How to solve $x^{x^{x^{x^{2010}}}} = 2010$

Question

So I know that there is a difference between $(x^2)^3$ and $x^{2^3}$.

But how do I use this knowledge to solve $$x^{x^{x^{x^{2010}}}} = 2010?$$

Answer 1

Here is a hint: if $x = a^{1/a}$, then $x^a = a$, $x^{x^a} = x^a = a$, and so on.

Answer 2

I assume that $x>0$ since the expression on the LHS is not generally defined for $x\leq 0$. Suppose $x\in(0,1]$. Then $x^{x^{x^{2010}}}$ is a positive number, so $f(x)=x^{x^{x^{x^{2010}}}}\in(0,1]$, and there are no solutions in this interval. Now suppose $x>1$. $f$ is strictly increasing. Let $x>y>1$. Then $x^{2010}>y^{2010}$ which implies $x^{x^{2010}}>y^{y^{2010}}$, and continuing this, we get $f(x)>f(y)$. This means it has no more than $1$ positive solution. Let $a=\sqrt[2010]{2010}$. Notice that $a^{2010}=2010$. So $$a^{a^{a^{a^{2010}}}}=a^{a^{a^{2010}}}=a^{a^{2010}}=a^{2010}=2010 $$ Hence, $x=a$ is the only (positive) solution. $b=-\sqrt[2010]{2010}$ also satisfies $b^{2010}=2010$ so that would be a negative solution.

Answer 3

The solution for this equation is $$x=\sqrt[2010]{2010}.$$ The proof is quite easy. Maybe now you can complete it yourself :)