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.
 A: Be careful: when you take the logarithm to the base $x-2$, you need to ensure that $x-2$ is positive. If one of the solutions is $x<2$, you'll be taking a logarithm to a negative base. I would suggest using the natural logarithm instead:
$$(x-2)^{x^2-6x+8}>1$$
$$(x-2)(x-4)\ln(x-2)>0$$
For this quantity to be positive, $(x-2)(x-4)$ and $\ln(x-2)$ must have the same sign. This happens when $2<x<3$ and $x>4$.
PS: Thank you to both Stephen Donovan and Asher2211 for help improving my answer. As I cannot find an expression for all solutions, I will leave my answer as covering the portion with positive $x$.
A: 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 $x\in (2,3) \cup (4,\infty).$
A: You have covered the case when $x\ge 2$.
It is easy to see that there are no solutions for $1<x<2$.
For $x<1$ all solutions for $x^2-6x+8=\frac{n}{m}$ (where $m$ is odd integer and $n$ is even integer) is a valid solution for the inequality (provided $x<1$) . This is because $a^b$ is real for negative values of $a$ only if $b$ is of the form
$\frac{n}{m}$ where $m$ is a odd number and when $n$ is even the $a^b$ attains positive values.
A: $(x-2)^{x^2-6x+8}\gt 1$
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 $x=\frac{5}{2}$
$(0.5)^{-0.75}=1.68179....$
