## Precalculus: Mathematics for Calculus, 7th Edition

$x=3$ and $x=-4$
$\log x+\log(x+1)=\log12$ Combine the sum on the left side of the equation as the logarithm of a product: $\log x(x+1)=\log12$ Evaluate the product inside the $\log$ on the left side: $\log(x^{2}+x)=\log12$ If $\log(x^{2}+x)=\log12$, then $x^{2}+x=12$ $x^{2}+x=12$ Take $12$ to the left side: $x^{2}+x-12=0$ Solve by factoring: $(x+4)(x-3)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+4=0$ $x=-4$ $x-3=0$ $x=3$ The solutions are $x=3$ and $x=-4$