Logarithm of Reciprocal

Theorem

Let $x, b \in \R$ be strictly positive real numbers such that $b > 1$.


Then:

$\map {\log_b} {\dfrac 1 x} = -\log_b x$

where $\log_b$ denotes the logarithm to base $b$.


Proof

\(\ds \map {\log_b} {\dfrac 1 x}\) \(=\) \(\ds \log_b 1 - \log_b x\) Difference of Logarithms
\(\ds \) \(=\) \(\ds 0 - \log_b x\) Logarithm of 1 is 0

Hence the result.

$\blacksquare$

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