# Thread: Find all positive integers a and b such that a^b = b^a .

1. ## Find all positive integers a and b such that a^b = b^a .

I have this last question from my homework,
Find all positive integers a and b such that a^b = b^a . Prove your answer.

My prof said you can use calculus, think f(x) = x^(1/x), find max and min. And you can also use Fundamental Theory of Arithmetic. This may be very easy for some of you, but I really have no clue right now. Please help me? Thank you!

2. ## Re: Please help! Find all positive integers a and b such that a^b = b^a .

Originally Posted by libzdolce

I have this last question from my homework,
Find all positive integers a and b such that a^b = b^a . Prove your answer.

My prof said you can use calculus, think f(x) = x^(1/x), find max and min. And you can also use Fundamental Theory of Arithmetic. This may be very easy for some of you, but I really have no clue right now. Please help me? Thank you!
Note that $a^b=b^a \Longleftrightarrow a^{1/a}=b^{1/b}$.

Let $f(x) = x^{1/x}$.

$\ln f(x)=\frac{\ln x}{x}$

$\frac{f'(x)}{f(x)}=\frac{x*\frac{1}{x}-\ln x}{x^2}=\frac{1-\ln x}{x^2}$

$f'(x)=0 \implies 1-\ln x=0 \implies x=e$

So, $f(x)$ attains its maximum at $x=e$. (You can verify this by calculating values of $f(x)$ near $x=e$, or by using the second-derivative test.)

For $x the function is increasing, and for $x>e$ the function is decreasing.

So if $a$ is the smaller of $a$ and $b$, $a.

$a=1\text{ or }2$

Clearly, $a=2$. (If $a=1$, we will never be able to find $b$.)

$b>e$

By trial and error, we see that $b^{1/b}=2^{1/2} \implies b=4$.

There are no other solutions.

3. ## Re: Please help! Find all positive integers a and b such that a^b = b^a .

i see nothing in the statement of the problem to preclude the solution (1,1), or more generally (n,n).

EDIT: yes, i caught that in an edit

4. ## Re: Please help! Find all positive integers a and b such that a^b = b^a .

Originally Posted by Deveno
i see nothing in the statement of the problem to preclude the solution (1,1).
Nothing precludes (2, 2), (3, 3), (4, 4) etc. either! So I assumed a and b to be distinct.