All about logarithmic inequalities. Analysis of examples

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 range of acceptable values. If you have forgotten the ODZ of a logarithm, I strongly recommend repeating it - see “What is a 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.

Task. Solve the inequality:

First, let’s write out the logarithm’s ODZ:

The first two inequalities are satisfied automatically, but the last one will have to be written out. Since the square of a number is zero if and only if the number itself is zero, we have:

x 2 + 1 ≠ 1;
x2 ≠ 0;
x ≠ 0.

It turns out that the ODZ of the logarithm is all numbers except zero: x ∈ (−∞ 0)∪(0; +∞). Now we solve the main inequality:

We make the transition from logarithmic inequality to 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)< 0;
(9 − x 2) x 2< 0;
(3 − x) · (3 + x) · x 2< 0.

The zeros of this expression are: x = 3; x = −3; x = 0. Moreover, x = 0 is a root of the second multiplicity, which means that when passing through it, the sign of the function does not change. We have:

We get x ∈ (−∞ −3)∪(3; +∞). This set is completely contained in the ODZ of the logarithm, which means this is the answer.

Converting logarithmic inequalities

Often the original inequality is different from the one above. This can be easily corrected using the standard rules for working with logarithms - see “Basic properties of 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, the general scheme for solving logarithmic inequalities is as follows:

Find the VA 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.

Task. Solve the inequality:

Let's find the domain of definition (DO) of the first logarithm:

We solve using the interval method. Finding the zeros of the numerator:

3x − 2 = 0;
x = 2/3.

Then - the zeros of the denominator:

x − 1 = 0;
x = 1.

We mark zeros and signs on the coordinate arrow:

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 we transform the second logarithm so that the base is two:

As you can see, the threes at the base and in front of the logarithm have been reduced. We got two logarithms with the same base. Let's add them up:

log 2 (x − 1) 2< 2;
log 2 (x − 1) 2< log 2 2 2 .

We obtained the standard logarithmic inequality. We get rid of logarithms using the formula. Since the original inequality contains a “less than” sign, the resulting rational expression must also be less than zero. We have:

(f (x) − g (x)) (k (x) − 1)< 0;
((x − 1) 2 − 2 2)(2 − 1)< 0;
x 2 − 2x + 1 − 4< 0;
x 2 − 2x − 3< 0;
(x − 3)(x + 1)< 0;
x ∈ (−1; 3).

We got two sets:

ODZ: x ∈ (−∞ 2/3)∪(1; +∞);
Candidate answer: x ∈ (−1; 3).

It remains to intersect these sets - we get the real answer:

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.

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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 range of 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, the general scheme for solving logarithmic inequalities is as follows: Find the VA 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:

$\log_3⁡x≥\log_3⁡9$
$\log_3⁡ ((x^2-3))< \log_3⁡{(2x)}$
$\log_(x+1)⁡((x^2+3x-7))>2$
$\lg^2⁡((x+1))+10≤11 \lg⁡((x+1))$

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.

Examples:

$\log_2⁡((8-x))<1$
ODZ: $8-x>0$
$-x>-8$
$x<8$

Solution:
$\log$$_2$ $(8-x)<\log$$_2$ ${2}$
$8-x$$<$ $2$
$8-2\(x>6$
Answer: $(6;8)$

$\log$$_(0.5⁡)$ $(2x-4)$≥$\log$$_(0.5)$ ⁡$((x+ 1))$
ODZ: $\begin(cases)2x-4>0\\x+1 > 0\end(cases)$
$\begin(cases)2x>4\\x > -1\end(cases)$ $\Leftrightarrow$ $\begin(cases)x>2\\x > -1\end(cases) $ $\Leftrightarrow$ $x\in(2;\infty)$

Solution:
$2x-4$$≤$ $x+1$
$2x-x≤4+1$
$x≤5$
Answer: $(2;5]$

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))$$≤-1$	Let's write out the ODZ.
ODZ: $\frac(3x-2)(2x-3)$ $>0$
$⁡\frac(3x-2-3(2x-3))(2x-3)$$≥$ $0$	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.
$⁡\frac(3x-7)(2x-3)$ $≤$ $0$
$⁡\frac(3(x-\frac(7)(3)))(2(x-\frac(3)(2)))$$≤$ $0$	Let's construct a number line and mark the points $\frac(7)(3)$ and $\frac(3)(2)$ on it. Please note that the dot is removed from the denominator, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituted into inequality it will lead us to division by zero.
$x∈($$\frac(3)(2)$ $;$$\frac(7)(3)]$	Now we plot the ODZ on the same numerical axis and write down in response the interval that falls into the ODZ.
	We write down the final answer.

Answer: $x∈($$\frac(3)(2)$ $;$$\frac(7)(3)]$

Example . Solve the inequality: $\log^2_3⁡x-\log_3⁡x-2>0$

Solution:

$\log^2_3⁡x-\log_3⁡x-2>0$	Let's write out the ODZ.
ODZ: $x>0$	Let's get to the solution.
Solution: $\log^2_3⁡x-\log_3⁡x-2>0$	Here we have a typical square-logarithmic inequality. Let's do it.
$t=\log_3⁡x$ $t^2-t-2>0$	We expand the left side of the inequality into .
$D=1+8=9$ $t_1= \frac(1+3)(2)=2$ $t_2=\frac(1-3)(2)=-1$ $(t+1)(t-2)>0$
	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.
$\left[ \begin(gathered) t>2 \\ t<-1 \end{gathered} \right.$ $\Leftrightarrow$ $\left[ \begin{gathered} \log_3⁡x>2\\\log_3⁡x<-1 \end{gathered} \right.$	Transform $2=\log_3⁡9$, $-1=\log_3⁡\frac(1)(3)$.
$\left[ \begin(gathered) \log_3⁡x>\log_39 \\ \log_3⁡x<\log_3\frac{1}{3} \end{gathered} \right.$	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.
	Let's write down the answer.

Answer: $(0; \frac(1)(3))∪(9;∞)$

We looked at solving the simplest logarithmic inequalities and inequalities where the base of the logarithm is fixed in the last lesson.

But what if there is a variable at the base of the logarithm?

Then it will come to our aid rationalization of inequalities. To understand how this works, let's consider, for example, the inequality:

$$\log_(2x) x^2 > \log_(2x) x.$$

As expected, let's start with ODZ.

ODZ

$$\left[ \begin(array)(l)x>0,\\ 2x ≠ 1. \end(array)\right.$$

Solution to inequality

Let's reason as if we were solving an inequality with a fixed base. If the base is greater than one, we get rid of logarithms, and the inequality sign does not change; if it is less than one, it changes.

Let's write this as a system:

$$\left[ \begin(array)(l) \left\( \begin(array)(l)2x>1,\\ x^2 > x; \end(array)\right. \\ \left\ ( \begin(array)(l)2x<1,\\ x^2 < x; \end{array}\right. \end{array} \right.$$

For further reasoning, let us move all the right-hand sides of the inequalities to the left.

$$\left[ \begin(array)(l) \left\( \begin(array)(l)2x-1>0,\\ x^2 -x>0; \end(array)\right. \ \ \left\( \begin(array)(l)2x-1<0,\\ x^2 -x<0; \end{array}\right. \end{array} \right.$$

What did we get? It turns out that we need the expressions `2x-1` and `x^2 - x` to be either positive or negative at the same time. The same result will be obtained if we solve the inequality:

$$(2x-1)(x^2 - x) >0.$$

This inequality, like the original system, is true if both factors are either positive or negative. It turns out that you can move from a logarithmic inequality to a rational one (taking into account the ODZ).

Let's formulate method for rationalizing logarithmic inequalities$$\log_(f(x)) g(x) \vee \log_(f(x)) h(x) \Leftrightarrow (f(x) - 1)(g(x)-h(x)) \ vee 0,$$ where `\vee` is any inequality sign. (For the `>` sign, we have just checked the validity of the formula. For the rest, I suggest you check it yourself - it will be remembered better).

Let's return to solving our inequality. Expanding it into brackets (to make the zeros of the function easier to see), we get

$$(2x-1)x(x - 1) >0.$$

The interval method will give the following picture:

(Since the inequality is strict and we are not interested in the ends of the intervals, they are not shaded.) As can be seen, the resulting intervals satisfy the ODZ. We received the answer: `(0,\frac(1)(2)) \cup (1,∞)`.

Example two. Solving a logarithmic inequality with a variable base

$$\log_(2-x) 3 \leqslant \log_(2-x) x.$$

ODZ

$$\left\(\begin(array)(l)2-x > 0,\\ 2-x ≠ 1,\\ x > 0. \end(array)\right.$$

$$\left\(\begin(array)(l)x< 2,\\ x ≠ 1, \\ x >0.\end(array)\right.$$

Solution to inequality

According to the rule we just received rationalization of logarithmic inequalities, we find that this inequality is identical (taking into account the ODZ) to the following:

$$(2-x -1) (3-x) \leqslant 0.$$

$$(1-x) (3-x) \leqslant 0.$$

Combining this solution with the ODZ, we get the answer: `(1,2)`.

Third example. Logarithm of a fraction

$$\log_x\frac(4x+5)(6-5x) \leqslant -1.$$

ODZ

$$\left\(\begin(array)(l) \dfrac(4x+5)(6-5x)>0, \\ x>0,\\ x≠ 1.\end(array) \right.$ $

Since the system is relatively complex, let's immediately plot the solution to the inequalities on the number line:

Thus, ODZ: `(0,1)\cup \left(1,\frac(6)(5)\right)`.

Solution to inequality

Let's represent `-1` as a logarithm with base `x`.

$$\log_x\frac(4x+5)(6-5x) \leqslant \log_x x^(-1).$$

By using rationalization of logarithmic inequality we get a rational inequality:

$$(x-1)\left(\frac(4x+5)(6-5x) -\frac(1)(x)\right)\leqslant0,$$

$$(x-1)\left(\frac(4x^2+5x - 6+5x)(x(6-5x))\right)\leqslant0,$$

$$(x-1)\left(\frac(2x^2+5x - 3)(x(6-5x))\right)\leqslant0.$$

\(\log\) \(_(\frac(1)(3))⁡(\frac(3x-2)(2x-3))\)\(≤-1\)	Let's write out the ODZ.
ODZ: \(\frac(3x-2)(2x-3)\) \(>0\)
\(⁡\frac(3x-2-3(2x-3))(2x-3)\)\(≥\) \(0\)	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.
\(⁡\frac(3x-7)(2x-3)\) \(≤\) \(0\)
\(⁡\frac(3(x-\frac(7)(3)))(2(x-\frac(3)(2)))\)\(≤\) \(0\)	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.
\(x∈(\)\(\frac(3)(2)\) \(;\)\(\frac(7)(3)]\)	Now we plot the ODZ on the same numerical axis and write down in response the interval that falls into the ODZ.
	We write down the final answer.

\(\log^2_3⁡x-\log_3⁡x-2>0\)	Let's write out the ODZ.
ODZ: \(x>0\)	Let's get to the solution.
Solution: \(\log^2_3⁡x-\log_3⁡x-2>0\)	Here we have a typical square-logarithmic inequality. Let's do it.
\(t=\log_3⁡x\) \(t^2-t-2>0\)	We expand the left side of the inequality into .
\(D=1+8=9\) \(t_1= \frac(1+3)(2)=2\) \(t_2=\frac(1-3)(2)=-1\) \((t+1)(t-2)>0\)
	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.
\(\left[ \begin(gathered) t>2 \\ t<-1 \end{gathered} \right.\) \(\Leftrightarrow\) \(\left[ \begin{gathered} \log_3⁡x>2\\\log_3⁡x<-1 \end{gathered} \right.\)	Transform \(2=\log_3⁡9\), \(-1=\log_3⁡\frac(1)(3)\).
\(\left[ \begin(gathered) \log_3⁡x>\log_39 \\ \log_3⁡x<\log_3\frac{1}{3} \end{gathered} \right.\)	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.
	Let's write down the answer.