How to expand the logarithm of a sum. Logarithmic equation: basic formulas and techniques. Inverse trigonometric function
Logarithmic equations and inequalities in the Unified State Examination in mathematics it is devoted to problem C3 . Every student must learn to solve C3 tasks from the Unified State Exam in mathematics if he wants to pass the upcoming exam with “good” or “excellent”. This article provides a brief overview of commonly encountered logarithmic equations and inequalities, as well as basic methods for solving them.
So, let's look at a few examples today. logarithmic equations and inequalities, which were offered to students in the Unified State Examination in mathematics of previous years. But it will begin with a brief summary of the main theoretical points that we will need to solve them.
Logarithmic function
Definition
Function of the form
0,\, a\ne 1 \]" title="Rendered by QuickLaTeX.com">!}
called logarithmic function.
Basic properties
Basic properties of the logarithmic function y=log a x:
The graph of a logarithmic function is logarithmic curve:
Properties of logarithms
Logarithm of the product two positive numbers is equal to the sum of the logarithms of these numbers:
Title="Rendered by QuickLaTeX.com">!}
Logarithm of the quotient two positive numbers is equal to the difference between the logarithms of these numbers:
Title="Rendered by QuickLaTeX.com">!}
If a And b a≠ 1, then for any number r equality is true:
Title="Rendered by QuickLaTeX.com">!}
Equality log a t=log a s, Where a > 0, a ≠ 1, t > 0, s> 0, valid if and only if t = s.
If a, b, c are positive numbers, and a And c are different from unity, then the equality ( formula for moving to a new logarithm base):
Title="Rendered by QuickLaTeX.com">!}
Theorem 1. If f(x) > 0 and g(x) > 0, then the logarithmic equation log a f(x) = log a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).
Solving logarithmic equations and inequalities
Example 1. Solve the equation:
Solution. The range of acceptable values includes only those x, for which the expression under the logarithm sign is greater than zero. These values are determined by the following system of inequalities:
Title="Rendered by QuickLaTeX.com">!}
Considering that
Title="Rendered by QuickLaTeX.com">!}
we obtain the interval that defines the range of permissible values of this logarithmic equation:
Based on Theorem 1, all conditions of which are satisfied here, we proceed to the following equivalent quadratic equation:
The range of acceptable values includes only the first root.
Answer: x = 7.
Example 2. Solve the equation:
Solution. The range of acceptable values of the equation is determined by the system of inequalities:
ql-right-eqno">
Solution. The range of acceptable values of the equation is determined here easily: x > 0.
We use substitution:
The equation becomes:
Reverse substitution:
Both answer are within the range of acceptable values of the equation because they are positive numbers.
Example 4. Solve the equation:
Solution. Let's start the solution again by determining the range of acceptable values of the equation. It is determined by the following system of inequalities:
ql-right-eqno">
The bases of the logarithms are the same, so in the range of acceptable values we can proceed to the following quadratic equation:
The first root is not within the range of acceptable values of the equation, but the second is.
Answer: x = -1.
Example 5. Solve the equation:
Solution. We will look for solutions in between x > 0, x≠1. Let's transform the equation to an equivalent one:
Both answer are within the range of acceptable values of the equation.
Example 6. Solve the equation:
Solution. The system of inequalities defining the range of permissible values of the equation this time has the form:
Title="Rendered by QuickLaTeX.com">!}
Using the properties of the logarithm, we transform the equation to an equation that is equivalent in the range of acceptable values:
Using the formula for moving to a new logarithm base, we get:
The range of acceptable values includes only one answer: x = 4.
Let's now move on to logarithmic inequalities . This is exactly what you will have to deal with on the Unified State Exam in mathematics. To solve further examples we need the following theorem:
Theorem 2. If f(x) > 0 and g(x) > 0, then:
at a> 1 logarithmic inequality log a f(x) > log a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x);
at 0< a < 1 логарифмическое неравенство log a f(x) > log a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).
Example 7. Solve the inequality:
Solution. Let's start by defining the range of acceptable values of the inequality. The expression under the sign of the logarithmic function must take only positive values. This means that the required range of acceptable values is determined by the following system of inequalities:
Title="Rendered by QuickLaTeX.com">!}
Since the base of the logarithm is a number less than one, the corresponding logarithmic function will be decreasing, and therefore, according to Theorem 2, the transition to the following quadratic inequality will be equivalent:
Finally, taking into account the range of acceptable values, we obtain answer:
Example 8. Solve the inequality:
Solution. Let's start again by defining the range of acceptable values:
Title="Rendered by QuickLaTeX.com">!}
On the set of admissible values of the inequality we carry out equivalent transformations:
After reduction and transition to the inequality equivalent by Theorem 2, we obtain:
Taking into account the range of acceptable values, we obtain the final answer:
Example 9. Solve logarithmic inequality:
Solution. The range of acceptable values of inequality is determined by the following system:
Title="Rendered by QuickLaTeX.com">!}
It can be seen that in the range of acceptable values, the expression at the base of the logarithm is always greater than one, and therefore, according to Theorem 2, the transition to the following inequality will be equivalent:
Taking into account the range of acceptable values, we obtain the final answer:
Example 10. Solve the inequality:
Solution.
The range of acceptable values of inequality is determined by the system of inequalities:
Title="Rendered by QuickLaTeX.com">!}
Method I Let us use the formula for transition to a new base of the logarithm and move on to an inequality that is equivalent in the range of acceptable values.
Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.
You definitely need to know these rules - without them, not a single serious logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Adding and subtracting logarithms
Consider two logarithms with the same bases: log a x and log a y. Then they can be added and subtracted, and:
- log a x+ log a y=log a (x · y);
- log a x− log a y=log a (x : y).
So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Please note: the key point here is identical grounds. If the reasons are different, these rules do not work!
These formulas will help you calculate a logarithmic expression even when its individual parts are not considered (see lesson “What is a logarithm”). Take a look at the examples and see:
Log 6 4 + log 6 9.
Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.
Task. Find the value of the expression: log 2 48 − log 2 3.
The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.
Task. Find the value of the expression: log 3 135 − log 3 5.
Again the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 27 = 3.
As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.
Extracting the exponent from the logarithm
Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself. This is what is most often required.
Task. Find the value of the expression: log 7 49 6 .
Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the meaning of the expression:
[Caption for the picture]
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:
[Caption for the picture] I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator contain the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:
Let the logarithm log be given a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:
[Caption for the picture]
In particular, if we put c = x, we get:
[Caption for the picture]
From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.
These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:
Task. Find the value of the expression: log 5 16 log 2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;
Now let’s “reverse” the second logarithm:
[Caption for the picture]Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.
Task. Find the value of the expression: log 9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
[Caption for the picture]Now let's get rid of the decimal logarithm by moving to a new base:
[Caption for the picture]Basic logarithmic identity
Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:
In the first case, the number n becomes an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.
The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.
In fact, what will happen if the number b raise to such a power that the number b to this power gives the number a? That's right: you get this same number a. Read this paragraph carefully again - many people get stuck on it.
Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the meaning of the expression:
[Caption for the picture]
Note that log 25 64 = log 5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:
[Caption for the picture]If anyone doesn't know, this was a real task from the Unified State Exam :)
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.
- log a a= 1 is a logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
- log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a 0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
In relation to
the task of finding any of the three numbers from the other two given ones can be set. If a and then N are given, they are found by exponentiation. If N and then a are given by taking the root of the degree x (or raising it to the power). Now consider the case when, given a and N, we need to find x.
Let the number N be positive: the number a be positive and not equal to one: .
Definition. The logarithm of the number N to the base a is the exponent to which a must be raised to obtain the number N; logarithm is denoted by
Thus, in equality (26.1) the exponent is found as the logarithm of N to base a. Posts
have the same meaning. Equality (26.1) is sometimes called the main identity of the theory of logarithms; in reality it expresses the definition of the concept of logarithm. By this definition, the base of the logarithm a is always positive and different from unity; the logarithmic number N is positive. Negative numbers and zero have no logarithms. It can be proven that any number with a given base has a well-defined logarithm. Therefore equality entails . Note that the condition is essential here; otherwise, the conclusion would not be justified, since the equality is true for any values of x and y.
Example 1. Find
Solution. To obtain a number, you must raise the base 2 to the power Therefore.
You can make notes when solving such examples in the following form:
Example 2. Find .
Solution. We have
In examples 1 and 2, we easily found the desired logarithm by representing the logarithm number as a power of the base with a rational exponent. In the general case, for example, for etc., this cannot be done, since the logarithm has an irrational value. Let us pay attention to one issue related to this statement. In paragraph 12, we gave the concept of the possibility of determining any real power of a given positive number. This was necessary for the introduction of logarithms, which, generally speaking, can be irrational numbers.
Let's look at some properties of logarithms.
Property 1. If the number and base are equal, then the logarithm is equal to one, and, conversely, if the logarithm is equal to one, then the number and base are equal.
Proof. Let By the definition of a logarithm we have and whence
Conversely, let Then by definition
Property 2. The logarithm of one to any base is equal to zero.
Proof. By definition of a logarithm (the zero power of any positive base is equal to one, see (10.1)). From here
Q.E.D.
The converse statement is also true: if , then N = 1. Indeed, we have .
Before formulating the next property of logarithms, let us agree to say that two numbers a and b lie on the same side of the third number c if they are both greater than c or less than c. If one of these numbers is greater than c, and the other is less than c, then we will say that they lie on opposite sides of c.
Property 3. If the number and base lie on the same side of one, then the logarithm is positive; If the number and base lie on opposite sides of one, then the logarithm is negative.
The proof of property 3 is based on the fact that the power of a is greater than one if the base is greater than one and the exponent is positive or the base is less than one and the exponent is negative. A power is less than one if the base is greater than one and the exponent is negative or the base is less than one and the exponent is positive.
There are four cases to consider:
We will limit ourselves to analyzing the first of them; the reader will consider the rest on his own.
Let then in equality the exponent can be neither negative nor equal to zero, therefore, it is positive, i.e., as required to be proved.
Example 3. Find out which of the logarithms below are positive and which are negative:
Solution, a) since the number 15 and the base 12 are located on the same side of one;
b) since 1000 and 2 are located on one side of the unit; in this case, it is not important that the base is greater than the logarithmic number;
c) since 3.1 and 0.8 lie on opposite sides of unity;
G) ; Why?
d) ; Why?
The following properties 4-6 are often called the rules of logarithmation: they allow, knowing the logarithms of some numbers, to find the logarithms of their product, quotient, and degree of each of them.
Property 4 (product logarithm rule). The logarithm of the product of several positive numbers to a given base is equal to the sum of the logarithms of these numbers to the same base.
Proof. Let the given numbers be positive.
For the logarithm of their product, we write the equality (26.1) that defines the logarithm:
From here we will find
Comparing the exponents of the first and last expressions, we obtain the required equality:
Note that the condition is essential; the logarithm of the product of two negative numbers makes sense, but in this case we get
In general, if the product of several factors is positive, then its logarithm is equal to the sum of the logarithms of the absolute values of these factors.
Property 5 (rule for taking logarithms of quotients). The logarithm of a quotient of positive numbers is equal to the difference between the logarithms of the dividend and the divisor, taken to the same base. Proof. We consistently find
Q.E.D.
Property 6 (power logarithm rule). The logarithm of the power of any positive number is equal to the logarithm of that number multiplied by the exponent.
Proof. Let us write again the main identity (26.1) for the number:
Q.E.D.
Consequence. The logarithm of a root of a positive number is equal to the logarithm of the radical divided by the exponent of the root:
The validity of this corollary can be proven by imagining how and using property 6.
Example 4. Take logarithm to base a:
a) (it is assumed that all values b, c, d, e are positive);
b) (it is assumed that ).
Solution, a) It is convenient to go to fractional powers in this expression:
Based on equalities (26.5)-(26.7), we can now write:
We notice that simpler operations are performed on the logarithms of numbers than on the numbers themselves: when multiplying numbers, their logarithms are added, when dividing, they are subtracted, etc.
That is why logarithms are used in computing practice (see paragraph 29).
The inverse action of logarithm is called potentiation, namely: potentiation is the action by which the number itself is found from a given logarithm of a number. Essentially, potentiation is not any special action: it comes down to raising a base to a power (equal to the logarithm of a number). The term "potentiation" can be considered synonymous with the term "exponentiation".
When potentiating, you must use the rules inverse to the rules of logarithmation: replace the sum of logarithms with the logarithm of the product, the difference of logarithms with the logarithm of the quotient, etc. In particular, if there is a factor in front of the sign of the logarithm, then during potentiation it must be transferred to the exponent degrees under the sign of the logarithm.
Example 5. Find N if it is known that
Solution. In connection with the just stated rule of potentiation, we will transfer the factors 2/3 and 1/3 standing in front of the signs of logarithms on the right side of this equality into exponents under the signs of these logarithms; we get
Now we replace the difference of logarithms with the logarithm of the quotient:
to obtain the last fraction in this chain of equalities, we freed the previous fraction from irrationality in the denominator (clause 25).
Property 7. If the base is greater than one, then the larger number has a larger logarithm (and the smaller one has a smaller one), if the base is less than one, then the larger number has a smaller logarithm (and the smaller one has a larger one).
This property is also formulated as a rule for taking logarithms of inequalities, both sides of which are positive:
When logarithming inequalities to a base greater than one, the sign of inequality is preserved, and when logarithming to a base less than one, the sign of inequality changes to the opposite (see also paragraph 80).
The proof is based on properties 5 and 3. Consider the case when If , then and, taking logarithms, we obtain
(a and N/M lie on the same side of unity). From here
Case a follows, the reader will figure it out on his own.
With this video I begin a long series of lessons about logarithmic equations. Now you have three examples in front of you, on the basis of which we will learn to solve the simplest problems, which are called - protozoa.
log 0.5 (3x − 1) = −3
log (x + 3) = 3 + 2 log 5
Let me remind you that the simplest logarithmic equation is the following:
log a f (x) = b
In this case, it is important that the variable x is present only inside the argument, that is, only in the function f (x). And the numbers a and b are just numbers, and in no case are functions containing the variable x.
Basic solution methods
There are many ways to solve such structures. For example, most teachers at school offer this method: Immediately express the function f (x) using the formula f ( x ) = a b . That is, when you come across the simplest construction, you can immediately move on to the solution without additional actions and constructions.
Yes, of course, the decision will be correct. However, the problem with this formula is that most students do not understand, where it comes from and why we raise the letter a to the letter b.
As a result, I often see very annoying mistakes when, for example, these letters are swapped. This formula must either be understood or crammed, and the second method leads to mistakes at the most inopportune and most crucial moments: during exams, tests, etc.
That is why I suggest to all my students to abandon the standard school formula and use the second approach to solve logarithmic equations, which, as you probably guessed from the name, is called canonical form.
The idea of the canonical form is simple. Let's look at our problem again: on the left we have log a, and by the letter a we mean a number, and in no case a function containing the variable x. Consequently, this letter is subject to all the restrictions that are imposed on the base of the logarithm. namely:
1 ≠ a > 0
On the other hand, from the same equation we see that the logarithm must be equal to the number b, and no restrictions are imposed on this letter, because it can take any value - both positive and negative. It all depends on what values the function f(x) takes.
And here we remember our wonderful rule that any number b can be represented as a logarithm to the base a of a to the power of b:
b = log a a b
How to remember this formula? Yes, very simple. Let's write the following construction:
b = b 1 = b log a a
Of course, in this case all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm and introduce the multiplier b as the power of a. We get:
b = b 1 = b log a a = log a a b
As a result, the original equation will be rewritten as follows:
log a f (x) = log a a b → f (x) = a b
That's all. The new function no longer contains a logarithm and can be solved using standard algebraic techniques.
Of course, someone will now object: why was it necessary to come up with some kind of canonical formula at all, why perform two additional unnecessary steps if it was possible to immediately move from the original design to the final formula? Yes, if only because most students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.
But this sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where the final formula comes from. By the way, this entry is called the canonical formula:
log a f (x) = log a a b
The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.
Examples of solutions
Now let's look at real examples. So, let's decide:
log 0.5 (3x − 1) = −3
Let's rewrite it like this:
log 0.5 (3x − 1) = log 0.5 0.5 −3
Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. Indeed, when you are already well trained in solving such problems, you can immediately perform this step.
However, if you are now just starting to study this topic, it is better not to rush anywhere in order to avoid making offensive mistakes. So, we have the canonical form. We have:
3x − 1 = 0.5 −3
This is no longer a logarithmic equation, but linear with respect to the variable x. To solve it, let's first look at the number 0.5 to the power of −3. Note that 0.5 is 1/2.
(1/2) −3 = (2/1) 3 = 8
Convert all decimal fractions to common fractions when solving a logarithmic equation.
We rewrite and get:
3x − 1 = 8
3x = 9
x = 3
That's it, we got the answer. The first problem has been solved.
Second task
Let's move on to the second task:
As we see, this equation is no longer the simplest. If only because there is a difference on the left, and not a single logarithm to one base.
Therefore, we need to somehow get rid of this difference. In this case, everything is very simple. Let's take a closer look at the bases: on the left is the number under the root:
General recommendation: in all logarithmic equations, try to get rid of radicals, i.e., from entries with roots and move on to power functions, simply because the exponents of these powers are easily taken out of the sign of the logarithm and, ultimately, such an entry significantly simplifies and speeds up calculations. Let's write it down like this:
Now let us remember the remarkable property of the logarithm: powers can be derived from the argument, as well as from the base. In the case of grounds, the following happens:
log a k b = 1/k loga b
In other words, the number that was in the base power is brought forward and at the same time inverted, that is, it becomes a reciprocal number. In our case, the base degree was 1/2. Therefore, we can take it out as 2/1. We get:
5 2 log 5 x − log 5 x = 18
10 log 5 x − log 5 x = 18
Please note: under no circumstances should you get rid of logarithms at this step. Remember 4th-5th grade math and the order of operations: multiplication is performed first, and only then addition and subtraction. In this case, we subtract one of the same elements from 10 elements:
9 log 5 x = 18
log 5 x = 2
Now our equation looks as it should. This is the simplest construction, and we solve it using the canonical form:
log 5 x = log 5 5 2
x = 5 2
x = 25
That's all. The second problem has been solved.
Third example
Let's move on to the third task:
log (x + 3) = 3 + 2 log 5
Let me remind you of the following formula:
log b = log 10 b
If for some reason you are confused by the notation log b , then when performing all the calculations you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take powers, add and represent any numbers in the form lg 10.
It is these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.
First, note that the factor 2 in front of lg 5 can be added and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.
Judge for yourself: any number can be represented as log to base 10:
3 = log 10 10 3 = log 10 3
Let's rewrite the original problem taking into account the obtained changes:
log (x − 3) = log 1000 + log 25
log (x − 3) = log 1000 25
log (x − 3) = log 25,000
We have before us the canonical form again, and we got it without going through the transformation stage, i.e. the simplest logarithmic equation did not appear anywhere.
This is exactly what I talked about at the very beginning of the lesson. The canonical form allows you to solve a wider class of problems than the standard school formula that most school teachers give.
Well, that’s it, we get rid of the sign of the decimal logarithm, and we get a simple linear construction:
x + 3 = 25,000
x = 24,997
All! The problem is solved.
A note on scope
Here I would like to make an important remark regarding the scope of definition. Surely now there will be students and teachers who will say: “When we solve expressions with logarithms, we must remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why did we not require this inequality to be satisfied in any of the problems considered?
Do not worry. In these cases, no extra roots will appear. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in one single argument of a single logarithm), and nowhere else in our case does the variable x appear, then write down the domain of definition no need, because it will be executed automatically.
Judge for yourself: in the first equation we got that 3x − 1, i.e. the argument should be equal to 8. This automatically means that 3x − 1 will be greater than zero.
With the same success we can write that in the second case x should be equal to 5 2, i.e. it is certainly greater than zero. And in the third case, where x + 3 = 25,000, i.e., again, obviously greater than zero. In other words, the scope is satisfied automatically, but only if x occurs only in the argument of only one logarithm.
That's all you need to know to solve the simplest problems. This rule alone, together with the transformation rules, will allow you to solve a very wide class of problems.
But let's be honest: in order to finally understand this technique, to learn how to apply the canonical form of the logarithmic equation, it is not enough to just watch one video lesson. Therefore, right now, download the options for independent solutions that are attached to this video lesson and start solving at least one of these two independent works.
It will take you literally a few minutes. But the effect of such training will be much higher than if you simply watched this video lesson.
I hope this lesson will help you understand logarithmic equations. Use the canonical form, simplify expressions using the rules for working with logarithms - and you won’t be afraid of any problems. That's all I have for today.
Taking into account the domain of definition
Now let's talk about the domain of definition of the logarithmic function, and how this affects the solution of logarithmic equations. Consider a construction of the form
log a f (x) = b
Such an expression is called the simplest - it contains only one function, and the numbers a and b are just numbers, and in no case a function that depends on the variable x. It can be solved very simply. You just need to use the formula:
b = log a a b
This formula is one of the key properties of the logarithm, and when substituting into our original expression we get the following:
log a f (x) = log a a b
f (x) = a b
This is a familiar formula from school textbooks. Many students will probably have a question: since in the original expression the function f (x) is under the log sign, the following restrictions are imposed on it:
f(x) > 0
This limitation applies because the logarithm of negative numbers does not exist. So, perhaps, as a result of this limitation, a check on answers should be introduced? Perhaps they need to be inserted into the source?
No, in the simplest logarithmic equations additional checking is unnecessary. And that's why. Take a look at our final formula:
f (x) = a b
The fact is that the number a is in any case greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But this doesn’t matter, because no matter what power we raise a positive number to, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is satisfied automatically.
What's really worth checking is the domain of the function under the log sign. There may be quite complex structures, and you definitely need to keep an eye on them during the solution process. Let's get a look.
First task:
First step: convert the fraction on the right. We get:
We get rid of the logarithm sign and get the usual irrational equation:
Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks are required to ensure that the expression under the logarithm sign is greater than 0, because it is not just greater than 0, but according to the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is satisfied automatically.
Let's move on to the second task:
Everything is the same here. We rewrite the construction, replacing the triple:
We get rid of the logarithm signs and get an irrational equation:
We square both sides taking into account the restrictions and get:
4 − 6x − x 2 = (x − 4) 2
4 − 6x − x 2 = x 2 + 8x + 16
x 2 + 8x + 16 −4 + 6x + x 2 = 0
2x 2 + 14x + 12 = 0 |:2
x 2 + 7x + 6 = 0
We solve the resulting equation through the discriminant:
D = 49 − 24 = 25
x 1 = −1
x 2 = −6
But x = −6 does not suit us, because if we substitute this number into our inequality, we get:
−6 + 4 = −2 < 0
In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:
−1 + 4 = 3 > 0
The only answer in our case will be x = −1. That's the solution. Let's go back to the very beginning of our calculations.
The main takeaway from this lesson is that you don't need to check constraints on a function in simple logarithmic equations. Because during the solution process all constraints are satisfied automatically.
However, this in no way means that you can forget about checking altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own restrictions and requirements for the right side, which we have seen today in two different examples.
Feel free to solve such problems and be especially careful if there is a root in the argument.
Logarithmic equations with different bases
We continue to study logarithmic equations and look at two more quite interesting techniques with which it is fashionable to solve more complex constructions. But first, let’s remember how the simplest problems are solved:
log a f (x) = b
In this entry, a and b are numbers, and in the function f (x) the variable x must be present, and only there, that is, x must only be in the argument. We will transform such logarithmic equations using the canonical form. To do this, note that
b = log a a b
Moreover, a b is precisely an argument. Let's rewrite this expression as follows:
log a f (x) = log a a b
This is exactly what we are trying to achieve, so that there is a logarithm to base a on both the left and the right. In this case, we can, figuratively speaking, cross out the log signs, and from a mathematical point of view we can say that we are simply equating the arguments:
f (x) = a b
As a result, we will get a new expression that will be much easier to solve. Let's apply this rule to our problems today.
So, the first design:
First of all, I note that on the right is a fraction whose denominator is log. When you see an expression like this, it’s a good idea to remember a wonderful property of logarithms:
Translated into Russian, this means that any logarithm can be represented as the quotient of two logarithms with any base c. Of course 0< с ≠ 1.
So: this formula has one wonderful special case, when the variable c is equal to the variable b. In this case we get a construction like:
This is exactly the construction we see from the sign on the right in our equation. Let's replace this construction with log a b , we get:
In other words, in comparison with the original task, we swapped the argument and the base of the logarithm. Instead, we had to reverse the fraction.
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. We are talking about expressions based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely through the canonical form.
First, let's remember how the simplest problems are solved, based on ordinary numbers. So, the simplest construction is called
log a f (x) = b
To solve such problems we can use the following formula:
b = log a a b
We rewrite our original expression and get:
log a f (x) = log a a b
Then we equate the arguments, i.e. we write:
f (x) = a b
Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained from the solution will be the roots of the original logarithmic equation. In addition, a record when both the left and the right are in the same logarithm with the same base is precisely called the canonical form. It is to such a record that we will try to reduce today's designs. So, let's go.
First task:
log x − 2 (2x 2 − 13x + 18) = 1
Replace 1 with log x − 2 (x − 2) 1 . The degree that we observe in the argument is actually the number b that stood to the right of the equal sign. Thus, let's rewrite our expression. We get:
log x − 2 (2x 2 − 13x + 18) = log x − 2 (x − 2)
What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:
2x 2 − 13x + 18 = x − 2
But the solution does not end there, because this equation is not equivalent to the original one. After all, the resulting construction consists of functions that are defined on the entire number line, and our original logarithms are not defined everywhere and not always.
Therefore, we must write down the domain of definition separately. Let's not split hairs and first write down all the requirements:
First, the argument of each of the logarithms must be greater than 0:
2x 2 − 13x + 18 > 0
x − 2 > 0
Secondly, the base must not only be greater than 0, but also different from 1:
x − 2 ≠ 1
As a result, we get the system:
But don’t be alarmed: when processing logarithmic equations, such a system can be significantly simplified.
Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required that it be greater than zero.
In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will automatically be satisfied. Therefore, we can safely cross out the inequality containing the quadratic function. Thus, the number of expressions contained in our system will be reduced to three.
Of course, with the same success we could cross out the linear inequality, that is, cross out x − 2 > 0 and require that 2x 2 − 13x + 18 > 0. But you will agree that solving the simplest linear inequality is much faster and simpler, than quadratic, even under the condition that as a result of solving this entire system we get the same roots.
In general, try to optimize calculations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.
Let's rewrite our system:
Here is a system of three expressions, two of which we, in fact, have already dealt with. Let's write out the quadratic equation separately and solve it:
2x 2 − 14x + 20 = 0
x 2 − 7x + 10 = 0
Before us is a reduced quadratic trinomial and, therefore, we can use Vieta’s formulas. We get:
(x − 5)(x − 2) = 0
x 1 = 5
x 2 = 2
Now we return to our system and find that x = 2 does not suit us, because we are required that x be strictly greater than 2.
But x = 5 suits us perfectly: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution to this system will be x = 5.
That's it, the problem is solved, including taking into account the ODZ. Let's move on to the second equation. More interesting and informative calculations await us here:
The first step: like last time, we bring this whole matter to canonical form. To do this, we can write the number 9 as follows:
You don’t have to touch the base with the root, but it’s better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write down:
Let me not rewrite our entire large logarithmic equation, but just immediately equate the arguments:
x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27
x 2 + 4x + 3 = 0
Before us is a newly reduced quadratic trinomial, let’s use Vieta’s formulas and write:
(x + 3)(x + 1) = 0
x 1 = −3
x 2 = −1
So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, the log signs impose additional restrictions (here we should have written down the system, but due to the cumbersome nature of the whole structure, I decided to calculate the domain of definition separately).
First of all, remember that the arguments must be greater than 0, namely:
These are the requirements imposed by the scope of definition.
Let us immediately note that since we equate the first two expressions of the system to each other, we can cross out any of them. Let's cross out the first one because it looks more threatening than the second one.
In addition, note that the solution to the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly, with a root of the third degree - these inequalities are completely analogous, so we can cross it out).
But with the third inequality this will not work. Let's get rid of the radical sign on the left by raising both parts to a cube. We get:
So we get the following requirements:
− 2 ≠ x > −3
Which of our roots: x 1 = −3 or x 2 = −1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (since our inequality is strict). So, returning to our problem, we get one root: x = −1. That's it, problem solved.
Once again, the key points of this task:
- 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 a variable base appears in a logarithm, the problem ceases to be the simplest. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they also must not be equal to 1.
The final requirements can be applied to the final answers in different ways. For example, you can solve an entire system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember the domain of definition, separately work it out in the form of a system and apply it to the obtained roots.
Which method to choose when solving a particular logarithmic equation is up to you. In any case, the answer will be the same.
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.
Loading...