Question

Solve: ${\log }_2{\log }_{\left(0.5\right)}(2^x-\frac{15}{16})\le 2$

• A. $0\le x<{\log }_2\frac{31}{16}$
• B. $0\le x<{\log }_5\frac{31}{16}$
• C. $0\le x<{\log }_2\frac{1}{16}$
• D. None of these

Answer 1

The correct option is A $0\le x<{\log }_2\frac{31}{16}$

Given expression is,

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

$\to loglo{g}_{(}0.5\left)\right(2^x-\frac{15}{16})\le 2$

$\to lo{g}_{(}0.5\left)\right(2^x-\frac{15}{16})\le$

$\to log(2^x-\frac{15}{16})=2\times 0.5$

$\to log(2^x-\frac{15}{16})=1$

$\to 2^x-\frac{15}{16}=1$

$\to 2^x=1+\frac{15}{16}$

$\to 0\le 2^x<le\frac{31}{16}\to )\le xlog2\le log\frac{31}{16}$

Or

$0\le x\le lo{g}_2\left(\frac{31}{16}\right)$