## Algebra 2 (1st Edition)

$x=2$.
We know that $x\log_ab=\log_ab^x$ and that $\log_ab+\log_ac=\log_a(bc)$ Hence here: $\log_4 x+\log_4(x+6)=2\\\log_4(x(x+6))=2\\x(x+6)=4^2=16\\x^2+6x-16=0\\(x+8)(x-2)=0$ Thus $x=-8$ or $x=2$. However, $x=-8$ is an extraneous solution, thus $x=2$.