If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be? If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be?
I have seen such problems many times but no specific method to solve them.
I hope I find the general rule to solve them here.

We have $$2^x = 9\implies x\ln(2)=\ln(2^x)=\ln(9)\implies x = \frac{\ln(9)}{\ln(2)}= 2 \frac{\ln(3)}{\ln(2)}$$
and 
$$4^y = 27\implies y = \frac{\ln(27)}{\ln(4)}= \frac{3}{2}\frac{\ln(3)}{\ln(2)}$$
So
$$\frac {x+2y}{2y-4x} =\frac {2 \frac{\ln(3)}{\ln(2)}+3\frac{\ln(3)}{\ln(2)}}{3\frac{\ln(3)}{\ln(2)}-8\frac{\ln(3)}{\ln(2)}} =-1$$
