Logarithmic inequalities with a logarithm as the base. Manov's work "logarithmic inequalities in the Unified State Exam"
Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school. The presentation presents solutions to tasks C3 of the Unified State Exam - 2014 in mathematics.
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Solving logarithmic inequalities containing a variable in the base of the logarithm: methods, techniques, equivalent transitions, mathematics teacher, Secondary School No. 143 Knyazkina T.V.
Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved using a special formula, which for some reason is rarely taught in school: log k (x) f (x) ∨ log k (x) g (x) ⇒ (f (x) − g (x)) (k ( x) − 1) ∨ 0 Instead of the “∨” checkbox, you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same. This way we get rid of logarithms and reduce the problem to a rational inequality. The latter is much easier to solve, but when discarding logarithms, extra roots may appear. To cut them off, it is enough to find the area acceptable values. Don't forget the ODZ of the logarithm! Everything related to the range of acceptable values must be written out and solved separately: f (x) > 0; g(x) > 0; k(x) > 0; k(x) ≠ 1. These four inequalities constitute a system and must be satisfied simultaneously. When the range of acceptable values has been found, all that remains is to intersect it with the solution of the rational inequality - and the answer is ready.
Solve the inequality: Solution First, let's write out the OD of the logarithm. The first two inequalities are satisfied automatically, but the last one will have to be written down. Since the square of a number is equal to zero if and only if the number itself is equal to zero, we have: x 2 + 1 ≠ 1; x2 ≠ 0; x ≠ 0. It turns out that the ODZ of a logarithm is all numbers except zero: x ∈ (−∞0)∪(0 ;+ ∞). Now we solve the main inequality: We make the transition from the logarithmic inequality to the rational one. The original inequality has a “less than” sign, which means the resulting inequality must also have a “less than” sign.
We have: (10 − (x 2 + 1)) · (x 2 + 1 − 1)
Transforming Logarithmic Inequalities Often the original inequality is different from the one above. This can be easily corrected using standard rules for working with logarithms. Namely: Any number can be represented as a logarithm with a given base; The sum and difference of logarithms with the same bases can be replaced by one logarithm. Separately, I would like to remind you about the range of acceptable values. Since there may be several logarithms in the original inequality, it is required to find the VA of each of them. Thus, general scheme solutions to logarithmic inequalities are as follows: Find the ODZ of each logarithm included in the inequality; Reduce the inequality to a standard one using the formulas for adding and subtracting logarithms; Solve the resulting inequality using the scheme given above.
Solve the inequality: Solution Let's find the domain of definition (DO) of the first logarithm: Solve by the method of intervals. Find the zeros of the numerator: 3 x − 2 = 0; x = 2/3. Then - the zeros of the denominator: x − 1 = 0; x = 1. Mark zeros and signs on the coordinate line:
We get x ∈ (−∞ 2/3) ∪ (1; +∞). The second logarithm will have the same VA. If you don't believe it, you can check it. Now let's transform the second logarithm so that there is a two at the base: As you can see, the threes at the base and in front of the logarithm have been canceled. We got two logarithms with the same base. Add them up: log 2 (x − 1) 2
(f (x) − g (x)) (k (x) − 1)
We are interested in the intersection of sets, so we select intervals that are shaded on both arrows. We get: x ∈ (−1; 2/3) ∪ (1; 3) - all points are punctured. Answer: x ∈ (−1; 2/3)∪(1; 3)
Solving USE-2014 tasks type C3
Solve the system of inequalities. Solution. ODZ: 1) 2)
Solve the system of inequalities 3) -7 -3 - 5 x -1 + + + − − (continued)
Solve the system of inequalities 4) General solution: and -7 -3 - 5 x -1 -8 7 log 2 129 (continued)
Solve the inequality (continued) -3 3 -1 + − + − x 17 + -3 3 -1 x 17 -4
Solve the inequality Solution. ODZ:
Solve the inequality (continued)
Solve the inequality Solution. ODZ: -2 1 -1 + − + − x + 2 -2 1 -1 x 2
With them are inside logarithms.
Examples:
\(\log_3x≥\log_39\)
\(\log_3 ((x^2-3))< \log_3{(2x)}\)
\(\log_(x+1)((x^2+3x-7))>2\)
\(\lg^2((x+1))+10≤11 \lg((x+1))\)
How to solve logarithmic inequalities:
We should strive to reduce any logarithmic inequality to the form \(\log_a(f(x)) ˅ \log_a(g(x))\) (the symbol \(˅\) means any of ). This type allows you to get rid of logarithms and their bases, making the transition to the inequality of expressions under logarithms, that is, to the form \(f(x) ˅ g(x)\).
But when making this transition there is one very important subtlety:
\(-\) if is a number and it is greater than 1, the inequality sign remains the same during the transition,
\(-\) if the base is a number greater than 0 but less than 1 (lies between zero and one), then the inequality sign should change to the opposite, i.e.
\(\log_2((8-x))<1\) Solution: |
\(\log\)\(_(0.5)\) \((2x-4)\)≥\(\log\)\(_(0.5)\) \(((x+ 1))\) Solution: |
Very important! In any inequality, the transition from the form \(\log_a(f(x)) ˅ \log_a(g(x))\) to comparing expressions under logarithms can be done only if:
Example . Solve inequality: \(\log\)\(≤-1\)
Solution:
\(\log\) \(_(\frac(1)(3))(\frac(3x-2)(2x-3))\)\(≤-1\) |
Let's write out the ODZ. |
ODZ: \(\frac(3x-2)(2x-3)\) \(>0\) |
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\(\frac(3x-2-3(2x-3))(2x-3)\)\(≥\) \(0\) |
We open the brackets and bring . |
\(\frac(-3x+7)(2x-3)\) \(≥\) \(0\) |
We multiply the inequality by \(-1\), not forgetting to reverse the comparison sign. |
\(\frac(3x-7)(2x-3)\) \(≤\) \(0\) |
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\(\frac(3(x-\frac(7)(3)))(2(x-\frac(3)(2)))\)\(≤\) \(0\) |
Let's construct a number line and mark the points \(\frac(7)(3)\) and \(\frac(3)(2)\) on it. Please note that the dot is removed from the denominator, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituted into inequality it will lead us to division by zero. |
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Now we plot the ODZ on the same numerical axis and write down in response the interval that falls into the ODZ. |
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We write down the final answer. |
Example . Solve the inequality: \(\log^2_3x-\log_3x-2>0\)
Solution:
\(\log^2_3x-\log_3x-2>0\) |
Let's write out the ODZ. |
ODZ: \(x>0\) |
Let's get to the solution. |
Solution: \(\log^2_3x-\log_3x-2>0\) |
Here we have a typical square-logarithmic inequality. Let's do it. |
\(t=\log_3x\) |
We expand the left side of the inequality into . |
\(D=1+8=9\) |
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Now we need to return to the original variable - x. To do this, let's go to , which has the same solution, and make the reverse substitution. |
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\(\left[ \begin(gathered) t>2 \\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3x>2\\\log_3x<-1 \end{gathered} \right.\) |
Transform \(2=\log_39\), \(-1=\log_3\frac(1)(3)\). |
\(\left[ \begin(gathered) \log_3x>\log_39 \\ \log_3x<\log_3\frac{1}{3} \end{gathered} \right.\) |
Let's move on to comparing arguments. The bases of logarithms are greater than \(1\), so the sign of the inequalities does not change. |
\(\left[ \begin(gathered) x>9 \\ x<\frac{1}{3} \end{gathered} \right.\) |
Let us combine the solution to the inequality and the ODZ in one figure. |
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Let's write down the answer. |
Do you think that there is still time before the Unified State Exam and you will have time to prepare? Perhaps this is so. But in any case, the earlier a student begins preparation, the more successfully he passes the exams. Today we decided to devote an article to logarithmic inequalities. This is one of the tasks, which means an opportunity to get extra credit.
Do you already know what a logarithm is? We really hope so. But even if you don't have an answer to this question, it's not a problem. Understanding what a logarithm is is very simple.
Why 4? You need to raise the number 3 to this power to get 81. Once you understand the principle, you can proceed to more complex calculations.
You went through inequalities a few years ago. And since then you have constantly encountered them in mathematics. If you have problems solving inequalities, check out the appropriate section.
Now that we have become familiar with the concepts individually, let's move on to considering them in general.
The simplest logarithmic inequality.
The simplest logarithmic inequalities are not limited to this example; there are three more, only with different signs. Why is this necessary? To better understand how to solve inequalities with logarithms. Now let's give a more applicable example, still quite simple; we'll leave complex logarithmic inequalities for later.
How to solve this? It all starts with ODZ. It’s worth knowing more about it if you want to always easily solve any inequality.
What is ODZ? ODZ for logarithmic inequalities
The abbreviation stands for the range of acceptable values. This formulation often comes up in tasks for the Unified State Exam. ODZ will be useful to you not only in the case of logarithmic inequalities.
Look again at the above example. We will consider the ODZ based on it, so that you understand the principle, and solving logarithmic inequalities does not raise questions. From the definition of a logarithm it follows that 2x+4 must be greater than zero. In our case this means the following.
This number, by definition, must be positive. Solve the inequality presented above. This can even be done orally; here it is clear that X cannot be less than 2. The solution to the inequality will be the definition of the range of acceptable values.
Now let's move on to solving the simplest logarithmic inequality.
We discard the logarithms themselves from both sides of the inequality. What are we left with as a result? Simple inequality.
It's not difficult to solve. X must be greater than -0.5. Now we combine the two obtained values into a system. Thus,
This will be the range of acceptable values for the logarithmic inequality under consideration.
Why do we need ODZ at all? This is an opportunity to weed out incorrect and impossible answers. If the answer is not within the range of acceptable values, then the answer simply does not make sense. This is worth remembering for a long time, since in the Unified State Examination there is often a need to search for ODZ, and it concerns not only logarithmic inequalities.
Algorithm for solving logarithmic inequality
The solution consists of several stages. First, you need to find the range of acceptable values. There will be two meanings in the ODZ, we discussed this above. Next we need to solve the inequality itself. The solution methods are as follows:
- multiplier replacement method;
- decomposition;
- rationalization method.
Depending on the situation, it is worth using one of the above methods. Let's move directly to the solution. Let us reveal the most popular method, which is suitable for solving Unified State Examination tasks in almost all cases. Next we will look at the decomposition method. It can help if you come across a particularly tricky inequality. So, an algorithm for solving logarithmic inequality.
Examples of solutions :
It’s not for nothing that we took exactly this inequality! Pay attention to the base. Remember: if it is greater than one, the sign remains the same when finding the range of acceptable values; otherwise, you need to change the inequality sign.
As a result, we get the inequality:
Now we reduce the left side to the form of the equation equal to zero. Instead of the “less than” sign we put “equals” and solve the equation. Thus, we will find the ODZ. We hope that you will not have problems solving such a simple equation. The answers are -4 and -2. That's not all. You need to display these points on the graph, placing “+” and “-”. What needs to be done for this? Substitute the numbers from the intervals into the expression. Where the values are positive, we put “+” there.
Answer: x cannot be greater than -4 and less than -2.
We have found the range of acceptable values only for the left side; now we need to find the range of acceptable values for the right side. This is much easier. Answer: -2. We intersect both resulting areas.
And only now are we beginning to address the inequality itself.
Let's simplify it as much as possible to make it easier to solve.
We again use the interval method in the solution. Let’s skip the calculations; everything is already clear with it from the previous example. Answer.
But this method is suitable if the logarithmic inequality has the same bases.
Solving logarithmic equations and inequalities with different bases requires an initial reduction to the same base. Next, use the method described above. But there is a more complicated case. Let's consider one of the most complex types of logarithmic inequalities.
Logarithmic inequalities with variable base
How to solve inequalities with such characteristics? Yes, and such people can be found in the Unified State Examination. Solving inequalities in the following way will also have a beneficial effect on your educational process. Let's look at the issue in detail. Let's discard theory and go straight to practice. To solve logarithmic inequalities, it is enough to familiarize yourself with the example once.
To solve a logarithmic inequality of the form presented, it is necessary to reduce the right-hand side to a logarithm with the same base. The principle resembles equivalent transitions. As a result, the inequality will look like this.
Actually, all that remains is to create a system of inequalities without logarithms. Using the rationalization method, we move on to an equivalent system of inequalities. You will understand the rule itself when you substitute the appropriate values and track their changes. The system will have the following inequalities.
When using the rationalization method when solving inequalities, you need to remember the following: one must be subtracted from the base, x, by definition of the logarithm, is subtracted from both sides of the inequality (right from left), two expressions are multiplied and set under the original sign in relation to zero.
The further solution is carried out using the interval method, everything is simple here. It is important for you to understand the differences in solution methods, then everything will start to work out easily.
There are many nuances in logarithmic inequalities. The simplest of them are quite easy to solve. How can you solve each of them without problems? You have already received all the answers in this article. Now you have a long practice ahead of you. Constantly practice solving a variety of problems in the exam and you will be able to get the highest score. Good luck to you in your difficult task!