Solving logarithmic equations is the final lesson. Logarithmic equations

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With this video I begin a long series of lessons about logarithmic equations. Now you have three examples in front of you, on the basis of which we will learn to solve the simplest problems, which are called - protozoa.

log 0.5 (3x − 1) = −3

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

Let me remind you that the simplest logarithmic equation is the following:

log a f (x) = b

In this case, it is important that the variable x is present only inside the argument, that is, only in the function f (x). And the numbers a and b are just numbers, and in no case are functions containing the variable x.

Basic solution methods

There are many ways to solve such structures. For example, most teachers at school offer this method: Immediately express the function f (x) using the formula f ( x) = a b . That is, when you come across the simplest construction, you can immediately move on to the solution without additional actions and constructions.

Yes, of course, the decision will be correct. However, the problem with this formula is that most students do not understand, where it comes from and why we raise the letter a to the letter b.

As a result, I often see very annoying mistakes when, for example, these letters are swapped. This formula must either be understood or crammed, and the second method leads to mistakes at the most inopportune and most crucial moments: during exams, tests, etc.

That is why I suggest to all my students to abandon the standard school formula and use the second approach to solve logarithmic equations, which, as you probably guessed from the name, is called canonical form.

The idea of ​​the canonical form is simple. Let's look at our problem again: on the left we have log a, and by the letter a we mean a number, and in no case a function containing the variable x. Consequently, this letter is subject to all the restrictions that are imposed on the base of the logarithm. namely:

1 ≠ a > 0

On the other hand, from the same equation we see that the logarithm must be equal to the number b, and no restrictions are imposed on this letter, because it can take any value - both positive and negative. It all depends on what values ​​the function f(x) takes.

And here we remember our wonderful rule that any number b can be represented as a logarithm to the base a of a to the power of b:

b = log a a b

How to remember this formula? Yes, very simple. Let's write the following construction:

b = b 1 = b log a a

Of course, in this case all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm and introduce the multiplier b as the power of a. We get:

b = b 1 = b log a a = log a a b

As a result, the original equation will be rewritten as follows:

log a f (x) = log a a b → f (x) = a b

That's all. The new function no longer contains a logarithm and can be solved using standard algebraic techniques.

Of course, someone will now object: why was it necessary to come up with some kind of canonical formula at all, why perform two additional unnecessary steps if it was possible to immediately move from the original design to the final formula? Yes, if only because most students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.

But this sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where the final formula comes from. By the way, this entry is called the canonical formula:

log a f (x) = log a a b

The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.

Examples of solutions

Now let's look at real examples. So, let's decide:

log 0.5 (3x − 1) = −3

Let's rewrite it like this:

log 0.5 (3x − 1) = log 0.5 0.5 −3

Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. Indeed, when you are already well trained in solving such problems, you can immediately perform this step.

However, if you are now just starting to study this topic, it is better not to rush anywhere in order to avoid making offensive mistakes. So, we have the canonical form. We have:

3x − 1 = 0.5 −3

This is no longer a logarithmic equation, but linear with respect to the variable x. To solve it, let's first look at the number 0.5 to the power of −3. Note that 0.5 is 1/2.

(1/2) −3 = (2/1) 3 = 8

Convert all decimal fractions to common fractions when solving a logarithmic equation.

We rewrite and get:

3x − 1 = 8
3x = 9
x = 3

That's it, we got the answer. The first problem has been solved.

Second task

Let's move on to the second task:

As we see, this equation is no longer the simplest. If only because there is a difference on the left, and not a single logarithm to one base.

Therefore, we need to somehow get rid of this difference. In this case, everything is very simple. Let's take a closer look at the bases: on the left is the number under the root:

General recommendation: in all logarithmic equations, try to get rid of radicals, i.e., from entries with roots and move on to power functions, simply because the exponents of these powers are easily taken out of the sign of the logarithm and, ultimately, such an entry significantly simplifies and speeds up calculations. Let's write it down like this:

Now let us remember the remarkable property of the logarithm: powers can be derived from the argument, as well as from the base. In the case of grounds, the following happens:

log a k b = 1/k loga b

In other words, the number that was in the base power is brought forward and at the same time inverted, that is, it becomes a reciprocal number. In our case, the base degree was 1/2. Therefore, we can take it out as 2/1. We get:

5 2 log 5 x − log 5 x = 18
10 log 5 x − log 5 x = 18

Please note: under no circumstances should you get rid of logarithms at this step. Remember 4th-5th grade math and the order of operations: multiplication is performed first, and only then addition and subtraction. In this case, we subtract one of the same elements from 10 elements:

9 log 5 x = 18
log 5 x = 2

Now our equation looks as it should. This is the simplest construction, and we solve it using the canonical form:

log 5 x = log 5 5 2
x = 5 2
x = 25

That's all. The second problem has been solved.

Third example

Let's move on to the third task:

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

Let me remind you of the following formula:

log b = log 10 b

If for some reason you are confused by the notation log b , then when performing all the calculations you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take powers, add and represent any numbers in the form lg 10.

It is these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.

First, note that the factor 2 in front of lg 5 can be added and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.

Judge for yourself: any number can be represented as log to base 10:

3 = log 10 10 3 = log 10 3

Let's rewrite the original problem taking into account the obtained changes:

log (x − 3) = log 1000 + log 25
log (x − 3) = log 1000 25
log (x − 3) = log 25,000

We have before us the canonical form again, and we got it without going through the transformation stage, i.e. the simplest logarithmic equation did not appear anywhere.

This is exactly what I talked about at the very beginning of the lesson. The canonical form allows you to solve a wider class of problems than the standard school formula that most school teachers give.

Well, that’s it, we get rid of the sign of the decimal logarithm, and we get a simple linear construction:

x + 3 = 25,000
x = 24,997

All! The problem is solved.

A note on scope

Here I would like to make an important remark regarding the scope of definition. Surely now there will be students and teachers who will say: “When we solve expressions with logarithms, we must remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why did we not require this inequality to be satisfied in any of the problems considered?

Do not worry. In these cases, no extra roots will appear. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in one single argument of a single logarithm), and nowhere else in our case does the variable x appear, then write down the domain of definition no need, because it will be executed automatically.

Judge for yourself: in the first equation we got that 3x − 1, i.e. the argument should be equal to 8. This automatically means that 3x − 1 will be greater than zero.

With the same success we can write that in the second case x should be equal to 5 2, i.e. it is certainly greater than zero. And in the third case, where x + 3 = 25,000, i.e., again, obviously greater than zero. In other words, the scope is satisfied automatically, but only if x occurs only in the argument of only one logarithm.

That's all you need to know to solve the simplest problems. This rule alone, together with the transformation rules, will allow you to solve a very wide class of problems.

But let's be honest: in order to finally understand this technique, to learn how to apply the canonical form of the logarithmic equation, it is not enough to just watch one video lesson. Therefore, right now, download the options for independent solutions that are attached to this video lesson and start solving at least one of these two independent works.

It will take you literally a few minutes. But the effect of such training will be much higher than if you simply watched this video lesson.

I hope this lesson will help you understand logarithmic equations. Use the canonical form, simplify expressions using the rules for working with logarithms - and you won’t be afraid of any problems. That's all I have for today.

Taking into account the domain of definition

Now let's talk about the domain of definition of the logarithmic function, and how this affects the solution of logarithmic equations. Consider a construction of the form

log a f (x) = b

Such an expression is called the simplest - it contains only one function, and the numbers a and b are just numbers, and in no case a function that depends on the variable x. It can be solved very simply. You just need to use the formula:

b = log a a b

This formula is one of the key properties of the logarithm, and when substituting into our original expression we get the following:

log a f (x) = log a a b

f (x) = a b

This is a familiar formula from school textbooks. Many students will probably have a question: since in the original expression the function f (x) is under the log sign, the following restrictions are imposed on it:

f(x) > 0

This limitation applies because the logarithm of negative numbers does not exist. So, perhaps, as a result of this limitation, a check on answers should be introduced? Perhaps they need to be inserted into the source?

No, in the simplest logarithmic equations additional checking is unnecessary. And that's why. Take a look at our final formula:

f (x) = a b

The fact is that the number a is in any case greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But this doesn’t matter, because no matter what power we raise a positive number to, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is satisfied automatically.

What's really worth checking is the domain of the function under the log sign. There may be quite complex structures, and you definitely need to keep an eye on them during the solution process. Let's get a look.

First task:

First step: convert the fraction on the right. We get:

We get rid of the logarithm sign and get the usual irrational equation:

Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks are required to ensure that the expression under the logarithm sign is greater than 0, because it is not just greater than 0, but according to the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is satisfied automatically.

Let's move on to the second task:

Everything is the same here. We rewrite the construction, replacing the triple:

We get rid of the logarithm signs and get an irrational equation:

We square both sides taking into account the restrictions and get:

4 − 6x − x 2 = (x − 4) 2

4 − 6x − x 2 = x 2 + 8x + 16

x 2 + 8x + 16 −4 + ​​6x + x 2 = 0

2x 2 + 14x + 12 = 0 |:2

x 2 + 7x + 6 = 0

We solve the resulting equation through the discriminant:

D = 49 − 24 = 25

x 1 = −1

x 2 = −6

But x = −6 does not suit us, because if we substitute this number into our inequality, we get:

−6 + 4 = −2 < 0

In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:

−1 + 4 = 3 > 0

The only answer in our case will be x = −1. That's the solution. Let's go back to the very beginning of our calculations.

The main takeaway from this lesson is that you don't need to check constraints on a function in simple logarithmic equations. Because during the solution process all constraints are satisfied automatically.

However, this in no way means that you can forget about checking altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own restrictions and requirements for the right side, which we have seen today in two different examples.

Feel free to solve such problems and be especially careful if there is a root in the argument.

Logarithmic equations with different bases

We continue to study logarithmic equations and look at two more quite interesting techniques with which it is fashionable to solve more complex constructions. But first, let’s remember how the simplest problems are solved:

log a f (x) = b

In this entry, a and b are numbers, and in the function f (x) the variable x must be present, and only there, that is, x must only be in the argument. We will transform such logarithmic equations using the canonical form. To do this, note that

b = log a a b

Moreover, a b is precisely an argument. Let's rewrite this expression as follows:

log a f (x) = log a a b

This is exactly what we are trying to achieve, so that there is a logarithm to base a on both the left and the right. In this case, we can, figuratively speaking, cross out the log signs, and from a mathematical point of view we can say that we are simply equating the arguments:

f (x) = a b

As a result, we will get a new expression that will be much easier to solve. Let's apply this rule to our problems today.

So, the first design:

First of all, I note that on the right is a fraction whose denominator is log. When you see an expression like this, it’s a good idea to remember a wonderful property of logarithms:

Translated into Russian, this means that any logarithm can be represented as the quotient of two logarithms with any base c. Of course 0< с ≠ 1.

So: this formula has one wonderful special case, when the variable c is equal to the variable b. In this case we get a construction like:

This is exactly the construction we see from the sign on the right in our equation. Let's replace this construction with log a b , we get:

In other words, in comparison with the original task, we swapped the argument and the base of the logarithm. Instead, we had to reverse the fraction.

Let us remember that any degree can be derived from the base according to the following rule:

In other words, the coefficient k, which is the power of the base, is expressed as an inverted fraction. Let's render it as an inverted fraction:

The fractional factor cannot be left in front, because in this case we will not be able to represent this notation as a canonical form (after all, in the canonical form there is no additional factor before the second logarithm). Therefore, let's add the fraction 1/4 to the argument as a power:

Now we equate arguments whose bases are the same (and our bases are really the same), and write:

x + 5 = 1

x = −4

That's all. We got the answer to the first logarithmic equation. Please note: in the original problem, the variable x appears in only one log, and it appears in its argument. Therefore, there is no need to check the domain, and our number x = −4 is indeed the answer.

Now let's move on to the second expression:

log 56 = log 2 log 2 7 − 3log (x + 4)

Here, in addition to the usual logarithms, we will have to work with log f (x). How to solve such an equation? To an unprepared student it may seem like this is some kind of tough task, but in fact everything can be solved in an elementary way.

Take a close look at the term lg 2 log 2 7. What can we say about it? The bases and arguments of log and lg are the same, and this should give some ideas. Let's remember once again how powers are taken out from under the sign of the logarithm:

log a b n = nlog a b

In other words, what was a power of b in the argument becomes a factor in front of log itself. Let's apply this formula to the expression lg 2 log 2 7. Don't be scared by lg 2 - this is the most common expression. You can rewrite it as follows:

All the rules that apply to any other logarithm are valid for it. In particular, the factor in front can be added to the degree of the argument. Let's write it down:

Very often, students do not see this action directly, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal about this. Moreover, we get a formula that is easy to calculate if you remember an important rule:

This formula can be considered both as a definition and as one of its properties. In any case, if you are converting a logarithmic equation, you should know this formula just like you would know the log representation of any number.

Let's return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will be simply equal to lg 7. We have:

lg 56 = lg 7 − 3lg (x + 4)

Let's move lg 7 to the left, we get:

lg 56 − lg 7 = −3lg (x + 4)

We subtract the expressions on the left because they have the same base:

lg (56/7) = −3lg (x + 4)

Now let's take a closer look at the equation we got. It is practically the canonical form, but there is a factor −3 on the right. Let's add it to the right lg argument:

log 8 = log (x + 4) −3

Before us is the canonical form of the logarithmic equation, so we cross out the lg signs and equate the arguments:

(x + 4) −3 = 8

x + 4 = 0.5

That's all! We solved the second logarithmic equation. In this case, no additional checks are required, because in the original problem x was present in only one argument.

Let me list the key points of this lesson again.

The main formula that is taught in all the lessons on this page dedicated to solving logarithmic equations is the canonical form. And don’t be scared by the fact that most school textbooks teach you to solve such problems differently. This tool works very effectively and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.

In addition, to solve logarithmic equations it will be useful to know the basic properties. Namely:

1. The formula for moving to one base and the special case when we reverse log (this was very useful to us in the first problem);
2. Formula for adding and subtracting powers from the logarithm sign. Here, many students get stuck and do not see that the degree taken out and introduced can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of the other and at the same time significantly simplify the solution of the problem, which is what we observe in the second case.

In conclusion, I would like to add that it is not necessary to check the domain of definition in each of these cases, because everywhere the variable x is present in only one sign of log, and at the same time is in its argument. As a consequence, all requirements of the scope are fulfilled automatically.

Problems with variable base

Today we will look at logarithmic equations, which for many students seem non-standard, if not completely unsolvable. We are talking about expressions based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely through the canonical form.

First, let's remember how the simplest problems are solved, based on ordinary numbers. So, the simplest construction is called

log a f (x) = b

To solve such problems we can use the following formula:

b = log a a b

We rewrite our original expression and get:

log a f (x) = log a a b

Then we equate the arguments, i.e. we write:

f (x) = a b

Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained from the solution will be the roots of the original logarithmic equation. In addition, a record when both the left and the right are in the same logarithm with the same base is precisely called the canonical form. It is to such a record that we will try to reduce today's designs. So, let's go.

First task:

log x − 2 (2x 2 − 13x + 18) = 1

Replace 1 with log x − 2 (x − 2) 1 . The degree that we observe in the argument is actually the number b that stood to the right of the equal sign. Thus, let's rewrite our expression. We get:

log x − 2 (2x 2 − 13x + 18) = log x − 2 (x − 2)

What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:

2x 2 − 13x + 18 = x − 2

But the solution does not end there, because this equation is not equivalent to the original one. After all, the resulting construction consists of functions that are defined on the entire number line, and our original logarithms are not defined everywhere and not always.

Therefore, we must write down the domain of definition separately. Let's not split hairs and first write down all the requirements:

First, the argument of each of the logarithms must be greater than 0:

2x 2 − 13x + 18 > 0

x − 2 > 0

Secondly, the base must not only be greater than 0, but also different from 1:

x − 2 ≠ 1

As a result, we get the system:

But don’t be alarmed: when processing logarithmic equations, such a system can be significantly simplified.

Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required that it be greater than zero.

In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will automatically be satisfied. Therefore, we can safely cross out the inequality containing the quadratic function. Thus, the number of expressions contained in our system will be reduced to three.

Of course, with the same success we could cross out the linear inequality, that is, cross out x − 2 > 0 and require that 2x 2 − 13x + 18 > 0. But you will agree that solving the simplest linear inequality is much faster and simpler, than quadratic, even under the condition that as a result of solving this entire system we get the same roots.

In general, try to optimize calculations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.

Let's rewrite our system:

Here is a system of three expressions, two of which we, in fact, have already dealt with. Let's write out the quadratic equation separately and solve it:

2x 2 − 14x + 20 = 0

x 2 − 7x + 10 = 0

Before us is a reduced quadratic trinomial and, therefore, we can use Vieta’s formulas. We get:

(x − 5)(x − 2) = 0

x 1 = 5

x 2 = 2

Now we return to our system and find that x = 2 does not suit us, because we are required that x be strictly greater than 2.

But x = 5 suits us perfectly: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution to this system will be x = 5.

That's it, the problem is solved, including taking into account the ODZ. Let's move on to the second equation. More interesting and informative calculations await us here:

The first step: like last time, we bring this whole matter to canonical form. To do this, we can write the number 9 as follows:

You don’t have to touch the base with the root, but it’s better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write down:

Let me not rewrite our entire large logarithmic equation, but just immediately equate the arguments:

x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27

x 2 + 4x + 3 = 0

Before us is a newly reduced quadratic trinomial, let’s use Vieta’s formulas and write:

(x + 3)(x + 1) = 0

x 1 = −3

x 2 = −1

So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, the log signs impose additional restrictions (here we should have written down the system, but due to the cumbersome nature of the whole structure, I decided to calculate the domain of definition separately).

First of all, remember that the arguments must be greater than 0, namely:

These are the requirements imposed by the scope of definition.

Let us immediately note that since we equate the first two expressions of the system to each other, we can cross out any of them. Let's cross out the first one because it looks more threatening than the second one.

In addition, note that the solution to the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly, with a root of the third degree - these inequalities are completely analogous, so we can cross it out).

But with the third inequality this will not work. Let's get rid of the radical sign on the left by raising both parts to a cube. We get:

So we get the following requirements:

− 2 ≠ x > −3

Which of our roots: x 1 = −3 or x 2 = −1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (since our inequality is strict). So, returning to our problem, we get one root: x = −1. That's it, problem solved.

Once again, the key points of this task:

1. Feel free to apply and solve logarithmic equations using canonical form. Students who make such a notation, rather than moving directly from the original problem to a construction like log a f (x) = b, make much fewer errors than those who rush somewhere, skipping intermediate steps of calculations;
2. As soon as a variable base appears in a logarithm, the problem ceases to be the simplest. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they also must not be equal to 1.

The final requirements can be applied to the final answers in different ways. For example, you can solve an entire system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember the domain of definition, separately work it out in the form of a system and apply it to the obtained roots.

Which method to choose when solving a particular logarithmic equation is up to you. In any case, the answer will be the same.

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.

Examples:

\(\log_(2)(⁡x) = 32\)
\(\log_3⁡x=\log_3⁡9\)
\(\log_3⁡((x^2-3))=\log_3⁡((2x))\)
\(\log_(x+1)((x^2+3x-7))=2\)
\(\lg^2⁡((x+1))+10=11 \lg⁡((x+1))\)

How to solve logarithmic equations:

When solving a logarithmic equation, you should strive to transform it to the form \(\log_a⁡(f(x))=\log_a⁡(g(x))\), and then make the transition to \(f(x)=g(x) \).

\(\log_a⁡(f(x))=\log_a⁡(g(x))\) \(⇒\) \(f(x)=g(x)\).

Example:\(\log_2⁡(x-2)=3\)

Very important! This transition can only be made if:

You have written for the original equation, and at the end you will check whether those found are included in the DL. If this is not done, extra roots may appear, which means a wrong decision.

The number (or expression) on the left and right is the same;

The logarithms on the left and right are “pure”, that is, there should be no multiplications, divisions, etc. – only single logarithms on either side of the equal sign.

For example:

Note that Equations 3 and 4 can be easily solved by applying the necessary properties of logarithms.

Example . Solve the equation \(2\log_8⁡x=\log_8⁡2.5+\log_8⁡10\)

Solution :

Answer : \(5\)

Example : Solve the equation \(\log^2_2⁡(x)-3 \log_2(⁡x)+2=0\)

Solution :

Answer : \(4\); \(2\).

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