How to find $x$ in $\log_9 (x)+\log_9 (x-2)=\log_9 (8)$

Question

How would I go about solving this problem? I know that you can remove the $\log$s if the have the same base resulting leaving $x(x-2)=8$, but what would be the next step to finding the value of $x$?

Answer 1

Hint: $x(x-2) = 8$ if we distribute, $x^2 - 2x = 8$, or $x^2-2x-8 = 0$. Can you proceed from here?

Answer 2

$$x(x-2)=8$$ $$x^2-2x=8$$ $$x^2-2x-8=0$$ $$(x-4)(x+2)=0$$

But as you can see we want $x>0 $ and $x-2>0$ because of the domain of $\log x $ is $x>0$

Thus $x>2$

Thus $x=4$ is the only solution to this equation