## Precalculus (6th Edition) Blitzer

Use the rule $\log (p q)=\log p+\log q$ with $p=x+4$ and $q=x+1$ to obtain $\ln[(x+4)(x+1)]=2 \ln (x+3)$ Simplify: $(x+3)^2=(x+4)(x+1)$ $x^2+6x+9=x^2+5x+4$ or, $x+9=5x+4 \implies x=-5$ But $x=-5$ is not in the domain. Thus, our result is: No solution