Logarithmic equation: basic formulas and techniques. Logarithmic equations

Solving logarithmic equations. Part 1.

Logarithmic equation is an equation in which the unknown is contained under the sign of the logarithm (in particular, in the base of the logarithm).

The simplest logarithmic equation has the form:

Solving any logarithmic equation involves a transition from logarithms to expressions under the sign of logarithms. However, this action expands the range of permissible values ​​of the equation and can lead to the appearance of extraneous roots. To avoid the appearance of foreign roots, you can do one of three ways:

1. Make an equivalent transition from the original equation to a system including

depending on which inequality or simpler.

If the equation contains an unknown in the base of the logarithm:

then we go to the system:

2. Separately find the range of acceptable values ​​of the equation, then solve the equation and check whether the solutions found satisfy the equation.

3. Solve the equation, and then check: substitute the found solutions into the original equation and check whether we get the correct equality.

A logarithmic equation of any level of complexity always ultimately reduces to the simplest logarithmic equation.

All logarithmic equations can be divided into four types:

1 . Equations that contain logarithms only to the first power. With the help of transformations and use, they are brought to the form

Example. Let's solve the equation:

Let's equate the expressions under the logarithm sign:

Let's check whether our root of the equation satisfies:

Yes, it satisfies.

Answer: x=5

2 . Equations that contain logarithms to powers other than 1 (particularly in the denominator of a fraction). Such equations can be solved using introducing a change of variable.

Example. Let's solve the equation:

Let's find the ODZ equation:

The equation contains logarithms squared, so it can be solved using a change of variable.

Important! Before introducing a replacement, you need to “pull apart” the logarithms that are part of the equation into “bricks”, using the properties of logarithms.

When “pulling apart” logarithms, it is important to use the properties of logarithms very carefully:

In addition, there is one more subtle point here, and in order to avoid a common mistake, we will use an intermediate equality: we will write the degree of the logarithm in this form:

Likewise,

Let's substitute the resulting expressions into the original equation. We get:

Now we see that the unknown is contained in the equation as part of . Let's introduce the replacement: . Since it can take any real value, we do not impose any restrictions on the variable.

Instructions

Write the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is the decimal logarithm. If the logarithm has the number e as its base, then write the expression: ln b – natural logarithm. It is understood that the result of any is the power to which the base number must be raised to obtain the number b.

When finding the sum of two functions, you simply need to differentiate them one by one and add the results: (u+v)" = u"+v";

When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function multiplied by the first function: (u*v)" = u"*v+v"*u;

In order to find the derivative of the quotient of two functions, it is necessary to subtract from the product of the derivative of the dividend multiplied by the divisor function the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;

If a complex function is given, then it is necessary to multiply the derivative of the internal function and the derivative of the external one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).

Using the results obtained above, you can differentiate almost any function. So let's look at a few examples:

y=x^4, y"=4*x^(4-1)=4*x^3;

y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also problems involving calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).

2) Calculate the value of the function at a given point y"(1)=8*e^0=8

Video on the topic

Helpful advice

Learn the table of elementary derivatives. This will significantly save time.

Sources:

• derivative of a constant

So, what is the difference between an irrational equation and a rational one? If the unknown variable is under the square root sign, then the equation is considered irrational.

Instructions

The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore this equation has no roots.

So, an irrational equation is solved using the method of squaring both its sides. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots into the original equation.

Consider another one.
2х+vх-3=0
Of course, this equation can be solved using the same equation as the previous one. Move Compounds equations, which do not have a square root, to the right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, an ordinary quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.

Solving identities is quite simple. To do this, it is necessary to carry out identical transformations until the set goal is achieved. Thus, with the help of simple arithmetic operations, the problem posed will be solved.

You will need

• - paper;
• - pen.

Instructions

The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many trigonometric formulas, which are essentially the same identities.

Indeed, the square of the sum of two terms is equal to the square of the first plus twice the product of the first by the second and plus the square of the second, that is, (a+b)^2= (a+b)(a+b)=a^2+ab +ba+b ^2=a^2+2ab+b^2.

Simplify both

General principles of the solution

Variable Replacement Method

Solving integrals of the second kind

Substitution of integration limits

Logarithmic equation is an equation in which the unknown (x) and expressions with it are under the sign of the logarithmic function. Solving logarithmic equations assumes that you are already familiar with and .
How to solve logarithmic equations?

The simplest equation is log a x = b, where a and b are some numbers, x is an unknown.
Solving a logarithmic equation is x = a b provided: a > 0, a 1.

It should be noted that if x is somewhere outside the logarithm, for example log 2 x = x-2, then such an equation is already called mixed and a special approach is needed to solve it.

The ideal case is when you come across an equation in which only numbers are under the logarithm sign, for example x+2 = log 2 2. Here it is enough to know the properties of logarithms to solve it. But such luck does not happen often, so get ready for more difficult things.

But first, let's start with simple equations. To solve them, it is advisable to have a very general understanding of the logarithm.

Solving simple logarithmic equations

These include equations of the type log 2 x = log 2 16. The naked eye can see that by omitting the sign of the logarithm we get x = 16.

To solve a more complex logarithmic equation, it is usually reduced to solving an ordinary algebraic equation or to solving a simple logarithmic equation log a x = b. In the simplest equations this happens in one movement, which is why they are called simplest.

The above method of dropping logarithms is one of the main ways to solve logarithmic equations and inequalities. In mathematics, this operation is called potentiation. There are certain rules or restrictions for this type of operation:

• logarithms have the same numerical bases
• The logarithms in both sides of the equation are free, i.e. without any coefficients or other various kinds of expressions.

Let's say in the equation log 2 x = 2log 2 (1 - x) potentiation is not applicable - the coefficient 2 on the right does not allow it. In the following example, log 2 x+log 2 (1 - x) = log 2 (1+x) also does not satisfy one of the restrictions - there are two logarithms on the left. If there was only one, it would be a completely different matter!

In general, you can remove logarithms only if the equation has the form:

log a (...) = log a (...)

Absolutely any expressions can be placed in brackets; this has absolutely no effect on the potentiation operation. And after eliminating logarithms, a simpler equation will remain - linear, quadratic, exponential, etc., which, I hope, you already know how to solve.

Let's take another example:

log 3 (2x-5) = log 3 x

We apply potentiation, we get:

log 3 (2x-1) = 2

Based on the definition of a logarithm, namely, that a logarithm is the number to which the base must be raised in order to obtain an expression that is under the logarithm sign, i.e. (4x-1), we get:

Again we received a beautiful answer. Here we did without eliminating logarithms, but potentiation is also applicable here, because a logarithm can be made from any number, and exactly the one we need. This method is very helpful in solving logarithmic equations and especially inequalities.

Let's solve our logarithmic equation log 3 (2x-1) = 2 using potentiation:

Let's imagine the number 2 as a logarithm, for example, this log 3 9, because 3 2 =9.

Then log 3 (2x-1) = log 3 9 and again we get the same equation 2x-1 = 9. I hope everything is clear.

So we looked at how to solve the simplest logarithmic equations, which are actually very important, because solving logarithmic equations, even the most terrible and twisted ones, in the end always comes down to solving the simplest equations.

In everything we did above, we lost sight of one very important point, which will play a decisive role in the future. The fact is that the solution to any logarithmic equation, even the most elementary one, consists of two equal parts. The first is the solution of the equation itself, the second is working with the range of permissible values ​​(APV). This is exactly the first part that we have mastered. In the above examples, ODZ does not affect the answer in any way, so we did not consider it.

Let's take another example:

log 3 (x 2 -3) = log 3 (2x)

Outwardly, this equation is no different from an elementary one, which can be solved very successfully. But it is not so. No, of course we will solve it, but most likely incorrectly, because it contains a small ambush, into which both C-grade students and excellent students immediately fall into it. Let's take a closer look.

Let's say you need to find the root of the equation or the sum of the roots, if there are several of them:

log 3 (x 2 -3) = log 3 (2x)

We use potentiation, it is acceptable here. As a result, we obtain an ordinary quadratic equation.

Finding the roots of the equation:

It turned out two roots.

Answer: 3 and -1

At first glance everything is correct. But let's check the result and substitute it into the original equation.

Let's start with x 1 = 3:

log 3 6 = log 3 6

The check was successful, now the queue is x 2 = -1:

log 3 (-2) = log 3 (-2)

Okay, stop! On the outside everything is perfect. One thing - there are no logarithms from negative numbers! This means that the root x = -1 is not suitable for solving our equation. And therefore the correct answer will be 3, not 2, as we wrote.

This is where ODZ played its fatal role, which we had forgotten about.

Let me remind you that the range of acceptable values ​​includes those values ​​of x that are allowed or make sense for the original example.

Without ODZ, any solution, even an absolutely correct one, of any equation turns into a lottery - 50/50.

How could we get caught solving a seemingly elementary example? But precisely at the moment of potentiation. Logarithms disappeared, and with them all restrictions.

What to do in this case? Refuse to eliminate logarithms? And completely refuse to solve this equation?

No, we just, like real heroes from one famous song, will take a detour!

Before we begin solving any logarithmic equation, we will write down the ODZ. But after that, you can do whatever your heart desires with our equation. Having received the answer, we simply throw out those roots that are not included in our ODZ and write down the final version.

Now let’s decide how to record ODZ. To do this, we carefully examine the original equation and look for suspicious places in it, such as division by x, even root, etc. Until we have solved the equation, we do not know what x is equal to, but we know for sure that those x that, when substituted, give division by 0 or the square root of a negative number, are obviously not suitable as an answer. Therefore, such x are unacceptable, while the rest will constitute ODZ.

Let's use the same equation again:

log 3 (x 2 -3) = log 3 (2x)

log 3 (x 2 -3) = log 3 (2x)

As you can see, there is no division by 0, there are also no square roots, but there are expressions with x in the body of the logarithm. Let us immediately remember that the expression inside the logarithm must always be >0. We write this condition in the form of ODZ:

Those. We haven’t solved anything yet, but we have already written down a mandatory condition for the entire sublogarithmic expression. The curly brace means that these conditions must be true simultaneously.

The ODZ is written down, but it is also necessary to solve the resulting system of inequalities, which is what we will do. We get the answer x > v3. Now we know for sure which x will not suit us. And then we begin to solve the logarithmic equation itself, which is what we did above.

Having received the answers x 1 = 3 and x 2 = -1, it is easy to see that only x1 = 3 suits us, and we write it down as the final answer.

For the future, it is very important to remember the following: we solve any logarithmic equation in 2 stages. The first is to solve the equation itself, the second is to solve the ODZ condition. Both stages are performed independently of each other and are compared only when writing the answer, i.e. discard everything unnecessary and write down the correct answer.

To reinforce the material, we strongly recommend watching the video:

The video shows other examples of solving log. equations and working out the interval method in practice.

To this question, how to solve logarithmic equations That's all for now. If something is decided by the log. equations remain unclear or incomprehensible, write your questions in the comments.

Note: The Academy of Social Education (ASE) is ready to accept new students.

Logarithmic equations. We continue to consider problems from Part B of the Unified State Examination in mathematics. We have already examined solutions to some equations in the articles “”, “”. In this article we will look at logarithmic equations. I’ll say right away that there will be no complex transformations when solving such equations on the Unified State Exam. They are simple.

It is enough to know and understand the basic logarithmic identity, to know the properties of the logarithm. Please note that after solving it, you MUST do a check - substitute the resulting value into the original equation and calculate, in the end you should get the correct equality.

Definition:

The logarithm of a number to base b is the exponent,to which b must be raised to obtain a.

For example:

Log 3 9 = 2, since 3 2 = 9

Properties of logarithms:

Special cases of logarithms:

Let's solve problems. In the first example we will do a check. In the future, check it yourself.

Find the root of the equation: log 3 (4–x) = 4

Since log b a = x b x = a, then

3 4 = 4 – x

x = 4 – 81

x = – 77

Examination:

log 3 (4–(–77)) = 4

log 3 81 = 4

3 4 = 81 Correct.

Answer: – 77

Decide for yourself:

Find the root of the equation: log 2 (4 – x) = 7

Find the root of the equation log 5(4 + x) = 2

We use the basic logarithmic identity.

Since log a b = x b x = a, then

5 2 = 4 + x

x =5 2 – 4

x = 21

Examination:

log 5 (4 + 21) = 2

log 5 25 = 2

5 2 = 25 Correct.

Answer: 21

Find the root of the equation log 3 (14 – x) = log 3 5.

The following property takes place, its meaning is as follows: if on the left and right sides of the equation we have logarithms with the same base, then we can equate the expressions under the signs of the logarithms.

14 – x = 5

x=9

Do a check.

Answer: 9

Decide for yourself:

Find the root of the equation log 5 (5 – x) = log 5 3.

Find the root of the equation: log 4 (x + 3) = log 4 (4x – 15).

If log c a = log c b, then a = b

x + 3 = 4x – 15

3x = 18

x=6

Do a check.

Answer: 6

Find the root of the equation log 1/8 (13 – x) = – 2.

(1/8) –2 = 13 – x

8 2 = 13 – x

x = 13 – 64

x = – 51

Do a check.

A small addition - the property is used here

degrees ().

Answer: – 51

Decide for yourself:

Find the root of the equation: log 1/7 (7 – x) = – 2

Find the root of the equation log 2 (4 – x) = 2 log 2 5.

Let's transform the right side. Let's use the property:

log a b m = m∙log a b

log 2 (4 – x) = log 2 5 2

If log c a = log c b, then a = b

4 – x = 5 2

4 – x = 25

x = – 21

Do a check.

Answer: – 21

Decide for yourself:

Find the root of the equation: log 5 (5 – x) = 2 log 5 3

Solve the equation log 5 (x 2 + 4x) = log 5 (x 2 + 11)

If log c a = log c b, then a = b

x 2 + 4x = x 2 + 11

4x = 11

x = 2.75

Do a check.

Answer: 2.75

Decide for yourself:

Find the root of the equation log 5 (x 2 + x) = log 5 (x 2 + 10).

Solve the equation log 2 (2 – x) = log 2 (2 – 3x) +1.

It is necessary to obtain an expression of the form on the right side of the equation:

log 2 (......)

We represent 1 as a base 2 logarithm:

1 = log 2 2

log c (ab) = log c a + log c b

log 2 (2 – x) = log 2 (2 – 3x) + log 2 2

We get:

log 2 (2 – x) = log 2 2 (2 – 3x)

If log c a = log c b, then a = b, then

2 – x = 4 – 6x

5x = 2

x = 0.4

Do a check.

Answer: 0.4

Decide for yourself: Next you need to solve the quadratic equation. By the way,

the roots are 6 and – 4.

Root "–4" is not a solution, since the base of the logarithm must be greater than zero, and with "– 4" it is equal to " – 5". The solution is root 6.Do a check.

Answer: 6.

R eat on your own:

Solve the equation log x –5 49 = 2. If the equation has more than one root, answer with the smaller one.

As you have seen, no complicated transformations with logarithmic equationsNo. It is enough to know the properties of the logarithm and be able to apply them. In USE problems related to the transformation of logarithmic expressions, more serious transformations are performed and more in-depth skills in solving are required. We will look at such examples, don’t miss them!I wish you success!!!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

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