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:
\(\ds z\) | \(=\) | \(\ds \map {\log_a} {b^y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y \log_a b\) | Logarithms of Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds \log_b x \log_a b\) |
Hence the result.
$\blacksquare$
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 7$: Change of Base of Logarithms: $7.13$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: $(14)$