How to solve inequalities with USE logarithms. All about logarithmic inequalities. Analysis of examples

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)))$$≤$ $0$	Let's construct a number line and mark the points $\frac(7)(3)$ and $\frac(3)(2)$ on it. Please note that the point from the denominator is removed, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituted into inequality it will lead us to division by zero.
$x∈($$\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: $(0; \frac(1)(3))∪(9;∞)$

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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. 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;
x 2 ≠ 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 me, 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.

An inequality is called logarithmic if it contains a logarithmic function.

Methods for solving logarithmic inequalities are no different from, except for two things.

Firstly, when moving from the logarithmic inequality to the inequality of sublogarithmic functions, one should follow the sign of the resulting inequality. It obeys the following rule.

If the base of the logarithmic function is greater than $1$, then when moving from the logarithmic inequality to the inequality of sublogarithmic functions, the sign of the inequality is preserved, but if it is less than $1$, then it changes to the opposite.

Secondly, the solution to any inequality is an interval, and, therefore, at the end of solving the inequality of sublogarithmic functions it is necessary to create a system of two inequalities: the first inequality of this system will be the inequality of sublogarithmic functions, and the second will be the interval of the domain of definition of the logarithmic functions included in the logarithmic inequality.

Practice.

Let's solve the inequalities:

1. $\log_(2)((x+3)) \geq 3.$

$D(y): \x+3>0.$

$x \in (-3;+\infty)$

The base of the logarithm is $2>1$, so the sign does not change. Using the definition of logarithm, we get:

$x+3 \geq 2^(3),$

$x \in )

\(\log\) \(_(\frac(1)(3))⁡(\frac(3x-2)(2x-3))\)\(≤-1\)	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 point from the denominator is removed, 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.