Let #x in ZZ^+# and #y=x^x# s.t. #x=floor(ln(y))#. For what value(s) of #x# will this hold?

This was a problem I came up with for my birthday a couple of weeks ago...

2 Answers
Jul 31, 2017

Void.

Jul 18, 2018

Answer:

#x=3#

Explanation:

I posted this question a year ago for fun. Since I have received no answers I will post my solution.

Since #x\in\ZZ^+# we know that #x>=0#.

Also, we know by properties of the floor function that

#x<=ln(y)<x+1#

Combining both statements we have the inequality

#x+1>ln(y)>=x>=0#

#<=># since #y=x^x#

#x+1>ln(x^x)>=x>=0#

#<=>#

#x+1>xln(x)>=x>=0#

#<=>#

#1/x+1>ln(x)>=1#

#<=>#

#e^(1/x+1)>x>=e#

Thus #x\in\ZZ\cap\[e,oo)#

Claim: #e^(1+1/x)< x # #\forall x>=4#

Since #e^(1/x)# is decreasing we know that

#e^(1/4)>e^(1/5)>…#

#<=>#

#e*e^(1/4)>e*e^(1/5)>…#

#<=>#

#e^(1+1/4)>e^(1+1/5)>…#

Thus is suffices to show that #e^(1+1/4)<4#

Since #e^(1+1/4)~~3.49<4# we now can see that

#x\in\ZZ\cap[e,4)={3}#

Thus,

#x=3#

I came up with this problem because last year I turned #y=3^3=27# years old.

So today I am #28#