## Precalculus (6th Edition) Blitzer

Consider the given equation, ${{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2$ Apply the quotient rule of logarithms: \begin{align} & {{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2 \\ & {{\log }_{3}}\frac{\left( x-1 \right)}{\left( x+2 \right)}=2 \end{align} The provided equation can be written as \begin{align} & \frac{\left( x-1 \right)}{\left( x+2 \right)}={{3}^{2}} \\ & \frac{\left( x-1 \right)}{\left( x+2 \right)}=9 \\ & \left( x-1 \right)=9\left( x+2 \right) \\ \end{align} $x-1=9x+18$ Now, add to both sides $-x-18$: \begin{align} & x-1-x-18=9x+18-x-18 \\ & 8x=-19 \\ & x=\frac{-19}{8} \\ \end{align} Here $x=\frac{-19}{8}$ is not the solution of the given expression because \begin{align} & {{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2 \\ & {{\log }_{3}}\left( \frac{-19}{8}-1 \right)-{{\log }_{3}}\left( \frac{-19}{8}+2 \right)=2 \\ & {{\log }_{3}}\left( \frac{-19}{8}-1 \right)-{{\log }_{3}}\left( \frac{-19}{8}+2 \right)=2 \\ & {{\log }_{3}}\left( -3.375 \right)-{{\log }_{3}}\left( -0.375 \right)=2 \end{align} By definition of ${{\log }_{a}}b$, $b>0$ but $-3.375$ and $-0.375$ is not greater than zero. Thus, $x=\frac{-19}{8}$ does not satisfy the equation ${{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+2 \right)=2$ Therefore, there is no solution of the given expression.