<s>[INST] 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.
 [/INST] since
$$2^{x+2y}=2^x\cdot 4^y=9\cdot 27=3^5$$
and note
$$2^{2y-4x}=4^y/(2^x)^4=27/9^4=3^{-5}$$
so
$$2^{x+2y}\cdot 2^{2y-4x}=2^{(x+2y)+(2y-4x)}=3^5\cdot 3^{-5}=1=2^{0}$$
so $$\Longrightarrow x+2y+2y-4x=0$$
$$\Longrightarrow  x+2y=-(2y-4x)$$
so
$$\Longrightarrow  \dfrac{x+2y}{2y-4x}=-1$$
</s>