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$\log_{8}x=\frac{1}{\log_{x}8} =\frac{1}{3\log_{x}2}=\frac{1}{3}\log_{2}x$

$\log_{8}x\neq \frac{1}{2}\log_{2}x$
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log8x = log2x/log28 = 1/3 *log2x . Am I missing something or this is what you asked?

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