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Prove or disprove: $\begin{align*} \log_8x = \frac{1}{2}.\log_{2}x \end{align*}$.
asked in Set Theory & Algebra by Veteran (56.9k points)
edited by | 106 views
+1

log8x = log2x/log28 = 1/3 *log2x . Am I missing something or this is what you asked?

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+4 votes
Best answer
$\log_{8}x=\frac{1}{\log_{x}8} =\frac{1}{3\log_{x}2}=\frac{1}{3}\log_{2}x$

$\log_{8}x\neq \frac{1}{2}\log_{2}x$
answered by Veteran (92k points)
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@Srestha, which log property have you applied?  How could you interchange the base x with tat of  8? Please specify. :). Thanx in advance as well. :)
+1
$\log _{y}x=\frac{1}{\log _{x}y}$
+1 vote

log8x = log2x/log28 = 1/3 *log2x . Am I missing something or this is what you asked?

answered by Active (3.8k points)
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