<s>[INST] System of exponential equations If $x,y,z \in \mathbb{R}$ and 
$$
\begin{cases}
2^x+3^y=5^z \\
2^y+3^z=5^x \\
2^z+3^x=5^y
\end{cases}
$$
does it imply that $x=y=z=1$?
 [/INST] I think so.
Let $f(x) = 5^x - 2^x - 3^x$, you have that
$$x < 1 \Leftrightarrow f(x) < 0$$
$$x = 1 \Leftrightarrow f(x) = 0$$
$$x > 1 \Leftrightarrow f(x) > 0$$
Suppose that $x \geq y \geq z$:
If $x > 1$, then $5^x > 2^x + 3^x \geq 2^y + 3^z$, therefore $5^x \neq 2^x + 3^y$.
This implies that $1 \geq x \geq y \geq z$.
If $z < 1$, then $5^z < 2^z + 3^z \leq 2^x + 3^y$, this implies that $1 \geq x \geq y \geq z \geq 1$ so $x = y = z = 1$.
By cyclicity of the equations you get the same result if $y \geq z \geq x$ and $z \geq x \geq y$.
Suppose $x \geq z \geq y$:
If $x > 1$ then $5^x > 2^x + 3^x \geq 2^y + 3^z$ therefore $1 \geq x \geq y \geq z$.
If $y < 1$ then $5^y < 2^y + 3^y \leq 2^z + 3^x$ therefore $1 \geq x \geq z \geq y \geq 1$ which implies that $x = y = z = 1$.
By cyclicity we have covered the other 3 orders, so the only solution should be $x = y = z = 1$.
</s>