#### Answer

2.4849

#### Work Step by Step

If $a\gt0$, $a\ne1$, $b\ne1$, and $x\gt0$, then we know that $log_{a}x=\frac{log_{b}x}{log_{b}a}$.
Therefore, $log_{e}12=\frac{log_{10}12}{log_{10}e}\approx\frac{1.0792}{0.4343}\approx2.4849$.
Recall that $b$ can be any positive number other than 1.