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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $18$