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(x-2)^{x-3}<(x-2)^{5-x} what is the x?

Guest Apr 18, 2015

Best Answer 

 #3
avatar+93616 
+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
 #1
avatar+27032 
+5

Plotting a graph often helps:

 inequality

.

Alan  Apr 18, 2015
 #2
avatar+93616 
+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
avatar+93616 
+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

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