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
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$