Change of Base of Logarithm/Base 2 to Base 8

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Theorem

Let $\log_8 x$ be the logarithm base $8$ of $x$.

Let $\lg x$ be the binary (base $2$) logarithm of $x$.


Then:

$\log_8 x = \dfrac {\lg x} 3$


Proof

From Change of Base of Logarithm:

$\log_a x = \dfrac {\log_b x} {\log_b a}$

Substituting $a = 8$ and $b = 2$ gives:

$\log_8 x = \dfrac {\log_2 x} {\log_2 8}$


We have that:

\(\ds 2^3\) \(=\) \(\ds 8\)
\(\ds \leadsto \ \ \) \(\ds \lg 8\) \(=\) \(\ds \log_2 2^3\)
\(\ds \) \(=\) \(\ds 3 \log_2 2\)
\(\ds \) \(=\) \(\ds 3\) Logarithm to Own Base equals 1

The result follows.

$\blacksquare$


Sources