PRMO 2014 question 7: exponents: power towers

7. If \( x^{(x^4)} = 4 \), what is the value of \( x^{(x^2)} + x^{(x^8)} \)?

Answer: 258

Solution 1: Guesswork
x^4 . log(x) = 2.log(2) = 4.log(sqrt(2)) = (sqrt 2)^4 . log(sqrt(2))
So x = sqrt(2)

Solution 2:

$x^{x^4} = 4$. Let us replace 4 of the exponent of left hand side by $x^{x^4}$ then equation becomes
$x^{x^{x^4}} = 4$. Repeating this process infinitely we get $x^{x^{x^{x^{\cdots}}}} = 4$. Now we replace the exponent of left hand side by 4 and equation now becomes $x^4 = 4$ hence $x = \sqrt{2}$ or $x = -\sqrt{2}$. So
$x^{x^2} + x^{x^8} = 258$

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Solution 3:

Given 

$x^{x^4} = 4$

Now raise to the power 4 on both sides.
$\left(x^{x^4}\right)^4 = 4^4$

We know that $(x^m)^n = x^{mn} = (x^n)^m$ ... (1)

In LHS, $m = x^4$ and $n = 4$

Hence LHS = $(x^4)^{(x^4)}$ (using the first and third terms of the equality in (1))

Let $y = x^4$

Hence LHS = $y^y$, RHS = $4^4$

Solving for real numbers only

y has to be positive as $y = x^4$

The function $y^y$ is increasing. This can be seen easily by finding the derivative and using the condition that $y > 1$ (If $0 < y < 1$, then $y^y < 1$).

$y = 4$ is hence the only positive solution. Hence $x^4 = 4$

$$x = \pm \sqrt[4]{4} = \pm \sqrt{2}$$