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....
As $\log_aa=1$ and $\log_a(a^m)=m\log_aa=m$

taking logarithm wrt $3,$

$$(x+2)\log_32+\frac{3x}{x-1}=2$$

$$\implies(x+2)\log_32=2-\frac{3x}{x-1}=-\frac{x+2}{x-1}$$

$$\implies (x+2)\left(\log_32+\frac1{x+1}\right)=0$$

We know if $a\cdot b\cdot c\cdots=0;$  at least one of $a,b,c,\cdots$ is zero
Think of something to multiply the entire equation by that will remove any $x$'s from the denominator.