Question

If $log_{2}\,6+\frac{1}{2x} = log_{2}\left(2^{\frac{1}{x}} + 8\right)$, 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)

$\frac{1}{4}, \frac{1}{2}$

Answer 2

$log_{2}\,6+log_{2}\, 2^{\frac{1}{2x}} = log_{2}\left(2^{\frac{1}{x}}+8\right)\,\Rightarrow 6\cdot2^{\frac{1}{2x}} = 2^{\frac{1}{x}}+8,\,\, let\, 2^{\frac{1}{2x}} = a$
$\Rightarrow a^{2}-6a+8 = 0 \,\Rightarrow\,a = 2\,\Rightarrow\,x = \frac{1}{4}, \frac{1}{2}$