Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$ 
Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$

In other words, simplify $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$ into $\log_2x$ 
I did:
$$\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12) \\
= \log_2(2x+x^2) + \log_{2^2}(36) - \log_2(6x+12) \\
= \log_2(2x+x^2) + \log_2(\sqrt{36}) - \log_2(6x+12) \\
= \log_2{(\frac{(2x+x^2)6}{6x+12})} \\
= \log_2(\frac{12x+6x^2}{6x+12}) \\
= \log_2(\frac{2+x}{12}) \\
= \log_2(\frac{1}{6}+\frac{x}{12}) \\
= \log_2(\frac{12+6x}{72}) \\
= \log_2(12+6x) - \log_2(72)
= ???$$
What do I do next?

Hint:
$$12x+6x^2=6x(x+2)$$
$$6x+12=6(x+2)$$
