## Intermediate Algebra for College Students (7th Edition)

The solution set is$\left\{ 9 \right\}$.
A logarithmic equation in one variable can be solved when the equation is written in exponential form. For example, the equation ${{\log }_{a}}x=b$ when written in exponential form is$x={{a}^{b}}$. Hence, solution for x is $x={{a}^{b}}$. Now, for the provided logarithmic equation, simplify by logarithmic properties: \begin{align} & {{\log }_{9}}\left( x \right)+{{\log }_{9}}\left( x-8 \right)=1 \\ & {{\log }_{9}}\left( x\left( x-8 \right) \right)=1 \end{align} Now, rewrite the logarithmic equation in the exponential form: \begin{align} & x\left( x-8 \right)=9 \\ & {{x}^{2}}-8x-9=0 \\ & {{x}^{2}}-9x+x-9=0 \\ & x\left( x-9 \right)+1\left( x-9 \right)=0 \end{align} Solve further as: $\left( x+1 \right)\left( x-9 \right)=0$ Case 1: $x=-1$ Argument of ${{\log }_{9}}\left( x \right)$ is negative in this case which is not possible. Case 2: $x=9$ Arguments of both logs are positive in this case. Hence, the solution of the equation is $x=9$.