## Precalculus (6th Edition) Blitzer

Solution set = $\{2\}$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{lll} x+2 \gt 0, & and & x-1 \gt 0\\ x \gt -2 & & x \gt 1 \end{array}\right]$ $x \gt 1 \qquad (*)$ This is the condition eventual solutions must satisfy. $\log_{4}(x+2)-\log_{4}(x-1)=1$ LHS: apply the quotient rule; RHS: write $1$ as $\log_{4}4$ $\displaystyle \log_{4}\frac{(x+2)}{(x-1)}=\log_{4}4\qquad$ (if $\log_{b}M=\log_{b}N,$ then M=N) $\displaystyle \frac{x+2}{x-1}=4\qquad$ ... multiply with $(x-1)$ $x+2=4(x-1)$ $x+2=4x-4\qquad$ ... add $4-x$ $6=3x$ $x=2\qquad$... satisfies (*) Solution set = $\{2\}$