## Precalculus (10th Edition)

a) $\{18\}$ b) $(2,18]$
We are given the function: $f(x)=\log_4 (x-2)$ a) Solve the equation: $f(x)=2$ $\log_4 (x-2)=2$ $x-2=4^2$ $x-2=16$ $x=18$ Check the solution: $\log_4 (18-2)\stackrel{?}{=} 2$ $\log_4 16\stackrel{?}{=} 2$ $2=2\checkmark$ The solution set is: $\{18\}$ b) Solve the inequality: $f(x)\leq 2$ $\log_4 (x-2)\leq 2$ $\log_4 (x-2)\leq \log_4 4^2$ $x-2\leq 16$ $x\leq 18$ As the function $f(x)=\log_4 (x-2)$ is defined for $x>2$, the solution set is: $(2,18]$