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.

$\blacksquare$