Question

The range of values of $x$ that satisfies the inequation ${\log }_2{\log }_{0.5}\left(2^x\frac{15}{16}\right)\le 2$ is

• A. $x\in [{\log }_2\frac{1}{15},{\log }_2\frac{16}{15})$
• B. $x\in ({\log }_2\frac{1}{15},{\log }_2\frac{16}{15})$
• C. $x\in [{\log }_2\frac{1}{15},{\log }_2\frac{16}{15}]$
• D. $x\in ({\log }_2\frac{1}{15},{\log }_2\frac{16}{15}]$

Answer 1

The correct option is A $x\in [{\log }_2\frac{1}{15},{\log }_2\frac{16}{15})$

For $\log$ to be defined,
${\log }_{0.5}\left(2^x\frac{15}{16}\right)>0$
$\Rightarrow 2^x\frac{15}{16}<1$
$\Rightarrow 2^x<\frac{16}{15}$
By taking $\log$ with base $2$, we get
$\Rightarrow x<{\log }_2\frac{16}{15}$
$\Rightarrow x\in (-\infty ,{\log }_2\frac{16}{15})\cdots \left(1\right)$

Given: ${\log }_2{\log }_{0.5}\left(2^x\frac{15}{16}\right)\le 2$
$\Rightarrow {\log }_{0.5}\left(2^x\frac{15}{16}\right)\le 4$
$\Rightarrow 2^x\left(\frac{15}{16}\right)\ge \frac{1}{16}$
$\Rightarrow 2^x\ge \frac{1}{15}$
By taking $\log$ with base $2$, we get
$\Rightarrow x\ge {\log }_2\frac{1}{15}$
$\Rightarrow x\in [{\log }_2\frac{1}{15},\infty )\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$x\in [{\log }_2\frac{1}{15},{\log }_2\frac{16}{15})$