How can I solve this using prime factors? I'm stuck with this problem: $2^x \cdot 3^3 \cdot 26^y = 39^z$ for $x, y, z \in \mathbb{N}$.
I know that there isn't a natural solution for the equation, but I need to "prove" it using prime factors. I'm stuck here:
$$
2^x \cdot 3^3 \cdot (2\cdot 13)^y = (3\cdot 13)^z \\
\implies 2^x \cdot 3^3 \cdot 2^y \cdot 13^y = 3^z \cdot 13^z \\
\implies 2^{x+y} \cdot 3^3 \cdot 13^y = 3^z \cdot 13^z
$$
Could anyone give me a hint on this problem? (Sorry for the formatting)

$2^x \cdot 3^3 \cdot 26^y = 39^z\iff 2^{x+y}\cdot3^3\cdot13^y=3^z\cdot13^z\Rightarrow2^{x+y}=3^{z-3}\cdot13^{z-y}$
This is possible just when you have $1=1\cdot1$ i. e. $x+y=z-3=z-y=0$. Thus $$(x,y,z)=(-3,3,3)$$
