TITLE: Validity of the change-of-base formula for all bases
QUESTION [8 upvotes]: Suppose I want to solve $3^x=10$. I convert it to logarithmic form
$$\log_{3}10=x$$
then change bases
$$\frac{\log10}{\log3}=x$$
and this will yield the same answer as if I had written
$$\frac{\ln10}{\ln3}=x$$
But log and ln have different bases. Why does this work both ways?

REPLY [3 votes]: Possibly the best way to see this is to write a change to an arbitrary base instead of choosing either $\log_{10}$ or $\ln$ initially.
(There is already an answer that does that.)
But I think it's also noteworthy that the change-of-base formula itself
provides an easy derivation of the fact that
$$\frac{\log10}{\log3}=\frac{\ln10}{\ln3}.$$
Starting with $\frac{\ln10}{\ln3}$ (for example), simply apply the
formula to change the base from $e$ to $10$ on both the numerator and denominator:
\begin{align}
\ln10 &= \log_e 10 = \frac{\log_{10} 10}{\log_{10} e}, \\
\ln3 &= \log_e 3 = \frac{\log_{10} 3}{\log_{10} e}, \\
\frac{\ln10}{\ln3} &=
  \frac{\left( \frac{\log_{10} 10}{\log_{10} e} \right)}
       {\left(\frac{\log_{10} 3}{\log_{10} e} \right)}
\end{align}
Multiply both the numerator and denominator of the right-hand side of the last equation by $log_{10} e$, and we find that
$$\frac{\ln10}{\ln3} = \frac{\log_{10} 10}{\log_{10} 3}
$$
In other words, every time we change the base of both the numerator and
denominator from base $b$ to base $c$ simultaneously, we
divide both the numerator and denominator by $\log_c b$,
and those two operations cancel each other out.