Question

If ${\log }_{2}6+\frac{1}{2x}={\log }_2(2^{\frac{1}{x}}+8),$ then the value of $x$ are

• A. $\frac{1}{4},\frac{1}{3}$
• B. $\frac{1}{4},\frac{1}{2}$
• C. $-\frac{1}{4},\frac{1}{2}$
• D. $\frac{1}{3},-\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{4},\frac{1}{2}$

${\log }_{2}6+\frac{1}{2x}={\log }_2(2^{\frac{1}{x}}+8)$
$\Rightarrow {\log }_2(2^{\frac{1}{x}}+8)-{\log }_{2}6=\frac{1}{2x}\Rightarrow {\log }_2\left(\frac{2^{\frac{1}{x}}+8}{6}\right)=\frac{1}{2x}$
$\Rightarrow \frac{2^{\frac{1}{x}}+8}{6}=2^{\frac{1}{2x}}\Rightarrow 2^{\frac{1}{x}}+8=6\cdot 2^{\frac{1}{2x}}$
Let $2^{\frac{1}{2x}}=u$
$u^2+8-6u=0$
$\Rightarrow u=2,4$
$\Rightarrow x=\frac{1}{2},\frac{1}{4}$