Question

If $lo{g}_{2}6+\frac{1}{2x}=lo{g}_2\left(2^{\frac{1}{x}}+8\right)$, then the values 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

Answer: (B)

$lo{g}_{2}6+lo{g}_2{2}^{\frac{1}{2x}}=lo{g}_2\left(2^{\frac{1}{x}}+8\right)\Rightarrow 6\cdot 2^{\frac{1}{2x}}=2^{\frac{1}{x}}+8,let{2}^{\frac{1}{2x}}=a$
$\Rightarrow a^2-6a+8=0\Rightarrow a=2\Rightarrow x=\frac{1}{4},\frac{1}{2}$