Change of Base of Logarithm/Base 2 to Base 8

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:

The result follows.