I want to ask about way of solving exponential inequalities, I am going to show you two similar examples, but their solving is kinda different.
First example: $3^{2x}-10\cdot3^x+9>0$ $(3^x-9)(3^x-1)>0$ $3^x\in(-\infty;1)\cup(9;+\infty)$ So to find x we need to solve:
$3^x>9$ and $3^x>1$ (1)
and we get: $x\in(-\infty;0)\cup(2;+\infty)$
Second example: $5^{2x}-6\cdot5^x+5<0$ $(5^x-5)(5^x-1)<0$ $5^x\in(1;5)$ So to find x we need to solve:
$\begin{cases}5^x<5\\5^x>1\end{cases}$ (2)
and we get: $x\in(0;1)$
My question is why in first example I get answer by solving 2 equations (1) seperately but in second example I get answer by solving system of equations (2)