## Algebra 2 (1st Edition)

Let $x=\log_b m\\y=\log_b n$ Rewrite as: $b^x=m\\b^y=n$ Obtain: $\log_b{mn}=\log_b b^xb^y=\log_b b^x b^y=\log_b b^{x+y}=(x+y)\log_b b$ Thus, $x+y=\log_b m+\log_b n\\ \rightarrow \log_b mm=\log_b m+\log_b n$