## Intermediate Algebra (12th Edition)

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, $log_{12}3=\frac{log_{10}3}{log_{10}12}\approx\frac{0.4771}{1.0792}\approx0.4421$. Recall that $b$ can be any positive number other than 1.