How to solve natural logarithm equations. Logarithmic equation: basic formulas and techniques. Logarithm of both sides of the equation
Logarithmic equations and inequalities in the USE variants in mathematics is devoted to task C3 . Every student should learn how to solve tasks C3 from the Unified State Examination in mathematics if he wants to pass the upcoming exam as “good” or “excellent”. This article provides a brief overview of commonly encountered logarithmic equations and inequalities, as well as the main methods for solving them.
So let's take a look at some examples today. logarithmic equations and inequalities, which were offered to students in the USE variants in mathematics of past years. But start with summary the main theoretical points that we need to solve them.
logarithmic function
Definition
View function
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called logarithmic function.
Basic properties
Basic properties of the logarithmic function y= log a x:
The graph of the 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:
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Logarithm of the quotient two positive numbers is equal to the difference of the logarithms of these numbers:
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If a and b a≠ 1, then for any number r fair equality:
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Equality log a t= log a s, where a > 0, a ≠ 1, t > 0, s> 0 is true if and only if t = s.
If a, b, c are positive numbers, and a and c are different from unity, then the equality ( conversion formula to the new base of the logarithm):
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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 sign of the logarithm is greater than zero. These values are determined by the following system of inequalities:
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Taking into account the fact that
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we obtain an interval that determines the area of admissible values of this logarithmic equation:
Based on Theorem 1, all the conditions of which are satisfied here, we pass to the following equivalent quadratic equation:
Only the first root is included in the range of acceptable values.
Answer: x=7.
Example 2 Solve the equation:
Solution. The range of admissible values of the equation is determined by the system of inequalities:
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Solution. The range of admissible values of the equation is easily defined here: x > 0.
We use substitution:
The equation takes the form:
Back substitution:
Both response enter the range of admissible values of the equation, since they are positive numbers.
Example 4 Solve the equation:
Solution. Let's start the solution again by determining the range of admissible values of the equation. It is defined by the following system of inequalities:
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The bases of the logarithms are the same, so in the range of valid values, you can go to the following quadratic equation:
The first root is not included in the range of admissible values of the equation, the second one is included.
Answer: x = -1.
Example 5 Solve the equation:
Solution. We will look for solutions in the interval x > 0, x≠1. Let's transform the equation to an equivalent one:
Both response are within the range of admissible values of the equation.
Example 6 Solve the equation:
Solution. The system of inequalities that defines the range of admissible values of the equation, this time has the form:
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Using the properties of the logarithm, we transform the equation to an equivalent equation in the range of permissible values:
Using the formula for the transition to a new base of the logarithm, we get:
Only one is within the allowed range. answer: x = 4.
Let's move on to logarithmic inequalities . This is exactly what you will have to deal with on the 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 of the opposite meaning: f(x) < g(x).
Example 7 Solve the inequality:
Solution. Let's start by defining the range of acceptable values of inequality. The expression under the sign of the logarithmic function must take only positive values. This means that the desired range of acceptable values is determined by the following system of inequalities:
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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 permissible values, we obtain answer:
Example 8 Solve the inequality:
Solution. Let's start again by defining the range of acceptable values:
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On the set of admissible values of the inequality, we carry out equivalent transformations:
After reduction and transition to an inequality equivalent by Theorem 2, we obtain:
Taking into account the range of permissible values, we obtain the final answer:
Example 9 Solve the logarithmic inequality:
Solution. The range of acceptable values of inequality is determined by the following system:
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It can be seen that in the region of admissible 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 area of acceptable values of inequality is determined by the system of inequalities:
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I way. Let us use the formula for the transition to a new base of the logarithm and proceed to an inequality that is equivalent in the region of admissible values.
With this video, I begin a long series of lessons about logarithmic equations. Now you have three examples at once, on the basis of which we will learn to solve the simplest tasks, which are called so - protozoa.
log 0.5 (3x - 1) = -3
lg (x + 3) = 3 + 2 lg 5
Let me remind you that the simplest logarithmic equation is the following:
log a f(x) = b
It is important that the variable x is present only inside the argument, i.e. 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 suggest this way: Immediately express the function f ( x ) using the formula f( x ) = a b . That is, when you meet the simplest construction, you can immediately proceed to the solution without additional actions and constructions.
Yes, of course, the decision will turn out to be correct. However, the problem with this formula is that most students do not understand, where does it come from and why exactly we raise the letter a to the letter b.
As a result, I often observe very offensive errors, when, for example, these letters are interchanged. This formula must either be understood or memorized, and the second method leads to errors at the most inopportune and most crucial moments: in 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 task again: on the left we have log a , while the letter a means exactly the number, and in no case the function containing the variable x. Therefore, this letter is subject to all 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 in base a from 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. And now let's use the basic property of the logarithm, and enter the factor 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 in the following form:
log a f (x) = log a a b → f (x) = a b
That's all. The new function no longer contains a logarithm and is solved by 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 go from the original construction 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 such a sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where that 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.
Solution examples
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. And indeed, when you are already well trained in solving such problems, you can immediately perform this step.
However, if now you are just starting to study this topic, it is better not to rush anywhere so as not to make offensive mistakes. So we have the canonical form. We have:
3x - 1 = 0.5 -3
This is no longer a logarithmic equation, but a linear one with respect to the variable x. To solve it, let's first deal with 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 decimals to fractions when you solve a logarithmic equation.
We rewrite and get:
3x − 1 = 8
3x=9
x=3
All we got the answer. The first task is solved.
Second task
Let's move on to the second task:
As you can see, this equation is no longer the simplest one. If only because the difference is on the left, and not a single logarithm in one base.
Therefore, you 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 a notation greatly simplifies and speeds up calculations. Let's write it like this:
Now we recall the remarkable property of the logarithm: from the argument, as well as from the base, you can take out degrees. In the case of bases, the following happens:
log a k b = 1/k loga b
In other words, the number that stood in the degree of the base is brought forward and at the same time turned over, that is, it becomes the reciprocal of the number. In our case, there was a degree of base with an indicator of 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: in no case should you get rid of logarithms at this step. Think back to grade 4-5 math and the order of operations: multiplication is performed first, and only then addition and subtraction are performed. 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 like 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 is solved.
Third example
Let's move on to the third task:
lg (x + 3) = 3 + 2 lg 5
Recall the following formula:
log b = log 10 b
If for some reason you are confused by writing lg b , then when doing all the calculations, you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take out powers, add, and represent any number as lg 10.
It is precisely 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.
To begin with, note that the factor 2 before lg 5 can be inserted 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 received changes:
lg (x − 3) = lg 1000 + lg 25
lg (x − 3) = lg 1000 25
lg (x - 3) = lg 25 000
Before us is again the canonical form, and we obtained it bypassing the stage of transformations, i.e., the simplest logarithmic equation did not come up anywhere with us.
That's what I was talking about at the very beginning of the lesson. The canonical form allows solving a wider class of problems than the standard school formula, which is given by most school teachers.
That's all, we get rid of the sign of the decimal logarithm, and we get a simple linear construction:
x + 3 = 25,000
x = 24997
Everything! Problem solved.
A note about scope
Here I would like to make an important remark about the domain of definition. Surely now there are students and teachers who will say: “When we solve expressions with logarithms, it is imperative to remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why in none of the considered problems did we require that this inequality be satisfied?
Do not worry. No extra roots will appear in these cases. 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 the one and only argument of the one and only logarithm), and nowhere else in our case does the variable x, then write the domain no need because it will run 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 must 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 automatic, but only if x occurs only in the argument of only one logarithm.
That's all you need to know to solve simple 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, in order to learn how to apply the canonical form of the logarithmic equation, it is not enough just to watch one video lesson. So download the options right now for independent solution, which are attached to this video tutorial and start solving at least one of these two independent works.
It will take you just a few minutes. But the effect of such training will be much higher compared to if you just watched this video tutorial.
I hope this lesson will help you understand logarithmic equations. Apply the canonical form, simplify expressions using the rules for working with logarithms - and you will not be afraid of any tasks. And that's all I have for today.
Scope consideration
Now let's talk about the domain of the logarithmic function, as well as how this affects the solution of logarithmic equations. Consider a construction of the form
log a f(x) = b
Such an expression is called the simplest - it has only one function, and the numbers a and b are just numbers, and in no case are a function that depends on the variable x. It is 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 already a familiar formula from school textbooks. Many students will probably have a question: since the function f ( x ) in the original expression is under the log sign, the following restrictions are imposed on it:
f(x) > 0
This restriction is valid because the logarithm of negative numbers does not exist. So, maybe because of this limitation, you should introduce a check for answers? Perhaps they need to be substituted in the source?
No, in the simplest logarithmic equations, an additional check is unnecessary. And that's why. Take a look at our final formula:
f(x) = a b
The fact is that the number a in any case is 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 does not matter, because no matter what degree we raise a positive number, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is fulfilled automatically.
What is really worth checking is the scope of the function under the log sign. There can be quite complex designs, and in the process of solving them, you must definitely follow them. Let's get a look.
First task:
First step: convert the fraction on the right. We get:
We get rid of the sign of the logarithm 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 that the expression under the logarithm sign is greater than 0 are required, because it is not just greater than 0, but by the condition of the equation it is equal to 2. Therefore, the requirement "greater than zero" is automatically satisfied.
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 signs of the logarithm and get an irrational equation:
We square both parts, taking into account the restrictions, and we get:
4 - 6x - x 2 = (x - 4) 2
4 - 6x - x 2 = x 2 + 8x + 16
x2 + 8x + 16 −4 + 6x + x2 = 0
2x2 + 14x + 12 = 0 |:2
x2 + 7x + 6 = 0
We solve the resulting equation through the discriminant:
D \u003d 49 - 24 \u003d 25
x 1 = -1
x 2 \u003d -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 is x = −1. That's all the solution. Let's go back to the very beginning of our calculations.
The main conclusion from this lesson is that it is not required to check the limits for a function in the simplest logarithmic equations. Because in the process of solving all the constraints are executed automatically.
However, this by no means means that you can forget about verification altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own limitations 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 analyze two more rather interesting tricks with which it is fashionable to solve more complex structures. But first, let's remember how the simplest tasks are solved:
log a f(x) = b
In this notation, a and b are just numbers, and in the function f (x) the variable x must be present, and only there, that is, x must be only in the argument. We will transform such logarithmic equations using the canonical form. For this, we note that
b = log a a b
And a b is just 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 both on the left and on the right there is a logarithm to the base a. In this case, we can, figuratively speaking, cross out the signs of log, and from the point of view of mathematics, we can say that we simply equate the arguments:
f(x) = a b
As a result, we get a new expression that will be solved much easier. Let's apply this rule to our tasks today.
So the first design:
First of all, I note that there is a fraction on the right, the denominator of which is log. When you see an expression like this, it's worth remembering the wonderful property of logarithms:
Translated into Russian, this means that any logarithm can be represented as a 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 of the form:
It is this construction that we observe 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 have swapped the argument and the base of the logarithm. Instead, we had to flip the fraction.
We recall that any degree can be taken out of the base according to the following rule:
In other words, the coefficient k, which is the degree of the base, is taken out as an inverted fraction. Let's take it out as an inverted fraction:
The fractional factor cannot be left in front, because in this case we will not be able to represent this entry as a canonical form (after all, in the canonical form, there is no additional factor in front of the second logarithm). Therefore, let's put the fraction 1/4 in the argument as a power:
Now we equate the arguments whose bases are the same (and we really have the same bases), and write:
x + 5 = 1
x = −4
That's all. We got the answer to the first logarithmic equation. Pay attention: in the original problem, the variable x occurs only in one log, and it is 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 − 3 log (x + 4)
Here, in addition to the usual logarithms, we will have to work with lg f (x). How to solve such an equation? It may seem to an unprepared student that this is some kind of tin, but in fact everything is solved elementarily.
Look closely 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 clues. Let's remember once again how the degrees are taken out from under the sign of the logarithm:
log a b n = nlog a b
In other words, what was the power of the number 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 afraid of lg 2 - this is the most common expression. You can rewrite it like this:
For him, all the rules that apply to any other logarithm are valid. In particular, the factor in front can be introduced into the power of the argument. Let's write:
Very often, students point blank do not see this action, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal in 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 convert a logarithmic equation, you should know this formula in the same way as the representation of any number in the form of log.
We return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will simply be 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've got. It is practically the canonical form, but there is a factor −3 on the right. Let's put it in the right lg argument:
lg 8 = lg (x + 4) −3
Before us is the canonical form of the logarithmic equation, so we cross out the signs of lg and equate the arguments:
(x + 4) -3 = 8
x + 4 = 0.5
That's all! We have 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 recap the key points of this lesson.
The main formula that is studied in all the lessons on this page devoted to solving logarithmic equations is the canonical form. And don't be put off by the fact that most school textbooks teach you how to solve these kinds of problems differently. This tool works very efficiently 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 a special case when we flip log (this was very useful to us in the first task);
- The formula for bringing in and taking out powers from under the sign of the logarithm. Here, many students get stuck and do not see point-blank that the power taken out and brought in can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of another 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 required to check the scope 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 domain requirements are met automatically.
Problems with variable base
Today we will consider logarithmic equations, which for many students seem non-standard, if not completely unsolvable. We are talking about expressions that are based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely, through the canonical form.
To begin with, let's recall how the simplest problems are solved, which are 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 in the solution will be the roots of the original logarithmic equation. In addition, the record, when both the left and the right are on the same logarithm with the same base, is called the canonical form. It is to this record that we will try to reduce today's constructions. 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, in fact, the number b , which was to the right of the equal sign. So 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:
2x2 - 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 be wiser 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 greatly 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 also be automatically satisfied. Therefore, we can safely cross out the inequality containing a quadratic function. Thus, the number of expressions contained in our system will be reduced to three.
Of course, we could just as well cross out the linear inequality, i.e. cross out x - 2 > 0 and require that 2x 2 - 13x + 18 > 0. But you must admit that solving the simplest linear inequality is much faster and easier, than quadratic, even if 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 such a system of three expressions, two of which we, in fact, have already figured out. Let's separately write out the quadratic equation and solve it:
2x2 - 14x + 20 = 0
x2 − 7x + 10 = 0
Before us is a reduced square trinomial and, therefore, we can use the Vieta formulas. We get:
(x − 5)(x − 2) = 0
x 1 = 5
x2 = 2
Now, back to our system, we find that x = 2 doesn't suit us, because we're required to have x strictly greater than 2.
But x \u003d 5 suits us quite well: 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 \u003d 5.
Everything, the task is solved, including taking into account the ODZ. Let's move on to the second equation. Here we are waiting for more interesting and meaningful calculations:
The first step: as well as last time, we bring all this business to a canonical form. To do this, we can write the number 9 as follows:
The base with the root can not be touched, but it is better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write:
Let me not rewrite our whole big 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 the again reduced square trinomial, we will use the Vieta 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, log signs impose additional restrictions (here we would have to write down the system, but due to the cumbersomeness of the whole construction, 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 domain of definition.
We note right away 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 menacing than the second one.
In addition, note that the solutions of 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 the root of the third degree - these inequalities are completely similar, so one of them we can cross it out).
But with the third inequality, this will not work. Let's get rid of the sign of the radical on the left, for which we raise 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 (because our inequality is strict). In total, 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 record, and do not go directly from the original problem to a construction like log a f ( x ) = b , make much fewer errors than those who are in a hurry somewhere, skipping intermediate steps of calculations;
- As soon as a variable base appears in the 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 must also not be equal to 1.
You can impose the last requirements on the final answers in different ways. For example, it is possible to solve the whole system containing all the domain requirements. On the other hand, you can first solve the problem itself, and then remember about the domain of definition, work it out separately in the form of a system and apply it to the obtained roots.
Which way 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 indicators. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where it is required to simplify cumbersome multiplication to simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. Simple and accessible language.
Definition in mathematics
The logarithm is an expression of the following form: log ab=c, that is, the logarithm of any non-negative number (that is, any positive) "b" by its base "a" is considered the power of "c", to which the base "a" must be raised, so that in the end 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 such a degree that from 2 to the required degree you get 8. Having done some calculations in your mind, we get the number 3! And rightly so, because 2 to the power of 3 gives the number 8 in the answer.
Varieties 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 distinct kinds of logarithmic expressions:
- Natural logarithm ln a, where the base is the Euler number (e = 2.7).
- Decimal a, where the base is 10.
- The logarithm of any number b to the base a>1.
Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to one logarithm using logarithmic theorems. To obtain the correct values of logarithms, one should remember their properties and the order of actions in their decisions.
Rules and some restrictions
In mathematics, there are several rules-limitations that are accepted as an axiom, that is, they are not subject to discussion and are true. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the root of an even degree from negative numbers. Logarithms also have their own rules, following which you can easily learn how to work even with long and capacious logarithmic expressions:
- the base "a" must always be greater than zero, and at the same time not be 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 be greater than zero.
How to solve logarithms?
For example, the task was given to find the answer to the equation 10 x \u003d 100. It is very easy, you need to choose such a power, raising the number ten to which we get 100. This, of course, is 10 2 \u003d 100.
Now let's represent this expression as a logarithmic one. We get log 10 100 = 2. When solving logarithms, all actions practically converge to finding the degree to which the base of the logarithm must be entered in order to obtain a given number.
To accurately determine the value of an unknown degree, you must 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 mindset and knowledge of the multiplication table. However, larger values will require a power table. It can be used even by those who do not understand anything at all in 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 in the cells, the values of the numbers are determined, which 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 real 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 equation. For example, 3 4 =81 can be written as the logarithm of 81 to base 3, which is four (log 3 81 = 4). For negative powers, the rules are the same: 2 -5 = 1/32 we write 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 consider examples and solutions of equations a little lower, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.
An expression of the following form is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number in 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 of 2 x = √9) imply one or more specific numerical values in the answer, while when solving the inequality, both the range of acceptable values and the points breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer of the equation, but a continuous series or set of numbers.
Basic theorems about logarithms
When solving primitive tasks on 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 get acquainted with examples of equations later, let's first analyze each property in more detail.
- The basic identity looks like this: a logaB =B. It only applies if 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 prerequisite is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this formula of logarithms, with examples and a solution. Let log as 1 = f 1 and log as 2 = f 2 , then a f1 = s 1 , a f2 = s 2. We get that s 1 *s 2 = a f1 *a f2 = a f1+f2 (degree properties ), and further by definition: log a (s 1 *s 2)= f 1 + f 2 = log a s1 + log as 2, which was to be proved.
- 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 "property of the degree of the logarithm". It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics rests on regular postulates. Let's look at the proof.
Let log a b \u003d t, it turns out a t \u003d b. If you raise both parts to the power m: a tn = b n ;
but since a tn = (a q) nt/q = b n , hence 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 logarithm problems are examples of equations and inequalities. They are found in almost all problem books, and are also included in the mandatory part of exams in mathematics. For admission to the university or passing entrance examinations in mathematics, you need to know how to solve such problems correctly.
Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, 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. Simplify long logarithmic expressions You can, if you use their properties correctly. Let's get to know them soon.
When solving logarithmic equations, it is necessary to determine what kind of logarithm we have before us: an example of an expression may contain a natural logarithm or a decimal one.
Here are examples ln100, ln1026. Their solution boils down to the fact that you need to determine the degree to which the base 10 will be equal to 100 and 1026, respectively. For solutions of natural logarithms, one must 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 main theorems on logarithms.
- The property of the logarithm of the product can be used in tasks where it is necessary to expand great importance numbers 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 degree of the logarithm, we managed to solve at first glance a complex and unsolvable expression. It is only necessary to factorize the base and then take the exponent values out of the sign of the logarithm.
Tasks from the exam
Logarithms are often found in entrance exams, especially a lot of logarithmic problems in the Unified State Exam (state exam for all school graduates). Usually these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most difficult and voluminous tasks). The exam implies an accurate and perfect knowledge of the topic "Natural logarithms".
Examples and problem solutions are taken from official USE options. 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.
- All logarithms are best reduced to the same base so that the solution is not cumbersome and confusing.
- All expressions under the sign of the logarithm are indicated as positive, therefore, when taking out the exponent of the exponent of the expression, which is under the sign of the logarithm and as its base, the expression remaining under the logarithm must be positive.
Preparation for the final test in mathematics includes an important section - "Logarithms". Tasks from this topic are necessarily contained in the exam. The experience of past years shows that the logarithmic equations caused difficulties for many schoolchildren. Therefore, students with different levels of training should understand how to find the correct answer and quickly cope with them.
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In preparation for the unified state exam high school graduates need a reliable source that provides the most complete and accurate information for the successful solution of test problems. However, the textbook is not always at hand, and searching for the necessary rules and formulas on the Internet often takes time.
The educational portal "Shkolkovo" allows you to prepare for the exam anywhere at any time. Our site offers the most convenient approach to repeating and mastering a large amount of information on logarithms, as well as on one and several unknowns. Start with easy equations. If you coped with them without difficulty, move on to more difficult ones. If you're having trouble solving a particular inequality, you can add it to your Favorites so you can come back to it later.
You can find the necessary formulas to complete the task, repeat special cases and methods for calculating the root of a standard logarithmic equation by looking at the "Theoretical Reference" section. Teachers of "Shkolkovo" collected, systematized and presented all the materials necessary for successful delivery in the most simple and understandable form.
In order to easily cope with tasks of any complexity, on our portal you can familiarize yourself with the solution of some typical logarithmic equations. To do this, go to the "Catalogs" section. We have presented a large number of examples, including those with equations of the profile level of the Unified State Examination in mathematics.
Students from schools all over Russia can use our portal. To get started, just register in the system and start solving equations. To consolidate the results, we advise you to return to the Shkolkovo website daily.
Before solving logarithmic equations, let's repeat the definition of the logarithm and the basic formulas.
Logarithm positive number b by reason a is an indicator of the degree to which it is necessary to raise a, To obtain b.
In this case, class="tex" alt="(!LANG:b> 0,\;a> 0,\;a\neq 1">.!}
Let's pay attention to the area of admissible values of the logarithm:
class="tex" alt="(!LANG:b> 0,\;a> 0,\;a\neq 1">. !}
Basic logarithmic identity:
Basic formulas for logarithms:
(The logarithm of the product is equal to the sum of the logarithms)
(The logarithm of the quotient is equal to the difference of the logarithms)
(Formula for the logarithm of the degree)
The formula for moving to a new base is:
We know what the graph of a logarithmic function looks like. This function is monotonic. If the base of the logarithm is greater than one, the logarithmic function is monotonically increasing. If the base is greater than zero and less than one, the logarithmic function decreases monotonically. And in any case, it takes each value only once. This means that if the logarithms of two numbers are equal in any base, then the numbers themselves are equal.
All this will be useful to us in solving logarithmic equations.
The simplest logarithmic equations
1. Solve the equation:
The bases of the logarithms are equal, the logarithms themselves are also equal, which means that the numbers from which they are taken are also equal.
Usually, students memorize this rule in a short jargon formulation: "Let's drop the logarithms!" Of course, we "discard" them not just like that, but using the monotonicity property of the logarithmic function.
We get:
When solving logarithmic equations, do not forget about tolerance range logarithm. Remember that the expression is defined with class="tex" alt="(!LANG:b> 0,\;a> 0,\;a\neq 1">.!}
It is very good if you, having found the root of the equation, just substitute it into the equation. If after such a substitution, the left or right side of the equation does not make sense, then the number found is not the root of the equation and cannot be the answer to the problem. This is a good way to test for the exam.
2. Solve the equation:
On the left side of the equation - the logarithm, on the right - the number 7. Applying the basic logarithmic identity, we represent the number 7 in the form. Then everything is simple.
Answer: -124
3. Solve the equation:
See the number 2 in front of the logarithm on the right side of the equation? Now it prevents you from "dropping logarithms." What can I do with it so that the left and right sides are just logarithms to base 5? Of course, the formula for the logarithm of the degree will help.
4. Solve the equation:
Valid range: class="tex" alt="(!LANG:4-x> 0."> Значит, class="tex" alt="x > -4.">!}
Let's represent 2 on the right side of the equation as - so that the left and right sides of the equation are logarithms to base 5.
The function is monotonically increasing and takes each of its values exactly once. Logarithms are equal, their bases are equal. Let's drop the logarithms! Of course, class="tex" alt="(!LANG:x> -4">.!}
5. Solve the equation:
We write the solution as a chain of equivalent transitions. We write down the ODZ and “remove” the logarithms:
Class="tex" alt="(!LANG:\log _(8)\left (x^(2)+x \right)=\log _(8)\left (x^(2)-4 \right )\Leftrightarrow \left\(\begin(matrix) x^(2)+x> 0\\ x^(2)-4> 0\\ x^(2)+x=x^(2)-4 \ end(matrix)\right.\Leftrightarrow \left\(\begin(matrix) x^(2)+x> 0\\ x^(2)-4> 0\\ x=-4 \end(matrix)\ right.\Leftrightarrow x=-4">!}
Answer: -4.
Note that solutions to logarithmic equations are best written as a chain of equivalent transitions. This will help us not to forget about the range of valid values.
6.Solve the equation:.
Let's move from the base 4 logarithm (in the exponent) to the base 2 logarithm. We do this using the base conversion formula:
We write the solution as a chain of equivalent transitions.
Class="tex" alt="(!LANG:2^(\log _(4)\left (4x+5 \right))=9\Leftrightarrow \left\(\begin(matrix) 2^\frac(( \log _(2)\left (4x+5 \right)))(2)=9\\ 4x+5> 0 \end(matrix)\right.\Leftrightarrow \left\(\begin(matrix) \left (2^(\log _(2)\left (4x+5 \right)) \right)^(\frac(1)(2))=9\\ x> -1\frac(1)(4) \end(matrix)\right.\Leftrightarrow \left\(\begin(matrix) \left (4x+5 \right)^(\frac(1)(2))=9\\ x> -1\frac( 1)(4) \end(matrix)\right.\Leftrightarrow \left\(\begin(matrix) \sqrt(4x+5)=9\\ x> -1\frac(1)(4) \end( matrix)\right.\Leftrightarrow \left\(\begin(matrix) 4x+5=81\\ x> -1\frac(1)(4) \end(matrix)\right.\Leftrightarrow \left\(\ begin(matrix) x=19\\ x> -1\frac(1)(4) \end(matrix)\right.">!}
7. Solve the equation:.
Please note: variable X both under the logarithm and at the base of the logarithm. We remember that the base of the logarithm must be positive and not equal to 1.
ODZ:
class="tex" alt="(!LANG:\left\(\begin(matrix) 12-x> 0\\ x> 0\\ x\neq 1 \end(matrix)\right.">!}
Now you can "remove" the logarithms.
Foreign root, because class="tex" alt="(!LANG:x> 0">.!}
8. Solve the equation.
ODZ equation: class="tex" alt="(!LANG:x> 0">!}
Let's make a replacement. As in algebraic equations, we make a change of variable whenever possible.
Back to the variable X:
9. Solve the equation:
The expression under the logarithm is always positive - since we add 25 to a non-negative value. The expression under the root on the right side is also positive. Means, X can be any real number.
We represent the sum of the logarithms on the left side as the logarithm of the product. On the right side - let's move on to the logarithm to the base 3. And use the formula for the logarithm of the degree.
We discard logarithms.
Such an equation is called biquadratic. It includes expressions and . Let's make a replacement
Back to the variable X. We get:
We have found all the roots of the original equation.
You can also meet logarithmic equations in task No. 5 of the Profile Unified State Examination in mathematics, and in task No. 13. And if in task No. 5 you need to solve the simplest equation, then in task 13 the solution consists of two points. The second point is the selection of roots on a given segment or interval.
