Change of Base of Logarithm

Theorem

Let $\log_a x$ be the logarithm to base $a$ of $x$.

Then:

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

Thus a convenient formula for calculating the logarithm of a number to a different base.

Proof

Let:

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

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle z\)	\(=\)	\(\displaystyle \log_a \left({b^y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle y \ \log_a b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by Logarithms of Powers
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \log_b x \ \log_a b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence the result.

$\blacksquare$

Note

Some people prefer to write this as:

$\log_a x = \log_a b \ \log_b x$

as it is delightfully easy to remember.

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.2$