Question

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...

Answer 1

Void.

Answer 2

`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`