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Equivalent transition of the logarithm. Solving simple logarithmic inequalities

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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:

Any logarithmic inequality should be reduced to the form $\log_a⁡(f(x)) ˅ \log_a(⁡g(x))$ (symbol $˅$ means any of ). This form allows us to get rid of logarithms and their bases by passing 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 - 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 (between zero and one), then the inequality sign must be reversed, 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)$ ⁡$((x+ one))$
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 only be done if:

Example . Solve the 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, give .
$⁡\frac(-3x+7)(2x-3)$ $≥$ $0$	We multiply the inequality by $-1$, remembering to reverse the comparison sign.
$⁡\frac(3x-7)(2x-3)$ $≤$ $0$
$⁡\frac(3(x-\frac(7)(3)))(2(x-\frac(3)(2)))$$≤$ $0$	Let's build a number line and mark the points $\frac(7)(3)$ and $\frac(3)(2)$ on it. Note that the point from the denominator is punctured, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituting into an inequality, it will lead us to division by zero.
$x∈($$\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.
	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: $\log^2_3⁡x-\log_3⁡x-2>0$	Before us is a typical square-logarithmic inequality. We do.
$t=\log_3⁡x$ $t^2-t-2>0$	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 you need to return to the original variable - x. To do this, we pass 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's combine the solution of the inequality and the ODZ in one figure.
	Let's write down the answer.

Answer: $(0; \frac(1)(3))∪(9;∞)$

Among the whole variety of logarithmic inequalities, inequalities with a variable base are studied separately. They are solved according to a special formula, which for some reason is rarely taught at school:

log k (x ) f (x ) ∨ log k (x ) g (x ) ⇒ (f (x ) − g (x )) (k (x ) − 1) ∨ 0

Instead of a jackdaw "∨", you can put any inequality sign: more or less. The main thing is that in both inequalities the signs are the same.

So 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 admissible values. If you forgot the ODZ of the 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 fulfilled simultaneously. When the range of acceptable values is found, it remains to cross it with the solution of a rational inequality - and the answer is ready.

A task. Solve the inequality:

First, let's write the ODZ of the logarithm:

The first two inequalities are performed automatically, and the last one will have to be written. 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 perform the transition from the logarithmic inequality to the rational one. In the original inequality there is a “less than” sign, so the resulting inequality should also be with a “less than” sign. We have:

(10 − (x 2 + 1)) (x 2 + 1 − 1)< 0;
(9 − x2) x2< 0;
(3 − x) (3 + x) x 2< 0.

Zeros of this expression: x = 3; x = -3; x = 0. Moreover, x = 0 is the 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 that this is the answer.

Transformation of logarithmic inequalities

Often the original inequality differs from the one above. This is easy to fix according to 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 base can be replaced by a single logarithm.

Separately, I want 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 DPV of each of them. Thus, the general scheme for solving logarithmic inequalities is as follows:

Find the ODZ of each logarithm included in the inequality;
Reduce the inequality to the standard one using the formulas for adding and subtracting logarithms;
Solve the resulting inequality according to the scheme above.

A task. Solve the inequality:

Find the domain of definition (ODZ) of the first logarithm:

We solve by 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 of the ODZ will be the same. If you don't believe me, you can check. Now we transform the second logarithm so that the base is two:

As you can see, the triples at the base and before the logarithm have shrunk. Get two logarithms with the same base. Let's put them together:

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

We have obtained the standard logarithmic inequality. We get rid of the logarithms by the formula. Since there is a less than sign in the original inequality, 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; +∞);
Answer candidate: x ∈ (−1; 3).

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

We are interested in the intersection of sets, so we choose the intervals 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.

First, when passing from the logarithmic inequality to the inequality of sublogarithmic functions, it follows 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 passing from the logarithmic inequality to the inequality of sublogarithmic functions, the inequality sign is preserved, and if it is less than $1$, then it is reversed.

Secondly, the solution of any inequality is an interval, and, therefore, at the end of the solution of the inequality of sublogarithmic functions, it is necessary to compose 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 the logarithm, we get:

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

$x \in )

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\(\log\) \(_(\frac(1)(3))⁡(\frac(3x-2)(2x-3))\)\(≤-1\)	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, give .
\(⁡\frac(-3x+7)(2x-3)\) \(≥\) \(0\)	We multiply the inequality by \(-1\), remembering 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 build a number line and mark the points \(\frac(7)(3)\) and \(\frac(3)(2)\) on it. Note that the point from the denominator is punctured, despite the fact that the inequality is not strict. The fact is that this point will not be a solution, since when substituting into an 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.
	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: \(\log^2_3⁡x-\log_3⁡x-2>0\)	Before us is a typical square-logarithmic inequality. We do.
\(t=\log_3⁡x\) \(t^2-t-2>0\)	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 you need to return to the original variable - x. To do this, we pass 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's combine the solution of the inequality and the ODZ in one figure.
	Let's write down the answer.