## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=\displaystyle \frac{83}{15}$
First, the solutions must satisfy $\left\{\begin{array}{l} x+3\gt 0\\ x-5\gt 0 \end{array}\right.\quad \Rightarrow x\gt$5$\qquad (*)$ in order for the equation to be defined. On the RHS, apply$\quad \log_{4}4^{2}=2$ $\log_{4}(x+3)=\log_{4}16+\log_{4}(x-5)$ ... apply $\qquad \log_{a}(MN)=\log_{a}M+\log_{a}N$ $\log_{4}(x+3)=\log[16(x-5)]$ ... apply the principle of logarithmic equality $x+3=16(x-5)$ $x+3=16x-80\qquad$... add $80-x$ $83=15x$ $x=\displaystyle \frac{83}{15}\quad$... satisfies (*), and is a valid solution.