Q: $2^{x-y} +2^{y-x} = 20$. What is the value of $ \frac{4^x + 4^y}{2^{x+y-3}}$? $$2^{x-y} +2^{y-x} = 20$$


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*What is the value of $\displaystyle \frac{4^x + 4^y}{2^{x+y-3}}$?


Let's call $2^{x-y} = t$, $2^{y-x} = t^{-1}$ 
Hence we have
$$t +t^{-1} = 20$$
However, there's no solution root as far as I can see. Where did I go wrong? 

A: The second expression is by inspection exactly $2^3$ times the first, i.e. $160$.

A: Dividing the numerator and denominator of the fraction by $4^y$, we see that
$$\frac{4^x+4^y}{2^{x+y-3}} = \frac{4^{x-y}+1}{2^{(x-y)-3}} = \frac{t^2+1}{t/8} = 8(t+t^{-1})$$

A: Let $a=2^x$, $b=2^y$. 
Then
$2^{x-y}+2^{y-x}=\frac{2^x}{2^y}+\frac{2^y}{2^x}=\frac{a}{b}+\frac{b}{a} =  \frac{a^2+b^2}{ab}=20$
Since $\frac{4^x + 4^y}{2^{x+y-3}}=\frac{a^2+b^2}{ab \times2^{-3}}=8\times\frac{a^2+b^2}{ab}=8\times20=160$
