# Problem Of The Week #385 Sep 26th, 2019

Status
Not open for further replies.

Staff member

#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solution!

1. kaliprasad
2. MegaMoh

Solution from kaliprasad :
We are given $a^{b^c} = a^{bc}$

Or $a^{b^c-bc} = 1$

This has got 2 sets of solutions

1) a = 1 and b and c can be any positive integer

2) a = any positive integer and

$b^c - bc = 0$

Or $b(b^{c-1}) = c$

As b is not zero so we have $b^{c-1} = c$

if c = 1 we get b any number

Or $b= \sqrt[c-1]c$ $c>=2$

The root can be taken to be integer if c-1 = 1 or c =2 giving b = 2

So solution set $(1,b,c)$ or $(a,2,2)$ or $(a,b,1)$ where a,b,c are any integers.

Alternate solution from MegaMoh :
$a^{b^c}=(a^b)^c$

$\vdash a^{b^c}=a^{bc}$

$\implies b^c\ln a=bc\ln a$

$\implies (b^c-bc)\ln a=0$

(1) $\ln a=0 \implies a=e^0=1$
so $1$ triple is ($1$, $x$, $y$) for any $x\in \mathbb{Z} ^+$ and $y \in \mathbb{Z} ^+$

(2) $b^c-bc=0$

$b^c=bc$

$\implies c\ln b=\ln(bc)=\ln b+\ln c$

$\implies (c-1)\ln b=\ln c$

$\implies \ln b=\frac1{c-1}\ln c=\ln(c^\frac1{c-1})$

$\implies b=c^\frac1{c-1}$

$\implies b^{c-1}=c$, here $c=1$ is a solution regardless of $b$ so another pair is (x, y, 1) for any $x\in \mathbb{Z}^+$ and $y \in \mathbb{Z}^+$

since $y=\sqrt[x-1]{x}$ has an asymptote at $y=1$ and $x=0$ the only positive integers that satisfy that equation are $(2, 2)$ and $(1, y^+)$ (which is a hole). since the asymptote is at $y=1$ so there are no solutions below $(2, 2)$ and another at $x=0$ so there are no solution to the left of $(1, y^+)$
so the only positive integer pairs are that satisfy $a^{b^c}=(a^b)^c$ are (1, x, y), (x, y, 1), and (x, 2, 2) for any $x\in \mathbb{Z} ^+$ and $y \in \mathbb{Z} ^+$

Status
Not open for further replies.