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 need to strive to convert 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:\(\log_2⁡(x-2)=3\)

Decision: \(\log_2⁡(x-2)=\log_2⁡8\) \(x-2=8\) \(x=10\) Examination:\(10>2\) - suitable for ODZ Answer:\(x=10\)

ODZ: \(x-2>0\) \(x>2\)

Very important! This transition can only be made if:

You wrote for the original equation, and at the end check if the found ones are included in the DPV. If this is not done, extra roots may appear, which means the wrong decision.

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

The logarithms on the left and right are "pure", that is, there should not be any, multiplications, divisions, etc. - only lone logarithms on both sides of the equals sign.

For example:

Note that equations 3 and 4 can be easily solved by applying desired properties logarithms.

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

Decision :

		Let's write ODZ: \(x>0\).
\(2\log_8⁡x=\log_8⁡2,5+\log_8⁡10\) ODZ: \(x>0\)		On the left in front of the logarithm is the coefficient, on the right is the sum of the logarithms. This bothers us. Let's transfer the two to the exponent \(x\) by the property: \(n \log_b(⁡a)=\log_b⁡(a^n)\). We represent the sum of logarithms as a single logarithm by the property: \(\log_a⁡b+\log_a⁡c=\log_a(⁡bc)\)
\(\log_8⁡(x^2)=\log_8⁡25\)		We brought the equation to the form \(\log_a⁡(f(x))=\log_a⁡(g(x))\) and wrote down the ODZ, which means that we can make the transition to the form \(f(x)=g(x)\ ).
		Happened . We solve it and get the roots.
\(x_1=5\) \(x_2=-5\)		We check whether the roots fit under the ODZ. To do this, in \(x>0\) instead of \(x\) we substitute \(5\) and \(-5\). This operation can be performed orally.
\(5>0\), \(-5>0\)		The first inequality is true, the second is not. So \(5\) is the root of the equation, but \(-5\) is not. We write down the answer.

Answer : \(5\)

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

Decision :

		Let's write ODZ: \(x>0\).
\(\log^2_2⁡(x)-3 \log_2(⁡x)+2=0\) ODZ: \(x>0\)		Typical Equation, solved with . Replace \(\log_2⁡x\) with \(t\).
\(t=\log_2⁡x\)
		Received the usual. Looking for its roots.
\(t_1=2\) \(t_2=1\)		Making a reverse substitution
\(\log_2(⁡x)=2\) \(\log_2(⁡x)=1\)		We transform the right parts, representing them as logarithms: \(2=2 \cdot 1=2 \log_2⁡2=\log_2⁡4\) and \(1=\log_2⁡2\)
\(\log_2(⁡x)=\log_2⁡4\) \(\log_2(⁡x)=\log_2⁡2 \)		Now our equations are \(\log_a⁡(f(x))=\log_a⁡(g(x))\) and we can jump to \(f(x)=g(x)\).
\(x_1=4\) \(x_2=2\)		We check the correspondence of the roots of the ODZ. To do this, instead of \(x\) we substitute \(4\) and \(2\) into the inequality \(x>0\).
\(4>0\) \(2>0\)		Both inequalities are true. So both \(4\) and \(2\) are the roots of the equation.

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

Today we will learn how to solve the simplest logarithmic equations, which do not require preliminary transformations and selection of roots. But if you learn how to solve such equations, then it will be much easier.

The simplest logarithmic equation is an equation of the form log a f (x) \u003d b, where a, b are numbers (a\u003e 0, a ≠ 1), f (x) is some function.

A distinctive feature of all logarithmic equations is the presence of the variable x under the sign of the logarithm. If such an equation is initially given in the problem, it is called the simplest one. Any other logarithmic equations are reduced to the simplest by special transformations (see "Basic properties of logarithms"). However, numerous subtleties must be taken into account: extra roots may appear, so complex logarithmic equations will be considered separately.

How to solve such equations? It is enough to replace the number to the right of the equal sign with a logarithm in the same base as on the left. Then you can get rid of the sign of the logarithm. We get:

log a f (x) \u003d b ⇒ log a f (x) \u003d log a a b ⇒ f (x) \u003d a b

We got the usual equation. Its roots are the roots of the original equation.

Pronouncement of degrees

Often, logarithmic equations, which outwardly look complicated and threatening, are solved in just a couple of lines without involving complex formulas. Today we will consider just such problems, where all that is required of you is to carefully reduce the formula to the canonical form and not get confused when searching for the domain of definition of logarithms.

Today, as you probably guessed from the title, we will solve logarithmic equations using the formulas for the transition to the canonical form. The main "trick" of this video lesson will be working with degrees, or rather, taking the degree from the base and argument. Let's look at the rule:

Similarly, you can take out the degree from the base:

As you can see, if when taking the degree out of the logarithm argument, we simply have additional multiplier in front, then when taking the degree out of the base - not just a factor, but an inverted factor. This must be remembered.

Finally, the most interesting. These formulas can be combined, then we get:

Of course, when performing these transitions, there are certain pitfalls associated with possible expansion domain of definition or, conversely, by narrowing the domain of definition. Judge for yourself:

log 3 x 2 = 2 ∙ log 3 x

If in the first case, x could be any number other than 0, i.e., the requirement x ≠ 0, then in the second case, we will only be satisfied with x, which are not only not equal, but strictly greater than 0, because the domain of the logarithm is that the argument be strictly greater than 0. Therefore, I will remind you of a wonderful formula from the course of algebra in grades 8-9:

That is, we must write our formula as follows:

log 3 x 2 = 2 ∙ log 3 |x |

Then no narrowing of the domain of definition will occur.

However, in today's video tutorial there will be no squares. If you look at our tasks, you will see only the roots. Therefore, apply this rule we won’t, but it still needs to be kept in mind in order to right moment when you see quadratic function in the argument or base of the logarithm, you will remember this rule and perform all the transformations correctly.

So the first equation is:

To solve this problem, I propose to carefully look at each of the terms present in the formula.

Let's rewrite the first term as a power with a rational exponent:

We look at the second term: log 3 (1 − x ). You don't need to do anything here, everything is already being transformed.

Finally, 0, 5. As I said in previous lessons, when solving logarithmic equations and formulas, I highly recommend moving from decimal fractions to ordinary ones. Let's do this:

0,5 = 5/10 = 1/2

Let's rewrite our original formula taking into account the obtained terms:

log 3 (1 − x ) = 1

Now let's move on to the canonical form:

log 3 (1 − x ) = log 3 3

Get rid of the sign of the logarithm by equating the arguments:

1 − x = 3

-x = 2

x = −2

That's it, we've solved the equation. However, let's still play it safe and find the domain of definition. For this, let's go back to original formula and see:

1 − x > 0

-x > -1

x< 1

Our root x = −2 satisfies this requirement, so x = −2 is a solution to the original equation. Now we have a strict clear justification. Everything, the task is solved.

Let's move on to the second task:

Let's deal with each term separately.

We write out the first:

We have modified the first term. We work with the second term:

Finally, the last term, which is to the right of the equal sign:

We substitute the resulting expressions for the terms in the resulting formula:

log 3 x = 1

We pass to the canonical form:

log 3 x = log 3 3

We get rid of the sign of the logarithm by equating the arguments, and we get:

x=3

Again, just in case, let's play it safe, go back to the original equation and see. In the original formula, the variable x is present only in the argument, therefore,

x > 0

In the second logarithm, x is under the root, but again in the argument, therefore, the root must be greater than 0, i.e. radical expression must be greater than 0. We look at our root x = 3. Obviously, it satisfies this requirement. Therefore, x = 3 is the solution to the original logarithmic equation. Everything, the task is solved.

There are two key points in today's video tutorial:

1) do not be afraid to convert logarithms and, in particular, do not be afraid to take degrees out of the sign of the logarithm, while remembering our basic formula: when taking the degree out of the argument, it is taken out simply without changes as a factor, and when taking the degree out of the base, this degree is reversed.

2) the second point is related to the self-canonical form. We performed the transition to the canonical form at the very end of the transformation of the formula of the logarithmic equation. Recall the following formula:

a = log b b a

Of course, by the expression "any number b", I mean those numbers that satisfy the requirements imposed on the base of the logarithm, i.e.

1 ≠ b > 0

For such b , and since we already know the base, this requirement will be fulfilled automatically. But for such b - any that satisfy this requirement- this transition can be performed, and we will get a canonical form in which we can get rid of the sign of the logarithm.

Extension of the domain of definition and extra roots

In the process of transforming logarithmic equations, an implicit extension of the domain of definition may occur. Often, students do not even notice this, which leads to errors and incorrect answers.

Let's start with the simplest designs. The simplest logarithmic equation is the following:

log a f(x) = b

Note that x is present in only one argument of one logarithm. How do we solve such equations? We use the canonical form. To do this, we represent the number b \u003d log a a b, and our equation will be rewritten in the following form:

log a f(x) = log a a b

This notation is called the canonical form. It is to it that any logarithmic equation that you will meet not only in today's lesson, but also in any independent and control work should be reduced.

How to come to the canonical form, what techniques to use - this is already a matter of practice. The main thing to understand: as soon as you receive such a record, we can assume that the problem is solved. because next step there will be an entry:

f(x) = a b

In other words, we get rid of the sign of the logarithm and simply equate the arguments.

Why all this talk? The fact is that the canonical form is applicable not only to the simplest problems, but also to any other. In particular, to those that we will address today. Let's get a look.

First task:

What is the problem with this equation? The fact that the function is in two logarithms at once. The problem can be reduced to the simplest by simply subtracting one logarithm from another. But there are problems with the domain of definition: extra roots may appear. So let's just move one of the logarithms to the right:

Here such a record is already much more similar to the canonical form. But there is one more nuance: in the canonical form, the arguments must be the same. And we have the logarithm to the base 3 on the left, and the logarithm to the base 1/3 on the right. You know, you need to bring these bases to the same number. For example, let's remember what negative exponents are:

And then we will use the exponent "-1" outside the log as a multiplier:

Please note: the degree that stood at the base is turned over and turns into a fraction. We got an almost canonical notation by getting rid of various bases, but instead we got the “−1” factor on the right. Let's put this factor into the argument by turning it into a power:

Of course, having received the canonical form, we boldly cross out the sign of the logarithm and equate the arguments. At the same time, let me remind you that when raised to the power of “−1”, the fraction simply turns over - a proportion is obtained.

Let's use the main property of the proportion and multiply it crosswise:

(x - 4) (2x - 1) = (x - 5) (3x - 4)

2x 2 - x - 8x + 4 = 3x 2 - 4x - 15x + 20

2x2 - 9x + 4 = 3x2 - 19x + 20

x2 − 10x + 16 = 0

Before us is the quadratic equation, so we solve it using the Vieta formulas:

(x − 8)(x − 2) = 0

x 1 = 8; x2 = 2

That's all. Do you think the equation is solved? Not! For such a solution, we will get 0 points, because in the original equation there are two logarithms with the variable x at once. Therefore, it is necessary to take into account the domain of definition.

And this is where the fun begins. Most students are confused: what is the domain of the logarithm? Of course, all arguments (we have two) must be greater than zero:

(x − 4)/(3x − 4) > 0

(x − 5)/(2x − 1) > 0

Each of these inequalities must be solved, marked on a straight line, crossed - and only then see what roots lie at the intersection.

I'll be honest: this technique has the right to exist, it is reliable, and you will get the right answer, but there are too many extra steps in it. So let's go through our solution again and see: where exactly do you want to apply scope? In other words, you need to clearly understand exactly when extra roots appear.

1. Initially, we had two logarithms. Then we moved one of them to the right, but this did not affect the definition area.
2. Then we remove the power from the base, but there are still two logarithms, and each of them contains the variable x .
3. Finally, we cross out the signs of log and get the classic fractional rational equation.

It is at the last step that the domain of definition is expanded! As soon as we switched to a fractional rational equation, getting rid of the signs of log, the requirements for the x variable changed dramatically!

Therefore, the domain of definition can be considered not at the very beginning of the solution, but only at the mentioned step - before we directly equate the arguments.

This is where the opportunity for optimization lies. On the one hand, we are required that both arguments be greater than zero. On the other hand, we further equate these arguments. Therefore, if at least one of them is positive, then the second one will also be positive!

So it turns out that requiring the fulfillment of two inequalities at once is an overkill. It is enough to consider only one of these fractions. Which one? The one that is easier. For example, let's look at the right fraction:

(x − 5)/(2x − 1) > 0

This is typical fractional rational inequality, we solve it by the interval method:

How to place signs? Let's take a number, obviously greater than all our roots. For example, 1 billion. And we substitute its fraction. We get a positive number, i.e. to the right of the root x = 5 there will be a plus sign.

Then the signs alternate, because there are no roots of even multiplicity anywhere. We are interested in intervals where the function is positive. Hence x ∈ (−∞; −1/2)∪(5; +∞).

Now let's remember the answers: x = 8 and x = 2. Strictly speaking, these are not answers yet, but only candidates for an answer. Which one belongs specified set? Of course, x = 8. But x = 2 does not suit us in terms of the domain of definition.

In total, the answer to the first logarithmic equation will be x = 8. Now we have a competent, reasonable solution, taking into account the domain of definition.

Let's move on to the second equation:

log 5 (x - 9) = log 0.5 4 - log 5 (x - 5) + 3

I remind you that if there is a decimal fraction in the equation, then you should get rid of it. In other words, we rewrite 0.5 as ordinary fraction. We immediately notice that the logarithm containing this base is easily considered:

This is a very important moment! When we have degrees in both the base and the argument, we can take out the indicators of these degrees using the formula:

We return to our original logarithmic equation and rewrite it:

log 5 (x - 9) = 1 - log 5 (x - 5)

We got a construction that is quite close to the canonical form. However, we are confused by the terms and the minus sign to the right of the equals sign. Let's represent unity as a logarithm to base 5:

log 5 (x - 9) = log 5 5 1 - log 5 (x - 5)

Subtract the logarithms on the right (while their arguments are divided):

log 5 (x − 9) = log 5 5/(x − 5)

Perfectly. So we got the canonical form! We cross out the log signs and equate the arguments:

(x − 9)/1 = 5/(x − 5)

This is a proportion that is easily solved by cross-multiplication:

(x − 9)(x − 5) = 5 1

x 2 - 9x - 5x + 45 = 5

x2 − 14x + 40 = 0

Obviously, we have a given quadratic equation. It is easily solved using the Vieta formulas:

(x − 10)(x − 4) = 0

x 1 = 10

x 2 = 4

We have two roots. But these are not final answers, but only candidates, because the logarithmic equation also requires checking the domain.

I remind you: do not look when everyone of the arguments will be greater than zero. It suffices to require that one argument, either x − 9 or 5/(x − 5) be greater than zero. Consider the first argument:

x − 9 > 0

x > 9

Obviously, only x = 10 satisfies this requirement. This is the final answer. All problem solved.

Again key thoughts today's lesson:

1. As soon as the variable x appears in several logarithms, the equation ceases to be elementary, and for it it is necessary to calculate the domain of definition. Otherwise, you can easily write extra roots in response.
2. Working with the domain of definition itself can be greatly simplified if the inequality is not written immediately, but exactly at the moment when we get rid of the signs of log. After all, when the arguments are equated to each other, it is enough to require that only one of them be greater than zero.

Of course, we ourselves choose from which argument to make an inequality, so it is logical to choose the simplest one. For example, in the second equation, we chose the argument (x − 9) − linear function, as opposed to the fractionally rational second argument. Agree, solving the inequality x − 9 > 0 is much easier than 5/(x − 5) > 0. Although the result is the same.

This remark greatly simplifies the search for ODZ, but be careful: you can use one inequality instead of two only when the arguments are precisely equate to each other!

Of course, someone will now ask: what happens differently? Yes, sometimes. For example, in the step itself, when we multiply two arguments containing a variable, there is a danger of extra roots.

Judge for yourself: at first it is required that each of the arguments be greater than zero, but after multiplication it is sufficient that their product be greater than zero. As a result, the case when each of these fractions is negative is missed.

Therefore, if you are just starting to deal with complex logarithmic equations, in no case do not multiply logarithms containing the variable x - too often this will lead to extra roots. Better take one extra step, transfer one term to the other side, make up the canonical form.

Well, what to do if you cannot do without multiplying such logarithms, we will discuss in the next video tutorial. :)

Once again about the powers in the equation

Today we will analyze a rather slippery topic regarding logarithmic equations, or rather, the removal of powers from the arguments and bases of logarithms.

I would even say that we will talk about taking out even powers, because it is with even powers that most of the difficulties arise when solving real logarithmic equations.

Let's start with the canonical form. Let's say we have an equation like log a f (x) = b. In this case, we rewrite the number b according to the formula b = log a a b . It turns out the following:

log a f(x) = log a a b

Then we equate the arguments:

f(x) = a b

The penultimate formula is called the canonical form. It is to her that they try to reduce any logarithmic equation, no matter how complicated and terrible it may seem at first glance.

Here, let's try. Let's start with the first task:

Preliminary remark: as I said, all decimals in a logarithmic equation, it is better to translate it into ordinary ones:

0,5 = 5/10 = 1/2

Let's rewrite our equation with this fact in mind. Note that both 1/1000 and 100 are powers of 10, and then we take the powers out of wherever they are: from the arguments and even from the base of the logarithms:

And here the question arises for many students: “Where did the module come from on the right?” Indeed, why not just write (x − 1)? Of course, now we will write (x − 1), but the right to such a record gives us the account of the domain of definition. After all, the other logarithm already contains (x − 1), and this expression must be greater than zero.

But when we take out the square from the base of the logarithm, we must leave the module at the base. I'll explain why.

The fact is that from the point of view of mathematics, taking a degree is tantamount to taking a root. In particular, when the expression (x − 1) 2 is squared, we are essentially extracting the root of the second degree. But the square root is nothing more than a modulus. Exactly module, because even if the expression x - 1 is negative, when squaring "minus" will still burn. Further extraction of the root will give us a positive number - already without any minuses.

In general, in order to avoid offensive mistakes, remember once and for all:

The root of an even degree from any function that is raised to the same power is equal not to the function itself, but to its modulus:

We return to our logarithmic equation. Speaking about the module, I argued that we can painlessly remove it. It's true. Now I will explain why. Strictly speaking, we had to consider two options:

1. x − 1 > 0 ⇒ |x − 1| = x − 1
2. x − 1< 0 ⇒ |х − 1| = −х + 1

Each of these options would need to be addressed. But there is one catch: the original formula already contains the function (x − 1) without any modulus. And following the domain of definition of logarithms, we have the right to immediately write down that x − 1 > 0.

This requirement must be satisfied regardless of any modules and other transformations that we perform in the solution process. Therefore, it is pointless to consider the second option - it will never arise. Even if, when solving this branch of the inequality, we get some numbers, they still will not be included in the final answer.

Now we are literally one step away from the canonical form of the logarithmic equation. Let's represent the unit as follows:

1 = log x − 1 (x − 1) 1

In addition, we introduce the factor −4, which is on the right, into the argument:

log x − 1 10 −4 = log x − 1 (x − 1)

Before us is the canonical form of the logarithmic equation. Get rid of the sign of the logarithm:

10 −4 = x − 1

But since the base was a function (and not a prime number), we additionally require that this function be greater than zero and not equal to one. Get the system:

Since the requirement x − 1 > 0 is automatically satisfied (because x − 1 = 10 −4), one of the inequalities can be deleted from our system. The second condition can also be crossed out because x − 1 = 0.0001< 1. Итого получаем:

x = 1 + 0.0001 = 1.0001

This is the only root that automatically satisfies all the requirements for the domain of definition of the logarithm (however, all requirements were eliminated as knowingly fulfilled in the conditions of our problem).

So the second equation is:

3 log 3 x x = 2 log 9 x x 2

How is this equation fundamentally different from the previous one? Already at least the fact that the bases of logarithms - 3x and 9x - are not natural degrees each other. Therefore, the transition we used in the previous solution is not possible.

Let's at least get rid of the degrees. In our case, the only power is in the second argument:

3 log 3 x x = 2 ∙ 2 log 9 x |x |

However, the modulus sign can be removed, because the variable x is also in the base, i.e. x > 0 ⇒ |x| = x. Let's rewrite our logarithmic equation:

3 log 3 x x = 4 log 9 x x

We got logarithms in which the arguments are the same, but different grounds. How to proceed? There are many options here, but we will consider only two of them, which are the most logical, and most importantly, these are quick and understandable tricks for most students.

We have already considered the first option: in any incomprehensible situation translate logarithms from variable base to some permanent foundation. For example, to a deuce. The conversion formula is simple:

Of course, a normal number should act as a variable c: 1 ≠ c > 0. In our case, let c = 2. Now we have an ordinary fractional rational equation. We collect all the elements on the left:

Obviously, the factor log 2 x is better to take out, since it is present in both the first and second fractions.

log 2 x = 0;

3 log 2 9x = 4 log 2 3x

We break each log into two terms:

log 2 9x = log 2 9 + log 2 x = 2 log 2 3 + log 2 x;

log 2 3x = log 2 3 + log 2 x

Let's rewrite both sides of the equality taking into account these facts:

3 (2 log 2 3 + log 2 x ) = 4 (log 2 3 + log 2 x )

6 log 2 3 + 3 log 2 x = 4 log 2 3 + 4 log 2 x

2 log 2 3 = log 2 x

Now it remains to add a deuce under the sign of the logarithm (it will turn into a power: 3 2 \u003d 9):

log 2 9 = log 2 x

Before us is the classical canonical form, we get rid of the sign of the logarithm and get:

As expected, this root turned out to be greater than zero. It remains to check the domain of definition. Let's look at the bases:

But the root x = 9 satisfies these requirements. Therefore, it is the final solution.

Conclusion from this decision simple: do not be afraid of long calculations! It's just that at the very beginning we chose a new base at random - and this significantly complicated the process.

But then the question arises: what basis is optimal? I will talk about this in the second way.

Let's go back to our original equation:

3 log 3x x = 2 log 9x x 2

3 log 3x x = 2 ∙ 2 log 9x |x |

x > 0 ⇒ |x| = x

3 log 3 x x = 4 log 9 x x

Now let's think a little: what number or function will be the optimal base? It's obvious that the best option will be c = x - what is already in the arguments. In this case log formula a b = log c b /log c a becomes:

In other words, the expression is simply reversed. In this case, the argument and the basis are reversed.

This formula is very useful and very often used in solving complex logarithmic equations. However, when using this formula, there is one very serious pitfall. If instead of the base we substitute the variable x, then restrictions are imposed on it that were not previously observed:

There was no such restriction in the original equation. Therefore, we should separately check the case when x = 1. Substitute this value in our equation:

3 log 3 1 = 4 log 9 1

We get the right numerical equality. Therefore, x = 1 is a root. We found exactly the same root in the previous method at the very beginning of the solution.

But now, when we separately considered this special case, we safely assume that x ≠ 1. Then our logarithmic equation will be rewritten in the following form:

3 log x 9x = 4 log x 3x

We expand both logarithms according to the same formula as before. Note that log x x = 1:

3 (log x 9 + log x x ) = 4 (log x 3 + log x x )

3 log x 9 + 3 = 4 log x 3 + 4

3 log x 3 2 − 4 log x 3 = 4 − 3

2 log x 3 = 1

Here we come to the canonical form:

log x 9 = log x x 1

x=9

We got the second root. It satisfies the requirement x ≠ 1. Therefore, x = 9 along with x = 1 is the final answer.

As you can see, the volume of calculations has slightly decreased. But when solving a real logarithmic equation, the number of steps will be much less also because you are not required to describe each step in such detail.

The key rule of today's lesson is as follows: if the task contains even degree, from which a root of the same degree is extracted, then at the output we get a module. However, this module can be removed if you pay attention to the domain of definition of logarithms.

But be careful: most students after this lesson think that they understand everything. But when deciding real tasks they cannot reproduce the entire logical chain. As a result, the equation acquires extra roots, and the answer is wrong.

basic properties.

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

same grounds

log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x >

Task. Find the value of the expression:

Transition to a new foundation

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

Task. Find the value of the expression:

See also:

Basic properties of the logarithm

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

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

Basic properties of logarithms

Knowing this rule you will know and exact value exhibitors, and the date of birth of Leo Tolstoy.

Examples for logarithms

Take the logarithm of expressions

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

By properties 3,5 we calculate

2.

3.

4. where .

Example 2 Find x if

Example 3. Let the value of logarithms be given

Calculate log(x) if

Basic properties of logarithms

Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not exactly regular numbers, there are rules here, which are called basic properties.

These rules must be known - without them, not a single serious logarithmic problem. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

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

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

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

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

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

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Based on this fact, many test papers. Yes, what are the control - similar expressions in all seriousness (sometimes practically unchanged) are offered at the exam.

Removing the exponent from the logarithm

It is easy to see that last rule follows the first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log7 496.

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

Task. Find the value of the expression:

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

I think to last example clarification is required. Where have logarithms gone? All the way last moment we work only with the denominator.

Formulas of logarithms. Logarithms are examples of solutions.

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

Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.

Transition to a new foundation

Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?

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

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

In particular, if we put c = x, we get:

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

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

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

Task. Find the value of the expression: log5 16 log2 25.

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

Now let's flip the second logarithm:

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

Task. Find the value of the expression: log9 100 lg 3.

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

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

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.

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

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

Like the formulas for moving to a new base, the main logarithmic identity sometimes it is the only possible solution.

Task. Find the value of the expression:

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

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

Logarithmic unit and logarithmic zero

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

1. logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
2. loga 1 = 0 is. The base a can be anything, but if the argument is one - the logarithm zero! Because a0 = 1 is a direct consequence of the definition.

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

See also:

The logarithm of the number b to the base a denotes the expression. To calculate the logarithm means to find such a power x () at which the equality is true

Basic properties of the logarithm

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

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

When calculating the formulas for the sum and difference of logarithms (3.4) are encountered quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.

Common cases of logarithms

Some of the common logarithms are those in which the base is even ten, exponential or deuce.
The base ten logarithm is usually called the base ten logarithm and is simply denoted lg(x).

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

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

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

And another important base two logarithm is

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

The integral or antiderivative logarithm is determined by the dependence

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

Examples for logarithms

Take the logarithm of expressions

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

By properties 3,5 we calculate

2.
By the difference property of logarithms, we have

3.
Using properties 3.5 we find

4. where .

By the look complex expression using a series of rules is simplified to the form

Finding Logarithm Values

Example 2 Find x if

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

Substitute in the record and mourn

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Calculate log(x) if

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

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

Basic properties of logarithms

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

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

Addition and subtraction of logarithms

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

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

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

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

Task. Find the value of the expression: log6 4 + log6 9.

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

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

It is easy to see that the last rule follows their first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself.

How to solve logarithms

This is what is most often required.

Task. Find the value of the expression: log7 496.

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

Task. Find the value of the expression:

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

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

Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.

Transition to a new foundation

Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?

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

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

In particular, if we put c = x, we get:

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

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

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

Task. Find the value of the expression: log5 16 log2 25.

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

Now let's flip the second logarithm:

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

Task. Find the value of the expression: log9 100 lg 3.

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

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

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.

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

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

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

Task. Find the value of the expression:

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

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

Logarithmic unit and logarithmic zero

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

1. logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
2. loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 1 is a direct consequence of the definition.

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

Let's consider some types of logarithmic equations that are not so often considered in mathematics lessons at school, but are widely used in the preparation of competitive tasks, including for the USE.

1. Equations solved by the logarithm method

When solving equations containing a variable both in the base and in the exponent, the logarithm method is used. If, in addition, the exponent contains a logarithm, then both sides of the equation must be logarithmized to the base of this logarithm.

Example 1

Solve the equation: x log 2 x + 2 = 8.

Decision.

We take the logarithm of the left and right sides of the equation in base 2. We get

log 2 (x log 2 x + 2) = log 2 8,

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

Let log 2 x = t.

Then (t + 2)t = 3.

t 2 + 2t - 3 = 0.

D \u003d 16. t 1 \u003d 1; t 2 \u003d -3.

So log 2 x \u003d 1 and x 1 \u003d 2 or log 2 x \u003d -3 and x 2 \u003d 1/8

Answer: 1/8; 2.

2. Homogeneous logarithmic equations.

Example 2

Solve the equation log 2 3 (x 2 - 3x + 4) - 3log 3 (x + 5) log 3 (x 2 - 3x + 4) - 2log 2 3 (x + 5) = 0

Decision.

Equation domain

(x 2 - 3x + 4 > 0,
(x + 5 > 0. → x > -5.

log 3 (x + 5) = 0 for x = -4. By checking, we determine that given value x not is the root of the original equation. Therefore, we can divide both sides of the equation by log 2 3 (x + 5).

We get log 2 3 (x 2 - 3x + 4) / log 2 3 (x + 5) - 3 log 3 (x 2 - 3x + 4) / log 3 (x + 5) + 2 = 0.

Let log 3 (x 2 - 3x + 4) / log 3 (x + 5) = t. Then t 2 - 3 t + 2 = 0. The roots of this equation are 1; 2. Returning to the original variable, we obtain a set of two equations

But taking into account the existence of the logarithm, only the values ​​\u200b\u200bof (0; 9] should be considered. This means that the expression on the left side takes highest value 2 for x = 1. Consider now the function y = 2 x-1 + 2 1-x. If we take t \u003d 2 x -1, then it will take the form y \u003d t + 1 / t, where t\u003e 0. Under such conditions, it has a unique critical point t = 1. This is the minimum point. Y vin \u003d 2. And it is achieved at x \u003d 1.

It is now obvious that the graphs of the considered functions can intersect only once at the point (1; 2). It turns out that x \u003d 1 is the only root of the equation being solved.

Answer: x = 1.

Example 5. Solve the equation log 2 2 x + (x - 1) log 2 x \u003d 6 - 2x

Decision.

We will decide given equation relative to log 2 x. Let log 2 x = t. Then t 2 + (x - 1) t - 6 + 2x \u003d 0.

D \u003d (x - 1) 2 - 4 (2x - 6) \u003d (x - 5) 2. t 1 \u003d -2; t 2 \u003d 3 - x.

We get the equation log 2 x \u003d -2 or log 2 x \u003d 3 - x.

The root of the first equation is x 1 = 1/4.

The root of the equation log 2 x \u003d 3 - x will be found by selection. This number is 2. This root is unique, since the function y \u003d log 2 x is increasing over the entire domain of definition, and the function y \u003d 3 - x is decreasing.

By checking it is easy to make sure that both numbers are the roots of the equation

Answer: 1/4; 2.

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Algebra Grade 11

Topic: "Methods for solving logarithmic equations"

Lesson Objectives:

educational: the formation of knowledge about different ways solving logarithmic equations, the ability to apply them in each specific situation and choose any method to solve;

developing: development of skills to observe, compare, apply knowledge in a new situation, identify patterns, generalize; formation of skills of mutual control and self-control;

educational: education of a responsible attitude to educational work, careful perception of the material in the lesson, accuracy of record keeping.

Lesson type: a lesson of familiarization with new material.

"The invention of logarithms, by shortening the work of the astronomer, has lengthened his life."
French mathematician and astronomer P.S. Laplace

During the classes

I. Setting the goal of the lesson

The studied definition of the logarithm, the properties of logarithms and the logarithmic function will allow us to solve logarithmic equations. All logarithmic equations, no matter how complex they are, are solved using unified algorithms. We will consider these algorithms today in the lesson. There are few of them. If you master them, then any equation with logarithms will be feasible for each of you.

Write in your notebook the topic of the lesson: "Methods for solving logarithmic equations." I invite everyone to cooperation.

II. Update basic knowledge

Let's get ready to study the topic of the lesson. You solve each task and write down the answer, you can not write the condition. Work in pairs.

1) For what values ​​of x does the function make sense:

(Answers are checked for each slide and errors are sorted out)

2) Do the function graphs match?

3) Rewrite the equalities as logarithmic equalities:

4) Write the numbers as logarithms with base 2:

5) Calculate:

6) Try to restore or complete the missing elements in these equalities.

III. Introduction to new material

The statement is shown on the screen:

"The equation is the golden key that unlocks all mathematical sesame."
Modern Polish mathematician S. Koval

Try to formulate the definition of a logarithmic equation. (An equation containing the unknown under the sign of the logarithm).

Consider the simplest logarithmic equation:logax = b(where a>0, a ≠ 1). As logarithmic function is increasing (or decreasing) on ​​the set positive numbers and takes all real values, then by the root theorem it follows that for any b, this equation has, and moreover, only one solution, and moreover, a positive one.

Remember the definition of a logarithm. (The logarithm of the number x to the base a is the exponent to which the base a must be raised to get the number x). It immediately follows from the definition of the logarithm that ain is such a solution.

Write down the title: Methods for solving logarithmic equations

1. By definition of the logarithm.

This is how simple equations of the form are solved.

Consider No. 514(a): Solve the equation

How do you propose to solve it? (By definition of logarithm)

Decision. , Hence 2x - 4 = 4; x = 4.

In this task, 2x - 4 > 0, since > 0, therefore extraneous roots cannot appear, and there is no need to check. The condition 2x - 4 > 0 is not necessary to write out in this task.

2. Potentiation(transition from logarithm given expression to this expression).

Consider No. 519(g): log5(x2+8)-log5(x+1)=3log5 2

What feature did you notice? (The bases are the same and the logarithms of the two expressions are equal). What can be done? (potentiate).

In this case, it should be taken into account that any solution is contained among all x for which the logarithm expressions are positive.

Solution: ODZ:

X2+8>0 extra inequality

log5(x2+8) = log5 23+ log5(x+1)

log5(x2+8)=log5(8x+8)

Potentiate the original equation

we get the equation x2+8= 8x+8

We solve it: x2-8x=0

Answer: 0; eight

AT general view transition to an equivalent system:

The equation

(The system contains a redundant condition - one of the inequalities can be ignored).

Question to the class: Which of these three solutions did you like the most? (Discussion of methods).

You have the right to decide in any way.

3. Introduction of a new variable.

Consider No. 520(g). .

What did you notice? (This is a quadratic equation for log3x) Any suggestions? (Introduce new variable)

Decision. ODZ: x > 0.

Let , then the equation will take the form:. Discriminant D > 0. Roots by Vieta's theorem:.

Let's return to the replacement: or .

Solving the simplest logarithmic equations, we get:

Answer: 27;

4. Logarithm of both sides of the equation.

Solve the equation:.

Solution: ODZ: x>0, take the logarithm of both sides of the equation in base 10:

Apply the property of the logarithm of the degree:

(lgx + 3) lgx = 4

Let lgx = y, then (y + 3)y = 4

, (D > 0) the roots according to the Vieta theorem: y1 = -4 and y2 = 1.

Let's return to the replacement, we get: lgx = -4,; logx = 1, .

Answer: 0.0001; ten.

5. Reduction to one base.

No. 523(c). Solve the equation:

Solution: ODZ: x>0. Let's move on to base 3.

6. Functional-graphical method.

№ 509(d). Solve graphically the equation: = 3 - x.

How do you propose to solve? (Construct graphs of two functions y \u003d log2x and y \u003d 3 - x by points and look for the abscissa of the intersection points of the graphs).

See your solution on the slide.

Is there a way to avoid plotting . It is as follows : if one of the functions y = f(x) increases and the other y = g(x) decreases on the interval X, then the equation f(x)=g(x) has at most one root on the interval X.

If there is a root, then it can be guessed.

In our case, the function increases for x>0, and the function y \u003d 3 - x decreases for all values ​​of x, including x>0, which means that the equation has no more than one root. Note that for x = 2, the equation turns into a true equality, since .

« Correct application methods can be learned
just applying them to various examples».
Danish historian of mathematics G. G. Zeiten

Iv. Homework

P. 39 consider example 3, solve No. 514 (b), No. 529 (b), No. 520 (b), No. 523 (b)

V. Summing up the lesson

What methods for solving logarithmic equations did we consider in the lesson?

In the next lesson, we'll look at more complex equations. To solve them, the studied methods are useful.

Showing the last slide:

“What is more than anything in the world?
Space.
What is the wisest?
Time.
What is the most enjoyable?
Achieve what you want."
Thales

I want everyone to achieve what they want. Thank you for your cooperation and understanding.