Question

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

Answer: (A)

$x\in [{\log }_2\frac{1}{15},{\log }_2\frac{16}{15})$

Given,

$0<{\log }_{2}lo{g}_{\left(0.5\right)}\left(2^x\frac{15}{16}\right)\le 2$

$=0<{\log }_{0.5}\left(2^x\frac{15}{16}\right)\le 4$

$=(\because lo{g}_a(b)=x\Rightarrow b=a^x)$

$=0>{\log }_2{2}^x\left(\frac{15}{16}\right)\ge -4$

$=1>2^x\left(\frac{15}{16}\right)\ge \frac{1}{16}$

$=\frac{16}{15}>2^x\ge \frac{1}{15}$

$takinglog,weget$

$x\in [{\log }_2\frac{1}{15},{\log }_2\frac{16}{15})$