Logarithm of Reciprocal
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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.
$\blacksquare$