Answer
$x=\frac{1- \sqrt{13}}{2}$ or $x=\frac{1+ \sqrt{13}}{2}$
Work Step by Step
$\log(x^2+3)=\log(x+6),x>-6$
$\log(x^2+3)-\log(x+6)=0,$
$\log(\frac{x^2+3}{x+6})=0,$ raising both sides to the power of $10.$
$10^{\log(\frac{x^2+3}{x+6})}=10^0,$
$\frac{x^2+3}{x+6}=1,$
$x^2+3=x+6,$
$x^2-x-3=0,$
Solving the quadratic equation using the quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},$
$x=\frac{1\pm \sqrt{1-4(-3)(1)}}{2}=\frac{1\pm \sqrt{13}}{2}$
Both solutions fit as they belong to the domain.
