Answer
$(1,3)$ and $(3,1)$
Work Step by Step
Let $m=2^x$ and $n=2^y$. Then, we have
$m+n=10$
$m^2+n^2=68$
Rewrite this system:
$n=10-m$
$m^2+n^2-68=0$
Substitute the first equation to the second one and then solve for $m$:
$m^2+(10-m)^2-68=0$
$m^2+m^2-20m+100-68=0$
$2m^2-20m+32=0$
$(2m-4)(m-8)=0$
$m=2$ and $m=8$
Now, obtaine $n$:
For $m=2$, $n=10-2=8$
For $m=8$, $n=10-8=2$.
Obtaine $x$ and $y$:
For $(m,n)=(2,8)=(2^1,2^3)$, we get $(x,y)=(1,3)$
For $(m,n)=(8,2)=(2^3,2^1)$, we get $(x,y)=(3,1)$
Therefore, the solutions are $(1,3)$ and $(3,1)$.
