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

Question

$$2^{x-y} +2^{y-x} = 20$$

• 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?

Answer 1

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 $\frac{4^x + 4^y}{2^{x+y-3}}=\frac{a^2+b^2}{ab \times2^{-3}}=8\times\frac{a^2+b^2}{ab}=8\times20=160$

Answer 2

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

Answer 3

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