Answer
$x=3$ and $x=-4$
Work Step by Step
$\log x+\log(x+1)=\log12$
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 $x=-4$
