Equal Powers

Algebra Level 1

$\Large x^{x^{x}}=(x^{x})(x^{x})$ Other than $1$ , what integer value could be $x$ ?

The answer is 2.

1 solution

Kaizen Cyrus
Dec 17, 2017

$\begin{aligned} x^{x^{x}} & = (x^{x})(x^{x}) \\ x^{x^{x}} & = x^{x+x} \\ x\color{#3D99F6}^{x^{x}} & = x\color{#3D99F6}^{2x} \\ \implies \color{#3D99F6}x^{x} & = \color{#3D99F6}2x \end{aligned}$

Only $x=2$ would satisfy the equality above.

$\begin{aligned} \color{#3D99F6}2^{2} & = \color{#3D99F6}2(2) \\ \color{#3D99F6}4 & = \color{#3D99F6}4 \end{aligned}$

So, the answer is $\boxed{2}$ .

$\begin{aligned} 2\color{#3D99F6}^4 & = 2\color{#3D99F6}^4 \\ 16 & = 16 \end{aligned}$

Nice explanation, but this only yields the "nice" solution to this problem (apart from 1). Unfortunately, there is a third solution, because x^x is continuous, positive and smaller or equal to 1 for 0<x<1,so there has to be an intersection with 2x at some point between 0 and 1.

By approximation, Wolfram Alpha yields something close to 0.3, and I don't see any "nicer" way of getting there, but I thought it would be interesting to add that there is one more solution, even if it's not as pretty as the others.

Felix H. - 11 months ago

I noticed that there's another solution. Although I DID say I was looking for an integer value.

Kaizen Cyrus - 11 months ago

Equal Powers

The answer is 2.

1 solution

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