Approximation to Binary Logarithm from Natural and Common Logarithm

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Theorem

The binary logarithm $\lg x$ can be approximated, to within $1 \%$, by the expression:

$\lg x \approx \ln x + \log_{10} x$

That is, by the sum of the natural logarithm and common logarithm.


Proof

\(\ds \lg x\) \(=\) \(\ds \frac {\ln x} {\ln 2}\) Change of Base of Logarithm
\(\ds \) \(=\) \(\ds \frac {\log_{10} x} {\log_{10} 2}\) Change of Base of Logarithm
\(\ds \leadsto \ \ \) \(\ds \lg x\) \(=\) \(\ds \frac {\ln x + \log_{10} x} {\ln 2 + \log_{10} 2}\)
\(\ds \) \(=\) \(\ds \frac {\ln x + \log_{10} x} {\ln 2 + \frac {\ln 2} {\ln 10} }\) Change of Base of Logarithm
\(\ds \) \(=\) \(\ds \frac {\ln x + \log_{10} x} {\ln 2 \left({1 + \frac 1 {\ln 10} }\right)}\)
\(\ds \) \(=\) \(\ds \frac {\ln x + \log_{10} x} {\ln 2 \left({1 + \log_{10} e}\right)}\) Reciprocal of Logarithm

We have that:

$\ln 2 = 0 \cdotp 69347 \ 1805 \ldots$
$\log_{10} e = 0 \cdotp 43429 \ 44819 \ldots$

Thus:

\(\ds \ln 2 \left({1 + \log_{10} e}\right)\) \(=\) \(\ds 0 \cdotp 69347 \ 1805 \ldots \left({1 + 0 \cdotp 43429 \ 44819 \ldots}\right)\)
\(\ds \) \(=\) \(\ds 0 \cdotp 99462 \ldots\)

Hence:

$\lg x + \log_{10} x \approx 0 \cdotp 994 \ln x$

That is, the approximation is $99.4 \%$ accurate.

$\blacksquare$


Historical Note

The useful Approximation to Binary Logarithm from Natural and Common Logarithm was observed by Richard Wesley Hamming.


Sources

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