If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution:  $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y+z =75$

Assume $x\geq y \geq z$. Then,
$$
3^{26}=9^{13}=3^x+3^y+3^z\leq 3(3^x)=3^{x+1}
$$
and so $x\geq 25$. Obviously, $x<26$ and so we must have $x=25$. But then
$$
2\times 3^{25}=3^{26}-3^x=3^y+3^z\leq 3^x+3^x=2\times 3^{25}.
$$
It must be the case that $y=x$ and $z=x$. We conclude that $x=y=z=25$ and their sum is $75$.
