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  4. If log2 6+(1/2x) = log2(2(1/x) + 8), then the values of x are

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Q. If $log_{2}\,6+\frac{1}{2x} = log_{2}\left(2^{\frac{1}{x}} + 8\right)$, then the values of x are

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Solution:

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