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 $2\log_{3}(-3) < 0$.

To get $x$ out of logarithm, you raise the base (3) with the logarithm (as log and exponent are inverses)
$\log_{3}(\frac{x^{2}}{x+6}) = 1$
$3^{\log_{3}(\frac{x^{2}}{x+6})} = 3^{1}$
$\frac{x^{2}}{x+6} = 3$
then solve the quadratic to get $x$ value
