## Precalculus (6th Edition) Blitzer

Solution set = $\{8\}$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{lllll} x-3 \gt 0, & and & x \gt 0 & and & x+2 \gt 0\\ x \gt 3 & & x \gt 0 & & x \gt -2 \end{array}\right]$ $x \gt$3 $\qquad (*)$ This is the condition eventual solutions must satisfy. $\log_{2}(x-3)+\log_{2}x-\log_{2}(x+2)=2$ LHS: apply the product rule; RHS write $2$ as $\log_{2}2^{2}$ $\log_{6}[(x-3)x]-\log_{2}(x+2)=\log_{2}2^{2}$ LHS: apply the quotient rule, $\displaystyle \log_{6}\frac{x(x-3)}{x+2}=\log_{2}4$ If $\log_{b}M=\log_{b}N,$ then M=N $\displaystyle \frac{x(x-3)}{x+2}=4\quad$ ... multiply with $(x+2)$ $x(x-3)=4(x+2)$ $x^{2}-3x= 4x+8$ $x^{2}-7x-8=0$ Factor: -8 and +1 are factors of -8 whose sum is -7 $(x+1)(x-8)=0$ $x=-1$ - does not satisfy (*), discard $x=8$ - satisfies (*) Thus, the solution set = $\{8\}$