Find an approximation of log₁₀ e

by RoRi

November 6, 2015

Using the fact that

$\log_e 10 = 2.302585,$

compute $\log_{10} e$ using the solution of the previous exercise (Section 6.9, Exercise #32 of Apostol).

Since we know from the solution of the previous exercise that

$\log_b x = \log_b a \log_a x$

we let $a = e$ , $x = 10$ , $b = 10$ to get

$\begin{align*} && \log_{10} 10 &= (\log_{10} e)(\log_e 10) \\ \implies && \log_{10} e &= \frac{1}{2.302585} \\ &&&= 0.434294 \end{align*}$

Stumbling Robot