X raised to power-X raised to power-3 equals to 3.

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The question is what are the possible values of $x$ when we have

$$x^{x^3} = 3$$

(that is $x^3$ in the exponent itself and not $x*3$).

I solved one answer by guessing that $x = \sqrt[3]3$. My work is below however I was only able to solve it assuming $x = \sqrt[3]3$, can anyone solve it without the assumption. I suspect that logarithms are involved which I have to review. Also there are supposed to be more solutions to this than just the one mentioned. Thanks in advance and enjoy. It's a fun problem]1

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2 Answers

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If one wishes to use logarithms to facilitate then we proceed accordingly. We have that

$$x^{x^3}=3\implies x^3\log x=\log 3\implies x^3\log x^3=\log 3^3 \implies (x^3)^{x^3}=3^3$$

Of course this result could be obtained instantly by simply cubing both sides of the original equation.

Therefore a solution is evidently $x^3=3$ or $x=3^{1/3}$.

Given that $x^x$ is monontoically increasing for $x>1/e$, then $x=3^{1/3}$ is the only real-valued solution.

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$$\begin{align*} x^{(x^3)}&=3\\ x^{(x^3)}&=\sqrt[3]{3}^{(\sqrt[3]{3}^3)} \end{align*}$$ So that base are same, hence $$x = \sqrt[3]{3}$$

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