Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation : 
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working : 
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2  \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$ 
Now how to proceed further in this problem... please suggest thanks....

Hint:
$$2^{x+2}\cdot3^{\frac{3x}{x-1}}=3^2\iff 2^{x+2}=3^{2-\frac{3x}{x-1}}=3^{-\frac{x+2}{x-1}}\iff \begin{cases}x+2=0\\{}\\x-1=-\log_23\end{cases} \;\;(\text{why?)}$$
