<s>[INST] 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 $2\log_{3}(-3) < 0$.
 [/INST] Proceeding from T. Bongers' concise solution: $$ x^2 = 3 \cdot (x + 6) \implies x^2 - 3x - 18 = 0. $$Now, we can try to factor this. We notice that it can be factored as follows: $$ (x+3)(x-6) = 0 \implies x \in \{ -3, 6 \}. $$Since $x=-3$ is extraneous, $x=6$ is the only solution.
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