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

$x=1$
First, the solutions must satisfy $\left\{\begin{array}{l} x+1\gt 0\\ x+2\gt 0 \end{array}\right.\quad \Rightarrow x\gt -1\qquad (*)$ in order for the equation to be defined. On the RHS, apply$\quad\log_{a}(MN)=\log_{a}M+\log_{a}N$ $\log_{7}[(x+1)(x+2)]=\log_{7}6$ ... apply the principle of logarithmic equality $(x+1)(x+2)=\log_{7}6$ $x^{2}+3x+2=6$ $x^{2}+3x-4=0\quad$to factor, find factors of -4 with sum =3 ... these are $+4$ and $-1$ $(x+4)(x-1)=0$ Possible solutions: $x=-4\qquad$... does not satisfy (*), not a solution. $x=1\qquad$... satisfies (*), and is a valid solution.