## Precalculus (6th Edition) Blitzer

The solution of the equation is $\underline{-2+{{\log }_{3}}11}$
Consider the given equation: ${{3}^{x+2}}=11$ Take the logarithm of both sides and obtain: $\log {{\left( 3 \right)}^{x+2}}=\log \left( 11 \right)$ Then use the multiplication and division rules of logarithmic functions to solve for x: \begin{align} & \log {{\left( 3 \right)}^{x+2}}=\log \left( 11 \right) \\ & \left( x+2 \right)\log \left( 3 \right)=\log \left( 11 \right) \\ & \left( x+2 \right)=\log \left( \frac{11}{3} \right) \\ & \left( x+2 \right)={{\log }_{3}}11 \\ & x={{\log }_{3}}11-2 \\ & x=-2+{{\log }_{3}}11 \end{align} Thus the solution of the equation is $x=-2+{{\log }_{3}}11$.