Change of Base of Logarithm

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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:

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

Hence the result.

$\blacksquare$


Also presented as

Some people prefer to write this as:

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

as it is delightfully easy to remember.


Examples

Base $10$ to Base $e$

Form 1

$\ln x = \paren {\ln 10} \paren {\log_{10} x} = 2 \cdotp 30258 \, 50929 \, 94 \ldots \log_{10} x$


Form 2

$\ln x = \dfrac {\log_{10} x} {\log_{10} e} = \dfrac {\log_{10} x} {0 \cdotp 43429 \, 44819 \, 03 \ldots}$


Base $e$ to Base $10$

Form 1

$\log_{10} x = \paren {\log_{10} e} \paren {\ln x} = 0 \cdotp 43429 \, 44819 \, 03 \ldots \ln x$


Form 2

$\log_{10} x = \dfrac {\ln x} {\ln 10} = \dfrac {\ln x} {2 \cdotp 30258 \, 50929 \, 94 \ldots}$


Base $2$ to Base $10$

$\log_{10} x = \left({\lg x}\right) \left({\log_{10} 2}\right) = 0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots \lg x$


Base $10$ to Base $2$

$\lg x = \dfrac {\log_{10} x} {\log_{10} 2} = \dfrac {\log_{10} x} {0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots}$


Base $2$ to Base $8$

$\log_8 x = \dfrac {\lg x} 3$


Sources

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