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.

Be careful with $f(x)=\log_a x$, apart from $x>0, a>0$, there are two cases when $0 <a<1$ and $a>1$. In the former $f(x)$ decreases and increases in the latter.
So take
Case 1: $0<(x-2)<1 \implies 2<x<3$, then
$$(x-2)(x-4)\log_{(x-2)} (x-2) < \log_{(x-2)} 1=0 \implies (x-2)(x-4) <0 \implies 2<x<4.$$ Eventually $2<x<3.$
Case 2: $(x-2)>1 \implies x>3$, then
$$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1=0 \implies (x-2)(x-4) >0 \implies x<2, or, x>4.$$
Notice the sign of inequalirt is reversed.So we get $x>4.$
So the total answer is $x\in (2,3) \cup (4,\infty).$
