View logarithmic equations examples of solutions online. Homogeneous logarithmic equations
Logarithmic equation is an equation in which the unknown (x) and expressions with it are under the sign logarithmic function. Solving logarithmic equations assumes that you are already familiar with and .
How to decide 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.
Note: The Academy of Social Education (ASE) is ready to accept new students.
We are all familiar with equations primary classes. There we also learned to solve the simplest examples, and we must admit that they find their application even in higher mathematics. Everything is simple with equations, including quadratic equations. If you are having trouble with this topic, we highly recommend that you review it.
You've probably already gone through logarithms too. However, we consider it important to tell what it is for those who do not yet know. A logarithm is equated to the power to which the base must be raised to obtain the number to the right of the logarithm sign. Let's give an example based on which everything will become clear to you.
If you raise 3 to the fourth power, you get 81. Now substitute the numbers by analogy, and you will finally understand how logarithms are solved. Now all that remains is to combine the two concepts discussed. Initially, the situation seems extremely complicated, but upon closer examination the weight falls into place. We are sure that after this short article you will not have problems in this part of the Unified State Exam.
Today there are many ways to solve such structures. We will tell you about the simplest, most effective and most applicable in the case of Unified State Examination tasks. Solving logarithmic equations must start from the very beginning. simple example. The simplest logarithmic equations consist of a function and one variable in it.
It's important to note that x is inside the argument. A and b must be numbers. In this case, you can simply express the function in terms of a number to a power. It looks like this.
Of course, solving a logarithmic equation using this method will lead you to the correct answer. The problem for the vast majority of students in this case is that they do not understand what comes from where. As a result, you have to put up with mistakes and not get the desired points. The most offensive mistake will be if you mix up the letters. To solve the equation this way, you need to memorize this standard school formula because it is difficult to understand.
To make it easier, you can resort to another method - the canonical form. The idea is extremely simple. Turn your attention back to the problem. Remember that the letter a is a number, not a function or variable. A is not equal to one and greater than zero. There are no restrictions on b. Now, of all the formulas, let us remember one. B can be expressed as follows.
It follows from this that all original equations with logarithms can be represented in the form:
Now we can drop the logarithms. The result is a simple design, which we have already seen earlier.
The convenience of this formula is that it can be used in the most different cases, and not just for the simplest designs.
Don't worry about OOF!
Many experienced mathematicians will notice that we have not paid attention to the domain of definition. The rule boils down to the fact that F(x) is necessarily greater than 0. No, we did not miss this point. Now we are talking about another serious advantage of the canonical form.
There will be no extra roots here. If a variable will only appear in one place, then a scope is not necessary. It is done automatically. To verify this judgment, try solving several simple examples.
How to solve logarithmic equations with different bases
These are already complex logarithmic equations, and the approach to solving them must be special. Here it is rarely possible to limit ourselves to the notorious canonical form. Let's begin our detailed story. We have the following construction.
Pay attention to the fraction. It contains the logarithm. If you see this in a task, it’s worth remembering one interesting trick.
What does it mean? Each logarithm can be represented as the quotient of two logarithms with a convenient base. And this formula has a special case that is applicable with this example (we mean if c=b).
This is exactly the fraction we see in our example. Thus.
Essentially, we turned the fraction around and got a more convenient expression. Remember this algorithm!
Now we need that the logarithmic equation did not contain different reasons. Let's represent the base as a fraction.
In mathematics there is a rule based on which you can derive a degree from a base. The following construction results.
It would seem that what is preventing us from now turning our expression into the canonical form and simply solving it? Not so simple. There should be no fractions before the logarithm. Let's fix this situation! Fractions are allowed to be used as degrees.
Respectively.
If the bases are the same, we can remove the logarithms and equate the expressions themselves. This way the situation will become much simpler than it was. What will remain is an elementary equation that each of us knew how to solve back in 8th or even 7th grade. You can do the calculations yourself.
We have obtained the only true root of this logarithmic equation. Examples of solving a logarithmic equation are quite simple, aren't they? Now you will be able to independently deal with even the most complex tasks for preparing for and passing the Unified State Exam.
What's the result?
In the case of any logarithmic equations, we start from one very important rule. It is necessary to act in such a way as to bring the expression to the maximum simple view. In this case, you will have a better chance of not only solving the task correctly, but also doing it in the simplest and most logical way possible. This is exactly how mathematicians always work.
We strongly advise you not to search difficult paths, especially in this case. Remember a few simple rules, which will allow you to transform any expression. For example, reduce two or three logarithms to the same base or derive a power from the base and win on this.
It is also worth remembering that solving logarithmic equations requires constant practice. Gradually you will move on to more and more complex structures, and this will lead you to confidently solving all variants of problems on the Unified State Exam. Prepare well in advance for your exams, and good luck!
Logarithmic equations. From simple to complex.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
What is a logarithmic equation?
This is an equation with logarithms. I’m surprised, right?) Then I’ll clarify. This is an equation in which the unknowns (x's) and expressions with them are found inside logarithms. And only there! It is important.
Here are some examples logarithmic equations:
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 = 11lg(x+1)
Well, you understand... )
Note! The most diverse expressions with X's are located exclusively within logarithms. If, suddenly, an X appears somewhere in the equation outside, For example:
log 2 x = 3+x,
this will be an equation mixed type. Such equations do not have clear rules for solving them. We will not consider them for now. By the way, there are equations where inside the logarithms only numbers. For example:
What can I say? You're lucky if you come across this! Logarithm with numbers is some number. That's all. It is enough to know the properties of logarithms to solve such an equation. Knowledge of special rules, techniques adapted specifically for solving logarithmic equations, not required here.
So, what is a logarithmic equation- we figured it out.
How to solve logarithmic equations?
Solution logarithmic equations- the thing is actually not very simple. So our section is a four... A decent amount of knowledge on all sorts of related topics is required. In addition, there is a special feature in these equations. And this feature is so important that it can safely be called the main problem in solving logarithmic equations. We will deal with this problem in detail in the next lesson.
For now, don't worry. We'll go the right way from simple to complex. On specific examples. The main thing is to delve into simple things and don’t be lazy to follow the links, I put them there for a reason... And everything will work out for you. Necessarily.
Let's start with the most elementary, simplest equations. To solve them, it is advisable to have an idea of the logarithm, but nothing more. Just no idea logarithm, take on a decision logarithmic equations - somehow even awkward... Very bold, I would say).
The simplest logarithmic equations.
These are equations of the form:
1. log 3 x = log 3 9
2. log 7 (2x-3) = log 7 x
3. log 7 (50x-1) = 2
Solution process any logarithmic equation consists in the transition from an equation with logarithms to an equation without them. In the simplest equations this transition is carried out in one step. That's why they are the simplest.)
And such logarithmic equations are surprisingly easy to solve. See for yourself.
Let's solve the first example:
log 3 x = log 3 9
To solve this example, you don’t need to know almost anything, yes... Purely intuition!) What do we need especially don't like this example? What-what... I don't like logarithms! Right. So let's get rid of them. We look closely at the example, and a natural desire arises in us... Downright irresistible! Take and throw out logarithms altogether. And what’s good is that Can do! Mathematics allows. Logarithms disappear the answer is:
Great, right? This can (and should) always be done. Eliminating logarithms in this manner is one of the main ways to solve logarithmic equations and inequalities. In mathematics this operation is called potentiation. Of course, there are rules for such liquidation, but they are few. Remember:
You can eliminate logarithms without any fear if they have:
a) the same numerical bases
c) logarithms from left to right are pure (without any coefficients) and are in splendid isolation.
Let me clarify the last point. In the equation, let's say
log 3 x = 2log 3 (3x-1)
Logarithms cannot be removed. The two on the right doesn't allow it. The coefficient, you know... In the example
log 3 x+log 3 (x+1) = log 3 (3+x)
It is also impossible to potentiate the equation. There is no lone logarithm on the left side. There are two of them.
In short, you can remove logarithms if the equation looks like this and only like this:
log a (.....) = log a (.....)
In parentheses, where there is an ellipsis, there may be any expressions. Simple, super complex, all kinds. Whatever. The important thing is that after eliminating logarithms we are left with simpler equation. It is assumed, of course, that you already know how to solve linear, quadratic, fractional, exponential and other equations without logarithms.)
Now you can easily solve the second example:
log 7 (2x-3) = log 7 x
Actually, it’s decided in the mind. We potentiate, we get:
Well, is it very difficult?) As you can see, logarithmic part of the solution to the equation is only in eliminating logarithms... And then comes the solution to the remaining equation without them. A trivial matter.
Let's solve the third example:
log 7 (50x-1) = 2
We see that there is a logarithm on the left:
Let us remember that this logarithm is a number to which the base must be raised (i.e. seven) to obtain a sublogarithmic expression, i.e. (50x-1).
But this number is two! According to Eq. That is:
That's basically all. Logarithm disappeared, What remains is a harmless equation:
We solved this logarithmic equation based only on the meaning of the logarithm. Is it still easier to eliminate logarithms?) I agree. By the way, if you make a logarithm from two, you can solve this example through elimination. Any number can be made into a logarithm. Moreover, the way we need it. A very useful technique in solving logarithmic equations and (especially!) inequalities.
Don't know how to make a logarithm from a number!? It's OK. Section 555 describes this technique in detail. You can master it and use it to the fullest! It greatly reduces the number of errors.
The fourth equation is solved in a completely similar way (by definition):
That's it.
Let's summarize this lesson. We looked at the solution of the simplest logarithmic equations using examples. It is very important. And not only because such equations appear in tests and exams. The fact is that even the most evil and complicated equations are necessarily reduced to the simplest!
Actually, the simplest equations are the final part of the solution any equations. And this final part must be understood strictly! And further. Be sure to read this page to the end. There's a surprise there...)
Now we decide for ourselves. Let's get better, so to speak...)
Find the root (or sum of roots, if there are several) of the equations:
ln(7x+2) = ln(5x+20)
log 2 (x 2 +32) = log 2 (12x)
log 16 (0.5x-1.5) = 0.25
log 0.2 (3x-1) = -3
ln(e 2 +2x-3) = 2
log 2 (14x) = log 2 7 + 2
Answers (in disarray, of course): 42; 12; 9; 25; 7; 1.5; 2; 16.
What, not everything works out? Happens. Don't worry! Section 555 explains the solution to all of these examples in a clear and detailed manner. You'll definitely figure it out there. You will also learn useful practical techniques.
Everything worked out!? All examples of “one left”?) Congratulations!
It's time to reveal the bitter truth to you. Successful solving of these examples does not guarantee success in solving all other logarithmic equations. Even the simplest ones like these. Alas.
The fact is that the solution to any logarithmic equation (even the most elementary!) consists of two equal parts. Solving the equation and working with ODZ. We have mastered one part - solving the equation itself. It's not that hard right?
For this lesson, I specially selected examples in which DL does not affect the answer in any way. But not everyone is as kind as me, right?...)
Therefore, it is imperative to master the other part. ODZ. This is the main problem in solving logarithmic equations. And not because it’s difficult - this part is even easier than the first. But because people simply forget about ODZ. Or they don't know. Or both). And they fall out of the blue...
In the next lesson we will deal with this problem. Then you can confidently decide any simple logarithmic equations and approach quite solid tasks.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
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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