A special exponent

Algebra Level 3

$\large \begin {cases} x = a^b = b^a\\ b = a^a\\ a \neq b\end {cases}$

The above system of equations is true for only one triplet of positive integers $(x, a, b)$ . What is $x$ ?

The answer is 16.

1 solution

Chew-Seong Cheong
Jun 23, 2016

$\begin{aligned} a^b & = b^a \\ a^{a^a} & = (a^a) ^a \\ \implies a^{a^a} & = a^{a^2} \end{aligned}$

$\implies \begin {cases} a = \pm 1 & \implies \color{#D61F06}{b = \pm 1^{\pm 1} = a} & \color{#D61F06}{\text{Unacceptable}} \\ a^a =a^2 & \implies \color{#3D99F6}{a =2 \implies b = 2^2 = 4 \ne a} & \color{#3D99F6}{\text{Acceptable}} \end {cases}$

$\implies x = 2^4 = 4^2 = \boxed{16}$

You probably should explain why $a \neq 1$ because it is another solution of $a^a=a^2$

Sam Bealing - 4 years, 11 months ago

If $a=1$ then $b=a^a=1=a$ contradict with $b \neq a$

Pham Khanh - 4 years, 11 months ago

I know, I just think it should be clarified in the solution.

Sam Bealing - 4 years, 11 months ago

Thanks, I have included it in the solution.

Chew-Seong Cheong - 4 years, 11 months ago

If that's the case then the solution should also exclude $a=-1$ , because

$\large(-1)^{(-1)^{-1}} = (-1)^{(-1)^2} = -1$

Hung Woei Neoh - 4 years, 11 months ago

Thanks, I will edit my solution.

Chew-Seong Cheong - 4 years, 11 months ago

Well the problem only allows three positive integers. Negative numbers are not allowed.

D C - 4 years, 11 months ago

@D C – Ah, I didn't read the question carefully. Sorry, it was my mistake

Hung Woei Neoh - 4 years, 11 months ago

A special exponent

The answer is 16.

1 solution

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