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.
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
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 $75$.
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 $x+y+z=75$.
Let $x\ge y\ge z$. Then $$3^x +3^y +3^z=3^{26}$$ $$3^z(3^{x-z}+3^{y-z}+1)=3^{z}\cdot3^{26-z}$$ $$3^{x-z}+3^{y-z}+1=3^{26-z} \Leftrightarrow x=z=y=25$$
If $x=y=z$ then clearly $x=y=z=25$ works.
Otherwise, this is a ternary number with sum of digits $3$. But the unique representation of this number in ternary is
$$1\underbrace{0 \ldots 0}_{26 \text{ times}}$$
which has a sum of digits as $1$.
