Logarithm of Base
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Theorem
Let $b \in \R_{>0}$ such that $b \ne 1$.
Then:
- $\log_b b = 1$
where $\log_b$ denotes the logarithm to base $b$.
Proof
Let $a \in \R_{>0}$ such that $a \ne 1$.
Then:
\(\ds \log_b b\) | \(=\) | \(\ds \dfrac {\log_a b} {\log_a b}\) | Change of Base of Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$