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

$x=\displaystyle \frac{2}{5}$
First, the solutions must satisfy $\left\{\begin{array}{l} x+6\gt 0\\ x \gt 0 \end{array}\right.\quad \Rightarrow x\gt 0\qquad (*)$ in order for the equation to be defined. LHS: Apply$\displaystyle \quad\log_{a}\frac{M}{N}=\log_{a}M-\log_{a}N$ RHS: Apply $\quad \log_{4}4^{2}=2$ $\displaystyle \log_{4}(\frac{x+6}{x})=\log_{4}16$ ... apply the principle of logarithmic equality $\displaystyle \frac{x+6}{x}=16$ $x+6=16x$ $6=15x$ $x=\displaystyle \frac{6}{15}=\frac{2}{5}\qquad$... satisfies (*), and is a valid solution.