I am going to try and do it by hand
$$\\(x-2)^{x-3}<(x-2)^{5-x}\\\\
(x-3)log(x-2)<(5-x)log(x-2)\\\\
(x-3)log(x-2)-(5-x)log(x-2)<0\\\\
(x-3-5+x)log(x-2)<0\\\\
(2x-8)log(x-2)<0\\\\
(x-4)log(x-2)<0$$
this will be true if one of the factors is negative and the other is positive.
log(x-2)>0 when x-2>1 that is when x>3
x-4>0 when x>4
so
If x4 then both are negative.
BUT the goldilocks moment comes
when x is between 3 and 4 because then x-4 is neg and log(x-2) is positive
so
This will be true when 3<x<4
Thanks Alan,
So you want the blue graph to be UNDER the green graph (because it is less than)
this appears to happen between x=3 and x=4 :))
3<x<4
I am going to try and do it by hand
$$\\(x-2)^{x-3}<(x-2)^{5-x}\\\\
(x-3)log(x-2)<(5-x)log(x-2)\\\\
(x-3)log(x-2)-(5-x)log(x-2)<0\\\\
(x-3-5+x)log(x-2)<0\\\\
(2x-8)log(x-2)<0\\\\
(x-4)log(x-2)<0$$
this will be true if one of the factors is negative and the other is positive.
log(x-2)>0 when x-2>1 that is when x>3
x-4>0 when x>4
so
If x4 then both are negative.
BUT the goldilocks moment comes
when x is between 3 and 4 because then x-4 is neg and log(x-2) is positive
so
This will be true when 3<x<4