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)

Assuming "natural solutions" means positive integers:
$2 \times 27 \times 26 = 702 = 2^2 \times 3^3 \times 13$ while $39 = 3 \times 13$. Therefore $x = y = z = 1$ is not a solution. In fact, no solution will work because the prime factorization of $2^x 3^3 26^y$ will always have that pesky oddly even prime while the factorization of $39^z$ will not, specifically, as $2^{x + y}$ as you have already figured.
Try seeing $39^z$ as $2^0 \times 3^z \times 13^z$. But since $x, y$ are positive integers, $x + y > 0$.
