#3**+10 **

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

Melody Apr 18, 2015

#2**+5 **

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

Melody Apr 18, 2015

#3**+10 **

Best Answer

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

Melody Apr 18, 2015