Change of Base of Logarithm/Proof 1

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Theorem

$\log_b x = \dfrac {\log_a x} {\log_a b}$


Proof

Let:

$y = \log_b x \iff b^y = x$
$z = \log_a x \iff a^z = x$

Then:

\(\ds z\) \(=\) \(\ds \map {\log_a} {b^y}\)
\(\ds \) \(=\) \(\ds y \log_a b\) Logarithms of Powers
\(\ds \) \(=\) \(\ds \log_b x \log_a b\)

Hence the result.

$\blacksquare$


Sources


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