TAOCP 1.2.2 Exercise 22

Using the change-of-base formula, \lg x=\frac{\ln x}{\ln 2}, \qquad \log_{10}x=\frac{\ln x}{\ln 10}.

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Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 22. ▶ [20] (R. W. Hamming.) Prove that

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

with less than 1% error! (Thus a table of natural logarithms and of common logarithms can be used to get approximate values of binary logarithms as well.)

Verified: no
Solve time: 1m31s

Solution

Using the change-of-base formula,

$$ \lg x=\frac{\ln x}{\ln 2}, \qquad \log_{10}x=\frac{\ln x}{\ln 10}. $$

Therefore

$$ \ln x+\log_{10}x = \ln x+\frac{\ln x}{\ln 10} = \left(1+\frac1{\ln 10}\right)\ln x. $$

On the other hand,

$$ \lg x=\frac1{\ln2}\ln x. $$

Hence the approximation

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

amounts to approximating the coefficient

$$ \frac1{\ln2} $$

by

$$ 1+\frac1{\ln10}. $$

The relative error is therefore

$$ \frac{ \left(1+\frac1{\ln10}\right)-\frac1{\ln2} }{ \frac1{\ln2} }. $$

Since

$$ \frac1{\ln10}\approx 0.43429448, \qquad \frac1{\ln2}\approx 1.44269504, $$

we obtain

$$ 1+\frac1{\ln10} \approx 1.43429448. $$

Thus

$$ \frac{ 1.43429448-1.44269504 }{ 1.44269504 } \approx -0.00582. $$

The magnitude of the relative error is therefore approximately

$$ 0.582%. $$

Since this is less than $1%$, the approximation

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

is valid with less than $1%$ relative error. ∎