log c (b) / log c (a);
As a consequence of the previous version: log a (b) \u003d 1 / log b (a).

To solve the inequalities it is useful to know:

The value of the logarithm will be positive only if the base and the argument are both greater 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 unity, then the inequality sign is preserved; otherwise it changes.

Examples of tasks

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

Consider the option of placing the logarithm in the degree:

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

Practical use

Being an exclusively mathematical tool, it seems far from real life that the logarithm unexpectedly acquired great importance for describing objects of the real world. It is difficult to find science where it is not used. This fully applies not only to natural, but also to humanitarian fields of knowledge.

Logarithmic dependencies

Here are a few examples of numerical dependencies:

Mechanics and physics

Historically, mechanics and physics have always developed using mathematical research methods and at the same time served as an incentive 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 the description of physical laws using the logarithm.

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

V \u003d I * ln (M1 / M2), where

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

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

S \u003d 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 will 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 with respect to the activity of substances and the equilibrium constant.
Calculation of such constants as an indicator of autoprolysis and acidity of a solution also cannot do without our function.

Psychology and biology

And it is completely incomprehensible what does psychology have to do with it. It turns out that the power of sensation is well described by this function as the inverse ratio of the stimulus intensity value 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 a connection with this function, and it rules all laws. Especially when the laws of nature are associated with geometric progression. It is worth referring to the MatProfi website, and there are many such examples in the following areas of activity:

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

The logarithm of a positive number b at the base a (a\u003e 0, a is not equal to 1) is a number c such that a c \u003d b: log a b \u003d c ⇔ a c \u003d b (a\u003e 0, a ≠ 1, b\u003e 0) & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp

Please note: the logarithm of a non-positive number is not defined. In addition, the base of the logarithm must have a positive number not equal to 1. For example, if we square -2 in the square, we get the number 4, but this does not mean that the logarithm on the base -2 of 4 is 2.

Basic logarithmic identity

a log a b \u003d b (a\u003e 0, a ≠ 1) (2)

It is important that the definition areas of the right and left sides of this formula are different. The left-hand side is defined only for b\u003e 0, a\u003e 0 and a ≠ 1. The right-hand side is defined for any b, but does not depend on a at all. Thus, the application of the basic logarithmic "identity" in solving equations and inequalities can lead to a change in the DLD.

Two obvious consequences of the definition of the logarithm

log a a \u003d 1 (a\u003e 0, a ≠ 1) (3)
log a 1 \u003d 0 (a\u003e 0, a ≠ 1) (4)

Indeed, when raising the number a to the first power, we get the same number, and when raising to the zero power, we get one.

The logarithm of the product and the logarithm of the quotient

log a (b c) \u003d log a b + log a c (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0) (5)

Log a b c \u003d log a b - log a c (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0) (6)

I would like to warn schoolchildren from the thoughtless use of these formulas in solving logarithmic equations and inequalities. When they are used “from left to right”, the ODZ is narrowed, and when moving from the sum or difference of the logarithms to the logarithm of the product or quotient, the ODZ is expanded.

Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f (x) and g (x) are both less than zero.

Transforming this expression into the sum log a f (x) + log a g (x), we are forced to confine ourselves only to the case when f (x)\u003e 0 and g (x)\u003e 0. There is a narrowing of the range of permissible values, and this is categorically unacceptable, since it can lead to loss of decisions. A similar problem exists for formula (6).

The degree can be taken out of the logarithm sign

log a b p \u003d p log a b (a\u003e 0, a ≠ 1, b\u003e 0) (7)

And again I would like to call for accuracy. Consider the following example:

Log a (f (x) 2 \u003d 2 log a f (x)

The left side of the equality is obviously defined for all values \u200b\u200bof f (x) except zero. The right side is only for f (x)\u003e 0! Taking the degree out of the logarithm, we again narrow down the ODZ. The inverse procedure expands the range of acceptable values. All these remarks apply not only to degree 2, but also to any even degree.

Transition Formula

log a b \u003d log c b log c a (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0, c ≠ 1) (8)

That rare case when the DLD does not change during the conversion. If you reasonably chose a base with (positive and not equal to 1), the formula for transition to a new base is absolutely safe.

If we choose the number b as the new base c, we obtain an important special case of formula (8):

Log a b \u003d 1 log b a (a\u003e 0, a ≠ 1, b\u003e 0, b ≠ 1) (9)

Some simple logarithm examples

Example 1. Calculate: lg2 + lg50.
Decision. log2 + log50 \u003d log100 \u003d 2. We used the formula for the sum of the logarithms (5) and the definition of the decimal logarithm.

Example 2. Calculate: lg125 / lg5.
Decision. log125 / log5 \u003d log 5 125 \u003d 3. We used the formula for the transition to a new base (8).

Logarithm Formula Table

a log a b \u003d b (a\u003e 0, a ≠ 1)

log a a \u003d 1 (a\u003e 0, a ≠ 1)

log a 1 \u003d 0 (a\u003e 0, a ≠ 1)

log a (b c) \u003d log a b + log a c (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0)

log a b c \u003d log a b - log a c (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0)

log a b p \u003d p log a b (a\u003e 0, a ≠ 1, b\u003e 0)

log a b \u003d log c b log c a (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0, c ≠ 1)

log a b \u003d 1 log b a (a\u003e 0, a ≠ 1, b\u003e 0, b ≠ 1)