Question

Solve the following inequality:
${\log _{0.5}}(4 - x) \geqslant {\log _{0.5}}2 - {\log _{0.5}}(x - 1)$

Answer 1

This type of question is done on the basis of the logarithm rule in which we have to use the logarithm formula.
This question also contains inequality in which we have to solve the question considering inequality sign.
Logarithm inequalities are the inequalities in which both sides or one side contain a logarithm.
FORMULA TO BE USED:
$\log m + \log n = \log mn$
$\log m - \log n = \log \dfrac{m}{n}$
Antilog formula:
$anti\log (y) = x$
${10^x} = y$

Complete answer:
In this question,
Firstly, we change the base $0.5 \to 2$
${\log _{\dfrac{1}{2}}}(4 - x) \geqslant {\log _{\dfrac{1}{2}}}2 - {\log _{\dfrac{1}{2}}}(x - 1)$
$- {\log _2}(4 - x) \geqslant - {\log _2}2 - {\log _2}(x - 1)$
Now, we have to analyse the inequality sign and we have to assemble the like logarithm on one side,
${\log _2}2 - {\log _2}(4 - x) \geqslant {\log _2}(x - 1)$
Now, we will take antilog both the side, we will get,
$\dfrac{2}{{4 - x}} \geqslant x - 1$
Now, bring RHS to LHS and solve the equation,
$\dfrac{2}{{4 - x}} - x - 1 \geqslant 0$
$\dfrac{{2 - (4 - x)(x - 1)}}{{4 - x}} \geqslant 0$
$\dfrac{{2 - (4x - 4 - {x^2} + x)}}{{4 - x}} \geqslant 0$
$\dfrac{{2 - 5x + {x^2} + 4}}{{4 - x}} \geqslant 0$
Now,
$\dfrac{{{x^2} - 5x + 6}}{{4 - x}} \leqslant 0$
$\dfrac{{(x - 3)(x - 2)}}{{4 - x}} \leqslant 0$
Therefore, when we will solve the equation, we get
$x \in ( - \infty ,2] \cup [3,4)$
It means that $x$contains a value from $- \infty$to $4$
Now, as $x < 4$ and $x > 1$
It gives that $x$belongs to open interval of $1$ and $4$i.e., $x \in (1,4)$
So, we can conclude that,
$x \in (1,2] \cup [3,4)$
Hence, the above given is the answer.

Note:
While doing this type of question, one should keep in mind all the formulas related to the logarithm and also keep in mind that how inequality is changing while changing signs so that there will be no mistake in the steps.
Also, one should know how to change base in logarithm, if you don’t know the knowledge of base change you will be unable to do the question.
You also need to know how to keep the value of $x$ in form of closed and open intervals because this question contains how the value of $x$ belongs in the form of interval.