Logarithmic equations examples of solutions. Variable Replacement Method
Let's consider some types of logarithmic equations, which are not so often discussed in mathematics lessons at school, but are widely used in the preparation of competitive tasks, including for the Unified State Exam.
1. Equations solved by the logarithm method
When solving equations containing a variable in both the base and the exponent, the logarithm method is used. If, at the same time, the exponent contains a logarithm, then both sides of the equation must be logarithmed to the base of this logarithm.
Example 1.
Solve the equation: x log 2 x+2 = 8.
Solution.
Let's take the logarithm of the left and right sides of the equation to 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 = 16. t 1 = 1; t 2 = -3.
So log 2 x = 1 and x 1 = 2 or log 2 x = -3 and x 2 =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
Solution.
Domain of the equation
(x 2 – 3x + 4 > 0,
(x + 5 > 0. → x > -5.
log 3 (x + 5) = 0 at 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, we need to consider only the values (0; 9]. This means that the expression on the left side takes highest value 2 for x = 1. Let us now consider the function y = 2 x-1 + 2 1-x. If we take t = 2 x -1, then it will take the form y = t + 1/t, where t > 0. Under such conditions, it has a single critical point t = 1. This is the minimum point. Y vin = 2. And it is achieved at x = 1.
Now it is obvious that the graphs of the functions under consideration can intersect only once at point (1; 2). It turns out that x = 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 = 6 – 2x
Solution.
Let's decide given equation relative to log 2 x. Let log 2 x = t. Then t 2 + (x – 1) t – 6 + 2x = 0.
D = (x – 1) 2 – 4(2x – 6) = (x – 5) 2. t 1 = -2; t 2 = 3 – x.
We get the equation log 2 x = -2 or log 2 x = 3 – x.
The root of the first equation is x 1 = 1/4. 
We will find the root of the equation log 2 x = 3 – x by selection. This is the number 2. This root is unique, since the function y = log 2 x is increasing throughout the entire domain of definition, and the function y = 3 – x is decreasing.
It is easy to check that both numbers are roots of the equation
Answer:1/4; 2.
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Preparation for the final test in mathematics includes an important section - “Logarithms”. Tasks from this topic are necessarily contained in the Unified State Examination. Experience from past years shows that logarithmic equations caused difficulties for many schoolchildren. Therefore, students with different levels of training must understand how to find the correct answer and quickly cope with them.
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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 most simple tasks, which are called so - 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 values - 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 take a look 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
All decimals convert to ordinary ones when you solve 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 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 a notation 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, i.e. it becomes 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.
That's it, let's get rid of the sign 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 bring important note 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 logarithmic function, as well as 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 degree we raise positive number, 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.
We recall 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:
- 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);
- 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. It's about 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, we could just as well cross out linear inequality, that is, cross out x − 2 > 0 and demand that 2x 2 − 13x + 18 > 0. But you must agree that solving the simplest linear inequality is much faster and easier than quadratic, even if as a result of solving the entire this system we will 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 it down separately quadratic equation and let's 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:
- 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;
- As soon as the logarithm appears variable base, the task 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.
Logarithmic equation is an equation in which the unknown (x) and expressions with it are under the sign of the logarithmic function. Solving logarithmic equations assumes that you are already familiar with and .
How to solve logarithmic equations?
The simplest equation is log a x = b, where a and b are some numbers, x is an unknown.
Solving a logarithmic equation is x = a b provided: a > 0, a 1.
It should be noted that if x is somewhere outside the logarithm, for example log 2 x = x-2, then such an equation is already called mixed and a special approach is needed to solve it.
The ideal case is when you come across an equation in which only numbers are under the logarithm sign, for example x+2 = log 2 2. Here it is enough to know the properties of logarithms to solve it. But such luck does not happen often, so get ready for more difficult things.
But first, let's start with simple equations. To solve them, it is advisable to have a very general understanding of the logarithm.
Solving simple logarithmic equations
These include equations of the type log 2 x = log 2 16. The naked eye can see that by omitting the sign of the logarithm we get x = 16.
To solve a more complex logarithmic equation, it is usually reduced to solving the usual algebraic equation or to the solution of the simplest logarithmic equation log a x = b. In the simplest equations this happens in one movement, which is why they are called simplest.
The above method of dropping logarithms is one of the main ways to solve logarithmic equations and inequalities. In mathematics, this operation is called potentiation. Exist certain rules or restrictions for this kind of operations:
- logarithms have the same numerical bases
- The logarithms in both sides of the equation are free, i.e. without any coefficients and other various kinds expressions.
Let's say in the equation log 2 x = 2log 2 (1 - x) potentiation is not applicable - the coefficient 2 on the right does not allow it. IN following example log 2 x+log 2 (1 - x) = log 2 (1+x) one of the restrictions is also not met - there are two logarithms on the left. If there was only one, it would be a completely different matter!
In general, you can remove logarithms only if the equation has the form:
log a (...) = log a (...)
Absolutely any expressions can be placed in brackets; this has absolutely no effect on the potentiation operation. And after eliminating logarithms, a simpler equation will remain - linear, quadratic, exponential, etc., which, I hope, you already know how to solve.
Let's take another example:
log 3 (2x-5) = log 3 x
We apply potentiation, we get:
log 3 (2x-1) = 2
Based on the definition of a logarithm, namely, that a logarithm is the number to which the base must be raised in order to obtain an expression that is under the logarithm sign, i.e. (4x-1), we get:
Again we received a beautiful answer. Here we did without eliminating logarithms, but potentiation is also applicable here, because a logarithm can be made from any number, and exactly the one we need. This method is very helpful in solving logarithmic equations and especially inequalities.
Let's solve our logarithmic equation log 3 (2x-1) = 2 using potentiation:
Let's imagine the number 2 as a logarithm, for example, this log 3 9, because 3 2 =9.
Then log 3 (2x-1) = log 3 9 and again we get the same equation 2x-1 = 9. I hope everything is clear.
So we looked at how to solve the simplest logarithmic equations, which are actually very important, because solving logarithmic equations, even the most terrible and twisted ones, in the end always comes down to solving the simplest equations.
In everything we did above, we missed one very important point, which will subsequently have decisive role. The fact is that the solution to any logarithmic equation, even the most elementary one, consists of two equal parts. The first is the solution of the equation itself, the second is working with the range of permissible values (APV). This is exactly the first part that we have mastered. In the above examples, ODZ does not affect the answer in any way, so we did not consider it.
Let's take another example:
log 3 (x 2 -3) = log 3 (2x)
Outwardly, this equation is no different from an elementary one, which can be solved very successfully. But it is not so. No, of course we will solve it, but most likely incorrectly, because it contains a small ambush, into which both C-grade students and excellent students immediately fall into it. Let's take a closer look.
Let's say you need to find the root of the equation or the sum of the roots, if there are several of them:
log 3 (x 2 -3) = log 3 (2x)
We use potentiation, it is acceptable here. As a result, we obtain an ordinary quadratic equation.
Finding the roots of the equation:
It turned out two roots.
Answer: 3 and -1
At first glance everything is correct. But let's check the result and substitute it into the original equation.
Let's start with x 1 = 3:
log 3 6 = log 3 6
The check was successful, now the queue is x 2 = -1:
log 3 (-2) = log 3 (-2)
Okay, stop! On the outside everything is perfect. One thing - there are no logarithms from negative numbers! This means that the root x = -1 is not suitable for solving our equation. And therefore the correct answer will be 3, not 2, as we wrote.
This is where ODZ played its fatal role, which we had forgotten about.
Let me remind you that the range of acceptable values includes those values of x that are allowed or make sense for the original example.
Without ODZ, any solution, even an absolutely correct one, of any equation turns into a lottery - 50/50.
How could we get caught solving a seemingly elementary example? But precisely at the moment of potentiation. Logarithms disappeared, and with them all restrictions.
What to do in this case? Refuse to eliminate logarithms? And completely refuse to solve this equation?
No, we just, like real heroes from one famous song, will take a detour!
Before we begin solving any logarithmic equation, we will write down the ODZ. But after that, you can do whatever your heart desires with our equation. Having received the answer, we simply throw out those roots that are not included in our ODZ and write down the final version.
Now let’s decide how to record ODZ. To do this, we carefully examine the original equation and look for suspicious places in it, such as division by x, even root, etc. Until we have solved the equation, we do not know what x is equal to, but we know for sure that there are x that, when substituted, will give division by 0 or extraction square root from negative number, are obviously not suitable as an answer. Therefore, such x are unacceptable, while the rest will constitute ODZ.
Let's use the same equation again:
log 3 (x 2 -3) = log 3 (2x)
log 3 (x 2 -3) = log 3 (2x)
As you can see, there is no division by 0, there are also no square roots, but there are expressions with x in the body of the logarithm. Let us immediately remember that the expression inside the logarithm must always be >0. We write this condition in the form of ODZ:
Those. We haven't decided anything yet, but we've already written it down required condition for the entire sublogarithmic expression. The curly brace means that these conditions must be true simultaneously.
The ODZ is written down, but it is also necessary to solve the resulting system of inequalities, which is what we will do. We get the answer x > v3. Now we know for sure which x will not suit us. And then we begin to solve the logarithmic equation itself, which is what we did above.
Having received the answers x 1 = 3 and x 2 = -1, it is easy to see that only x1 = 3 suits us, and we write it down as the final answer.
For the future, it is very important to remember the following: we solve any logarithmic equation in 2 stages. The first is to solve the equation itself, the second is to solve the ODZ condition. Both stages are performed independently of each other and are compared only when writing the answer, i.e. discard everything unnecessary and write down the correct answer.
To reinforce the material, we strongly recommend watching the video:
The video shows other examples of solving log. equations and working out the interval method in practice.
To this question, how to solve logarithmic equations That's all for now. If something is decided by the log. equations remain unclear or incomprehensible, write your questions in the comments.
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