Validity of the change-of-base formula for all bases 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?
 A: You can use whatever basis: the equality $3^x=10$ is equivalent to
$$
\log_a 3^x=\log_a 10
$$
for any $a>0$, $a\ne1$. Since $\log_a 3^x=x\log_a 3$, we get
$$
x=\frac{\log_a 10}{\log_a 3}
$$
so you see that the final result is independent of the base.
You can get the change of base formula by considering $b^c=k$ that says $c=\log_b k$; but we also have $c\log_a b=\log_a k$ and therefore
$$
\log_b k=\frac{\log_a k}{\log_a b}
$$
In the case of $b=3$, $k=10$ and $a=e$, you get
$$
\log_3 10=\frac{\ln 10}{\ln 3}
$$
A: 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.
A: In your question note that the definition of the logarithm implies that $$\log_310=x \iff 3^x=10 \tag 1.$$ In general  $$\log_bw=x \iff b ^x=w \tag 2$$
where $b>0, b\neq 1$. 
Now take the logarithm of both sides to some base $c>0$, $c\neq 1$ and use the fact that $\log_c(b^x)=x\log_cb $ to get that $$x\log_cb=\log_cw\implies x=\frac{\log_cw}{\log_cb}.\tag 3$$
Equations $(2)$ and $(3)$ imply that$$\log_bw=\frac{\log_cw}{\log_cb}. \tag 4$$
Take  $c=10$ and $c=e$ in equation $(4)$ to get the equality $$\log_{3}10=\frac{\log_{10}10}{\log_{10}3}=\frac{\ln10}{\ln3}. \tag 5$$
