$\text{Show that }{{\log }_{a}}{{(a+b)}^{2}}=2+{{\log }_{a}}\left( 1+\frac{2b}{a}+\frac{{{b}^{2}}}{{{a}^{2}}} \right)$
$\begin{align} & {{\log }_{a}}{{(a+b)}^{2}}={{\log }_{a}}({{a}^{2}}+2ab+{{b}^{2}}) \\ & \text{Factor out }{{a}^{2}}\text{ in }{{a}^{2}}+2ab+{{b}^{2}} \\ & {{\log }_{a}}{{(a+b)}^{2}}={{\log }_{a}}{{a}^{2}}\left( \frac{{{a}^{2}}}{{{a}^{2}}}+\frac{2ab}{{{a}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}} \right) \\ & {{\log }_{a}}{{(a+b)}^{2}}={{\log }_{a}}{{a}^{2}}\left( 1+\frac{2b}{{{a}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}} \right) \\ & {{\log }_{a}}{{(a+b)}^{2}}={{\log }_{a}}{{a}^{2}}+{{\log }_{a}}\left( 1+\frac{2b}{a}+\frac{{{b}^{2}}}{{{a}^{2}}} \right) \\ & {{\log }_{a}}{{(a+b)}^{2}}=2{{\log }_{a}}a+{{\log }_{a}}\left( 1+\frac{2b}{a}+\frac{{{b}^{2}}}{{{a}^{2}}} \right) \\ & {{\log }_{a}}{{(a+b)}^{2}}=2+{{\log }_{a}}\left( 1+\frac{2b}{a}+\frac{{{b}^{2}}}{{{a}^{2}}} \right) \\\end{align}$
