Discrete Math

Question

Prove or disprove: $\begin{align*} \log_8x = \frac{1}{2}.\log_{2}x \end{align*}$.

Comments

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

Answer 1

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

$\log_{8}x\neq \frac{1}{2}\log_{2}x$

Comments

@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. :)

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$\log _{y}x=\frac{1}{\log _{x}y}$

Answer 2

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