<s>[INST] 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 $x+y+z =75$
 [/INST] You can do this if $x,y,z$ are integers, otherwise there are infinitely many solutions. 
First suppose that $x \geq y \geq z$, this is possible because of symmetry.
Suppose that $x \geq 26$, then $3^x\geq3^{26}$ and hence $3^x+3^y+3^z>3^{26}$. Hence $x \leq 25$. 
If $z \leq 24$, then $3^x+3^y+3^z \leq 3^{25}+3^{25}+3^{24}<3^{26}$ hence $z\geq 25$. Hence $x=y=z=25$ and $x+y+z=75$. 
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