Solve the equation $x^{x^5} = 5$ over $\Bbb R.$

Question

Find real numbers $x$ (if any) such that $x^{x^5} = 5.$

I have shown that $x \notin \Bbb Z.$ Does there exist any $x \in \Bbb R \setminus \Bbb Z$ which satisfies the above equation? Any help will be highly appreciated.

Answer 1

As $x^{x^5} = 5$, we can always substitute the $5$ on the index with $x^{x^5}$ $$\begin{array}{rcll} x^{x^5} &=& 5\\ x^{x^{x^{x^5}}} &=& 5\\ \end{array}$$

Repeat infinitely many times and you will get:

$$x^{x^{x^{x^{\cdots}}}} = 5$$

The index that $x$ is raised to, is same as the entire "power tower", which is 5:

$$x^{\boxed{x^{x^{x^{\cdots}}}}} = 5$$ $$x^5 = 5$$

We get $x = \sqrt[5]{5}$.

Answer 2

The hint.

$\sqrt[5]5$ is the root.

Now, prove that it's an unique root for which consider two cases:

1. $0<x\leq1$;

2. $x>1$.

Answer 3

Here are graphs that show the solution, $x = \sqrt[5]{5}$:

Answer 4

\begin{align} x^{x^5}&=5 \tag{1}\label{1} \end{align}

Equation \eqref{1} can be solved formally with the help of the Lambert W function:

\begin{align} x^5\ln x&=\ln 5 ,\\ 5x^5\ln x&=5\ln 5 ,\\ x^5\ln (x^5)&=5\ln 5 ,\\ \ln (x^5)\,\exp(\ln (x^5))&=\ln 5\exp(\ln5) . \end{align}

At this point we can apply the Lambert W function: \begin{align} \operatorname{W}(\ln (x^5)\,\exp(\ln (x^5))) &= \operatorname{W}(\ln 5\exp(\ln5)) , \end{align} and since the argument $\ln 5\exp(\ln5)>0$, there is just one solution

\begin{align} \ln (x^5)&=\ln5 ,\\ x^5&=5 ,\\ x&=5^{\tfrac15}\approx 1.379729661 . \end{align}

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Edit

Or, less formally, without Lambert W function:

\begin{align} (x^{x^5})^5&=5^5 ,\\ x^{5x^5}&=5^5 ,\\ (x^5)^{(x^5)}&=(5)^{(5)} ,\\ x^5&=5 . \end{align}