<s>[INST] 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....
 [/INST] we could argue that 9 is divisible by 3 and not 2
so wlog 
$\large2^{x+2}3^{\frac{3x}{x-1}}=9$
$2^{x+2}$ should be one,so $x=-2$
also
$ \large 3^{\frac{3x}{x-1}}=3^2$
${\frac{3x}{x-1}}=2\implies x=-2$
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