Sum of Logarithms/General Logarithm/Proof 2

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Theorem

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


Then:

$\log_b x + \log_b y = \map {\log_b} {x y}$

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


Proof

\(\ds \log_b x + \log_b y\) \(=\) \(\ds \frac {\ln x} {\ln b} + \frac {\ln y} {\ln b}\) Change of Base of Logarithm
\(\ds \) \(=\) \(\ds \frac {\ln x + \ln y} {\ln b}\)
\(\ds \) \(=\) \(\ds \frac {\map \ln {x y} } {\ln b}\) Sum of Logarithms: Proof for Natural Logarithm
\(\ds \) \(=\) \(\ds \map {\log_b} {x y}\) Change of Base of Logarithm

$\blacksquare$