Solve: $2\log_{3}(x)-\log_{3}(x+6)=1$ just getting going with logarithms, having trouble with this question. 
$$2\log_{3}(x)-\log_{3}(x+6)=1$$
$$\log_{3}x^2-\log_{3}(x+6)=1$$
Stuck at this point: What do I do next?
$$\log_{3}\bigg({\dfrac{x^2}{x+6}}\bigg)=1$$
Edit: Got it!
$$\dfrac{x^2}{x+6}=3^1$$
$$\dfrac{x^2}{x+6}=3$$
$${x^2}=3(x+6)$$
$$x^2=3x+18$$
$$x^2-3x-18=0$$
$$(x-6)(x+3)=0$$
$x=6 $ and cannot equal $\ne3$ as $2\log_{3}(-3) < 0$.

Just use exponents. If $\log_xy=z$, then $y=z^x$.
