An algebra problem by Matheus Yuzawa

Solution 1 (if you don't have to explain your solution):

$2^{x+2}+8^x = 2^{x+2}(1+4^{x-1}) = 5\times2^{2y} \rightarrow$

$\rightarrow 1+4^{x-1} = 5\times2^{2y-x-2}$ ........... (1)

Because in (1) the LHS is odd, we need the RHS to be odd, so $1+4^{x-1} = 5 \rightarrow x=2$ With this result in (1), we get $y=2$ . Because this solution is allowed by both equations of the system, $2+2 = \boxed{4}$

Solution 2 (the smartest method):

$\sqrt{x}-\sqrt{y} = \log_3{\frac{y}{x}}$

i) $y > x$ , RHS is greater than LHS, no solution

ii) $x > y$ , LHS is greater than RHS, no solution

iii) $x = y \rightarrow 2^{x+2}+8^x = 2^{x+2}(1+4^{x-1}) = 5\times2^{2x} \rightarrow 1+4^{x-1} = 5\times 2^{x-2}$ .

Solving the last equation, we have $x=0$ or $x=2$ . Because $x=0$ is undefined (look at the first equation of the problem), $y=x=2$ leads to the answer.