Solving decimal logarithmic equations. Methods for solving logarithmic equations. Problems with variable base

Solving logarithmic equations. Part 1.

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

The simplest logarithmic equation has the form:

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

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

depending on which inequality or simpler.

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

then we go to the system:

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

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

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

All logarithmic equations can be divided into four types:

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

Example. Let's solve the equation:

Let's equate the expressions under the logarithm sign:

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

Yes, it satisfies.

Answer: x=5

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

Example. Let's solve the equation:

Let's find the ODZ equation:

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

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

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

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

Likewise,

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

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

We are all familiar with equations from primary school. 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 should start with the simplest 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 lies in the fact that it can be used in a wide variety of 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 it is necessary that the logarithmic equation does not contain different bases. 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 proceed from one very important rule. It is necessary to act in such a way as to reduce the expression to the simplest possible form. 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 do not recommend that you look for difficult paths, especially in this case. Remember a few simple rules that 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!

1. The solution is standard - let’s use multiplication rule by 1:

Now we remove logarithms:

Let's multiply crosswise:

Examination

Fits!

Examination

And it fits here! Maybe I was wrong, and roots are always suitable? Let's look at the next example!

Example No. 2

Let's represent the triple using our favorite method in the form

On the left and right we will use the formula for the sum of logarithms.

Example No. 3

The solution is similar to the previously discussed example: Let’s turn the unit on the right into (let me remind you that - a decimal logarithm, or a logarithm to the base), and perform operations between the logarithms on the left and right:

Now let’s remove the logarithms on the left and right:

\left((x) -2 \right)\left((x) -3 \right)=2

Examination:

Again, both logarithms on the left are undefined, since they are taken from negative numbers. Then it is not a root.

since then

Answer:

I hope that the examples just given will forever wean you from skipping checks when solving logarithmic equations. It is necessary!

Logarithmic equation with variable base

Now I would like to look at another (slightly more complex) type of logarithmic equations with you. These will be equations with variable base.

Before this, we only considered cases where the bases were constant: etc. But nothing prevents them from being some functions of, for example, etc.

But don't be scared! If, when solving logarithmic inequalities, a variable base causes quite a lot of inconvenience, then This has virtually no effect on the complexity of solving the equation! Judge for yourself:

Example No. 1

We proceed as before: apply the “multiply by one” method to the number:

Then the original equation is transformed to the form:

I will apply square difference formula:

Examination:

What conclusion do we draw? Wrong! The number is not the root of the equation because the base of the logarithm cannot be a negative number or equal to one!

Answer: .

As you can see, in the case of equations there is no fundamental difference whether our bases are variable or not. In this regard, we can say that decide logarithmic equation usually much easier than solving a logarithmic inequality!

Let's now try to solve another “strange” example.

Example No. 2

We will act as always - we will turn the right side into a logarithm, like this tricky one:

Then the original logarithmic equation will be equivalent to this equation (though logarithmic again)

I will solve this equation again using the difference of squares:

Let's solve the first one first, the second one will be solved in approximately the same way:

Will use again "multiplying by 1":

Similarly for the second equation:

Now comes the fun part: verification. Let's start with the first root

The base of the "large" logarithm is equal to

Therefore it is not a root.

Let's check the second number:

that number is the root of the original equation.

Answer:

I deliberately gave a rather complex example to show you that you should not be afraid of large and scary logarithms.

It is enough to know a few formulas (which I have already given you above) and you can find a way out of any (almost) situation!

Well, I’ve given you the basic methods for solving logarithmic equations (“no frills” methods), which will allow you to cope with most examples (primarily on the Unified State Exam).

Now it's your time to show what you've learned. Try to solve the following yourself logarithmic equations, and then we will compare the results with you.

Seven examples for independent work

The techniques discussed in this work, of course, do not exhaust all possible ways to solve logarithmic equations.

In some cases, we need to get really creative to figure out a way to find the roots of a tricky equation.

However, no matter how complex the initial equation is, as a result it will be reduced to an equation of the type that you and I have just learned to solve!

Answers to examples for independent work

1. A fairly simple task: let’s use the property:

in subtrahend:

Then we get:

Let's check:

(I already explained this transition to you above)

Answer: 9

2. Also nothing supernatural: I don’t want to divide, so I’ll move the term with the “minus” to the right: now I have decimal logarithms on the left and right, and I get rid of them:

I'm checking:

the expression under the logarithm sign cannot be negative, so the number is not the root of the equation.

Examination

Answer:

Here we need to do a little work: it is clear that I will again use the (isn’t it very useful?) formula:

What do I need to do before applying the logarithm addition formula? Yes, I need to get rid of the multiplier. There are two ways: the first is to enter it directly into a logarithm using the formula:

In principle, this method has a right to exist, but what’s bad about it? It’s bad to deal with an expression of the form (a “non-integer degree” is always unpleasant. So what else can we do? How can we get rid of such “non-integer degree”? Let’s multiply by our equation:

Well, now let’s put both factors into logarithms:

then I'll replace the zero with

And finally I get:

Do you remember what this “unloved” school formula is called? This cube difference! Maybe this is more clear?

Let me remind you that the difference of cubes is factorized like this:

and here's another one just in case:

In relation to our situation, this will give:

The first equation has a root, but the second has no roots (see for yourself!).

I'll leave it to you to check for yourself and make sure that the number is actually the root of our equation.

As in the previous example, we rewrite

Again, I don’t want any subtractions (and subsequent divisions) and therefore I will move the resulting expression to the right:

Now I remove the logarithms on the left and right:

We got an irrational equation, which I hope you already know how to solve. Let me just remind you that we square both sides:

Your task now is to make sure that it is not a root, but is.

Answer:

Everything is transparent: we apply the formula for the sum of logarithms on the left:

then we remove the logarithms on both sides:

Examination:

Answer: ;

Everything couldn’t be simpler: the equation has already been reduced to its simplest form. All we have to do is equalize

Let's check:

But when the base of logarithms is equal to:

And it is not a root.

Answer:

I left this example for dessert. Although there is nothing very complicated about it either.

Let's imagine zero as

Then you and I will get this logarithmic equation:

And we remove the first “skin” - external logarithms.

Let's represent the unit as

Then our equation will take the form:

Now we remove the “second skin” and get to the core:

Let's check:

Answer: .

3 METHODS FOR SOLVING LOGARITHMIC EQUATIONS. ADVANCED LEVEL

Now, after reading the first article on logarithmic equations, you have mastered the necessary minimum of knowledge necessary to solve the simplest examples.

Now I can move on to parsing some more three methods solving logarithmic equations:

method of introducing a new variable (or replacement)
logarithm method
method of transition to a new foundation.

First method- one of the most frequently used in practice. It solves most “difficult” problems related to solving logarithmic (and not only) equations.

Second method serves to solve mixed exponential-logarithmic equations, ultimately reducing the problem to choosing a good replacement variable (that is, to the first method).

Third method suitable for solving some equations in which logarithms with different bases occur.

I'll start by looking at the first method.

Method for introducing a new variable (4 examples)

As you already understood from the name, the essence of this method is to introduce such a change of variable that your logarithmic equation will miraculously transform into one that you can easily solve.

All that remains for you after solving this very “simplified equation” is to do "reverse replacement": that is, to return from the replaced to the replaced.

Let's illustrate what we just said with a very simple example:

In this example, the replacement suggests itself! After all, it is clear that if we replace with, then our logarithmic equation will turn into a rational one:

You can easily solve it by reducing it to a square:

(so that the denominator does not accidentally reset to zero!)

Simplifying the resulting expression, we finally get:

Now we make the reverse substitution: , then it follows from that, and from we get

Now, as before, it’s time to check:

Let it be in the beginning, because then it’s true!

Now, then, everything is correct!

Thus, the numbers are the roots of our original equation.

Answer: .

Here's another example with an obvious replacement:

In fact, let's replace it right away

then our original logarithmic equation will turn into a quadratic one:

Reverse replacement:

Check it yourself, make sure that in this case both numbers we found are roots.

I think you got the main idea. It is not new and applies not only to logarithmic equations.

Another thing is that sometimes it is quite difficult to immediately “see” the replacement. This requires some experience, which will come to you after some effort on your part.

In the meantime, practice solving the following examples:

Ready? Let's check what you got:

Let's solve the second example first.

He just demonstrates to you that it is not always possible to make a replacement, as they say, “head-on”.

First, we need to transform our equation a little: apply the formula for the difference of logarithms in the numerator of the first fraction, and take the power in the numerator of the second.

By doing this, you will receive:

Now the replacement has become obvious, hasn't it? Let's make it: .

Now let's bring the fractions to a common denominator and simplify.

Then we get:

Having solved the last equation, you will find its roots: where.

Do the check yourself and make sure that these are indeed the roots of our original equation.

Now let's try to solve the third equation.

Well, first of all, it's clear that it won't hurt us to multiply both sides of the equation by. There is no harm, but the benefits are obvious.

Now let's make a replacement. You guessed what we will be replacing, right? That's right, let's say . Then our equation will take the following form:

(both roots suit us!)

Now the reverse replacement: , from, from. Our original equation has as many as four roots! Make sure of this, let's substitute the obtained values into the equation. We write down the answer:

Answer: .

I think that now the idea of replacing a variable is completely clear to you? Okay, then let's not stop there and move on to another method for solving logarithmic equations: method of transition to a new foundation.

Method of transition to a new base

Let's consider the following equation:

What do we see? The two logarithms are supposedly “opposite” to each other. What do we have to do? Everything is easy: we just need to resort to one of two formulas:

In principle, nothing prevents me from using either of these two formulas, but due to the structure of the equation, it will be more convenient for me to use the first: I will get rid of the variable base of the logarithm in the second term by replacing it with. Now it’s easy to see that the task has been reduced to the previous one: choosing a replacement. Substituting, I get the following equation:

From here. All you have to do is substitute the found numbers into the original equation and make sure that they are, in fact, roots.

Here's another example where it makes sense to move to a new foundation:

However, as you can easily check, if you and I move to a new foundation right away, this will not give the desired effect. What do we need to do in this case? Let's simplify everything as much as possible, and then whatever happens.
So what I want to do is imagine how to, how to, take these powers out in front of the logarithms, and also take out the square of X in the first logarithm. We'll see later.

Remember, it can be much more difficult to make friends with the base than with the expression under the sign of the logarithm!

Following this rule, I will replace with and with. Then I will get:

Well, the next steps are already familiar to you. Replace and look for roots!

As a result, you will find two roots of the original equation:

It's time to show you what you've learned!

First try to solve the following (not the easiest) examples on your own:

1. Everything here is quite standard: I will try to reduce my original equation to such that the replacement is convenient. What do I need for this? First, transform the first expression on the left (remove the fourth power of two before the logarithm) and remove the power of two from the base of the second logarithm. Then I will get:

All that’s left is to “turn over” the first logarithm!

\frac(12)(\log_(2)(x))=3((\log )_(2))x

(for convenience, I moved the second logarithm from the left to the right side of the equation)

The problem is almost solved: you can make a replacement. After reduction to a common denominator, I get the following equation:

Having made the reverse substitution, it will not be difficult for you to calculate that:

Make sure that the values obtained are the roots of our equation.

2. Here I will also try to “fit” my equation to an acceptable replacement. Which one? Perhaps it will suit me.

So let's not waste time and start transforming!

((\log )_(x))5((x)^(2))\cdot \log \frac(2)(5)x=1

Well, now you can safely replace it! Then, with respect to the new variable, we get the following equation:

Where. Again, making sure that both of these numbers are actually roots is left to you as an exercise.

3. Here it’s not even immediately obvious what we will replace. There is one golden rule - If you don’t know what to do, do what you can! That's what I'll use!

Now I will “turn over” all the logarithms and apply the difference logarithm formula to the first one, and the sum logarithm to the last two:

Here I also used the fact that (at) and the property of taking a power out of a logarithm. Well, now we can apply a suitable replacement: . I'm sure that you already know how to solve rational equations, even this monstrous type. Therefore, I will allow myself to immediately write down the result:

It remains to solve two equations: . You have already familiarized yourself with the methods for solving such “almost simplest” equations in the previous section. So I'll write down the final solutions right away:

Make sure that only two of these numbers are the roots of my equation! Namely, it is and, while it is not a root!

This example is a bit trickier, however, I will try to solve it without resorting to variable replacement at all! Let's do it again, let's do what we can: first, we can expand the logarithm on the left according to the formula for the logarithm of a ratio, and also put the two in front of the logarithm in parentheses. In the end I will get:

Well, now the same formula that we have already used! So, let's shorten the right side! Now there’s just a deuce there! Let's move one to it from the left, and we finally get:

You already know how to solve such equations. The root is found without difficulty, and it is equal. I remind you to check!

Well, now, as I hope, you have learned to solve quite complex problems that you cannot overcome “head-on”! But logarithmic equations can be even more insidious! Here are some examples:

Here, alas, the previous solution will not give tangible results. How do you think why? Yes, there is no longer any “reciprocity” of logarithms here. This most general case, of course, can also be solved, but we already use the following formula:

This formula doesn’t care whether you have the “opposite” or not. You may ask, why choose a base? My answer is that it doesn't matter. The answer ultimately will not depend on this. Traditionally, either the natural or decimal logarithm is used. Although this is not important. For example, I will use decimal:

To leave an answer in this form is a complete disgrace! Let me first write down by definition that

Now it’s time to use: inside the brackets - the main logarithmic identity, and outside (to the degree) - turn the ratio into one logarithm: then we finally get this “strange” answer: .

Further simplifications, alas, are no longer available to us.

Let's check together:

Right! By the way, remember once again what the penultimate equality in the chain follows from!

In principle, the solution to this example can also be reduced to the transition to a logarithm based on a new base, but you should already be scared of what will happen in the end. Let's try to do something more reasonable: transform the left side as best as possible.

By the way, how do you think I got the last decomposition? That's right, I applied the theorem about factoring a quadratic trinomial, namely:

If, are the roots of the equation, then:

Well, now I will rewrite my original equation in this form:

But we are quite capable of solving such a problem!

So, let's introduce a replacement.

Then my initial equation will take this simple form:

Its roots are equal to: , then

Where does this equation come from? has no roots.

All you have to do is check!

Try to solve the following equation yourself. Take your time and be careful, then luck will be on your side!

Ready? Let's see what we got.

In fact, the example is solved in two steps:

1. Transform

2. now on the right I have an expression that is equal to

Thus, the original equation was reduced to the simplest:

The test shows that this number is indeed the root of the equation.

Logarithm method

And finally, I will very briefly discuss methods for solving some mixed equations. Of course, I do not undertake to cover all mixed equations, but will show methods for solving the simplest ones.

For example,

Such an equation can be solved using the logarithm method. All you have to do is take the logarithm of both sides.

It is clear that since we already have a logarithm to the base, I will take the logarithm to the same base:

Now I'll take the power out of the expression on the left:

and factorize the expression using the difference of squares formula:

Checking, as always, is on your conscience.

Try to solve the last example in this article yourself!

Let’s check: take the logarithm to the base of both sides of the equation:

I take out the degree on the left and split it using the sum formula on the right:

We guess one of the roots: it is a root.

In the article on solving exponential equations, I talked about how to divide one polynomial by a “corner” by another.

Here we need to divide by.

As a result we get:

If possible, carry out the check yourself (although in this case, especially with the last two roots, it will not be easy).

LOGARITHMIC EQUATIONS. SUPER LEVEL

In addition to the material already presented, I suggest that you and I consider another way to solve mixed equations containing logarithms, but here I will consider equations that cannot be solved by the previously discussed method of taking logarithms of both sides. This method is called mini-max.

Mini-max method

This method is applicable not only to solving mixed equations, but also turns out to be useful when solving some inequalities.

So, first we introduce the following basic definitions that are necessary to apply the mini-max method.

Simple pictures illustrate these definitions:

The function in the figure on the left is monotonically increasing, and on the right is monotonically decreasing. Now let's turn to the logarithmic function, it is known that the following is true:

The figure shows examples of a monotonically increasing and monotonically decreasing logarithmic function.

Let's describe it directly mini-max method. I think you understand what words this name comes from?

That's right, from the words minimum and maximum. Briefly, the method can be represented as:

Our most important goal is to find this very constant in order to further reduce the equation to two simpler ones.

For this purpose, the monotonicity properties of the logarithmic function formulated above can be useful.

Now let's look at specific examples:

1. Let's look at the left side first.

There is a logarithm with a base less. According to the theorem formulated above, what is the function? It's decreasing. At the same time, which means . On the other hand, by definition of a root: . Thus, the constant is found and equal. Then the original equation is equivalent to the system:

The first equation has roots, and the second: . Thus, the common root is equal, and this root will be the root of the original equation. Just in case, do a check to make sure.

Answer:

Let's immediately think about what is written here?

I mean the general structure. It says here that the sum of two squares is zero.

When it's possible?

Only when both of these numbers are individually equal to zero. Then let's move on to the following system:

The first and second equations do not have common roots, then the original equation has no roots.

Answer: no solutions.

Let's look at the right side first - it's simpler. By definition of sine:

From where, and then Therefore

Now let's return to the left side: consider the expression under the logarithm sign:

Trying to find the roots of an equation will not lead to a positive result. But nevertheless, I need to somehow evaluate this expression. You, of course, know a method like selecting a complete square. I will use it here.

Since is an increasing function, it follows that. Thus,

Then our original equation is equivalent to the following system:

I don’t know if you’re familiar or not with solving trigonometric equations, so I’ll do this: I’ll solve the first equation (it has a maximum of two roots), and then I’ll substitute the result into the second:

(you can check and make sure that this number is the root of the first equation of the system)

Now I'll substitute it into the second equation:

Answer:

Well, now the technique of using the mini-max method has become clear to you? Then try to solve the following example yourself.

Ready? Let's check:

The left side is the sum of two non-negative quantities (unity and modulus) and therefore the left side is not less than one, and it is equal to one only when

At the same time, the right side is the modulus (meaning greater than zero) of the product of two cosines (meaning no more than one), then:

Then the original equation is equivalent to the system:

I again propose to solve the first equation and substitute the result into the second:

This equation has no roots.

Then the original equation also has no roots.

Answer: there are no solutions.

BRIEFLY ABOUT THE MAIN THINGS. 6 METHODS FOR SOLVING LOGARITHMIC EQUATIONS

Logarithmic equation- an equation in which the unknown variables are inside logarithms.

The simplest logarithmic equation is an equation of the form.

The process of solving any logarithmic equation comes down to reducing the logarithmic equation to the form , and moving from an equation with logarithms to an equation without them: .

ODZ for a logarithmic equation:

Basic methods for solving logarithmic equations:

1 method. Using the definition of logarithm:

Method 2. Using the properties of the logarithm:

Method 3. Introduction of a new variable (replacement):

the substitution allows us to reduce the logarithmic equation to a simpler algebraic equation for t.

Method 4 Transition to a new base:

5 method. Logarithm:

take the logarithm of the right and left sides of the equation.

6 method. Mini-max:

Now we want to hear from you...

We tried to write as simply and thoroughly as possible about logarithmic equations.

Now it's your turn!

Write how you rate our article? Did you like her?

Maybe you already know how to solve logarithmic equations?

Perhaps you have questions. Or suggestions.

Write about it in the comments.

And good luck on your exams!

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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As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.

Definition in mathematics

A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number (that is, any positive) “b” to its base “a” is considered to be the power “c” to which the base “a” must be raised in order to ultimately get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.

Types of logarithms

For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three separate types of logarithmic expressions:

Natural logarithm ln a, where the base is the Euler number (e = 2.7).
Decimal a, where the base is 10.
Logarithm of any number b to base a>1.

Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. To obtain the correct values of logarithms, you should remember their properties and the sequence of actions when solving them.

Rules and some restrictions

In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the even root of negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:

The base “a” must always be greater than zero, and not equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
if a > 0, then a b >0, it turns out that “c” must also be greater than zero.

How to solve logarithms?

For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.

Now let's represent this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.

To accurately determine the value of an unknown degree, you need to learn how to work with a table of degrees. It looks like this:

As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However, for larger values you will need a power table. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!

Equations and inequalities

It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equality. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.

The following expression is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value “x” is under the logarithmic sign. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.

The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm 2 x = √9) imply one or more specific numerical values in the answer, while when solving an inequality, both the range of acceptable values and the points are determined breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or set of numbers.

Basic theorems about logarithms

When solving primitive tasks of finding the values of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.

The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case, the mandatory condition is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.

This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.

Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;

but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.

Examples of problems and inequalities

The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also a required part of mathematics exams. To enter a university or pass entrance examinations in mathematics, you need to know how to correctly solve such tasks.

Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, but certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, you should find out whether the expression can be simplified or reduced to a general form. You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them quickly.

When solving logarithmic equations, we must determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.

Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. To solve natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.

How to Use Logarithm Formulas: With Examples and Solutions

So, let's look at examples of using the basic theorems about logarithms.

The property of the logarithm of a product can be used in tasks where it is necessary to decompose a large value of the number b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values out of the sign of the logarithm.

Assignments from the Unified State Exam

Logarithms are often found in entrance exams, especially many logarithmic problems in the Unified State Exam (state exam for all school graduates). Typically, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge of the topic “Natural logarithms”.

Examples and solutions to problems are taken from the official versions of the Unified State Exam. Let's see how such tasks are solved.

Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.

It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.