Solving $(x-2)^{x^2-6x+8} >1$ Solving this equation:
$(x-2)^{x^2-6x+8} >1$, by taking log  on base $(x-2)$ both the sides, I get the solution as $x>4$. My work: Let $(x-2)>0$
$$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1 =0\implies x<2, or,  x>4$$
But this doesn't appear to be the complete solution for instance $x=5/2$ is also a solution. I would like to know how to solve it completely.

$(x-2)^{x^2-6x+8}\gt 1$
If
$x^2 -6x + 8=0$ $\implies$ $(x-2)^{x^2-6x+8}=1$
$x\ne4$ and $x\ne2$
At $x=3$
$(x-2)^{x^2-6x+8}=1$
$x\in (2,3)\cup (4,\infty)$
For $x=\frac{5}{2}$
$(0.5)^{-0.75}=1.68179....$
