# 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...

Jul 31, 2017

Void.

Jul 18, 2018

$x = 3$

#### Explanation:

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

Since $x \setminus \in \setminus {\mathbb{Z}}^{+}$ we know that $x \ge 0$.

Also, we know by properties of the floor function that

$x \le \ln \left(y\right) < x + 1$

Combining both statements we have the inequality

$x + 1 > \ln \left(y\right) \ge x \ge 0$

$\iff$ since $y = {x}^{x}$

$x + 1 > \ln \left({x}^{x}\right) \ge x \ge 0$

$\iff$

$x + 1 > x \ln \left(x\right) \ge x \ge 0$

$\iff$

$\frac{1}{x} + 1 > \ln \left(x\right) \ge 1$

$\iff$

${e}^{\frac{1}{x} + 1} > x \ge e$

Thus $x \setminus \in \setminus \mathbb{Z} \setminus \cap \setminus \left[e , \infty\right)$

Claim: ${e}^{1 + \frac{1}{x}} < x$ $\setminus \forall x \ge 4$

Since ${e}^{\frac{1}{x}}$ is decreasing we know that

e^(1/4)>e^(1/5)>…

$\iff$

e*e^(1/4)>e*e^(1/5)>…

$\iff$

e^(1+1/4)>e^(1+1/5)>…

Thus is suffices to show that ${e}^{1 + \frac{1}{4}} < 4$

Since ${e}^{1 + \frac{1}{4}} \approx 3.49 < 4$ we now can see that

$x \setminus \in \setminus \mathbb{Z} \setminus \cap \left[e , 4\right) = \left\{3\right\}$

Thus,

$x = 3$

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

So today I am $28$