# How do you solve log_6(x + 19) + log_6(x) = 3?

We must have $x > 0$ hence
log_6(x + 19) + log_6(x) = 3=>log_6 x(x+19)=3=> x(x+19)=6^3=>x^2+19x-216=0=>x^2+27x-8x-216=0=> (x-8)*(x+27)=0=>x=8 or x=-27
But $x = - 27$ is rejected so $x = 8$ is the only solution