Question

How do you solve `3^ { n + 2} = n ^ { n + 2}`?

Answer 1

`n=3` or `n=-2`

Explanation:

Given:

`3^(n+2) = n^(n+2)`

Note that if we put `3` in place of every `n`, then the left hand side and right hand side are identical:

`3^(3+2) = 3^(3+2)`

So `n=3` is a solution.

Also if we put `n=-2`, then both exponents are zero and so both sides are equal to `1`...

`3^0 = 1 = (-2)^0`

So `n=0` is a solution.

Are these the only solutions?

We can divide both sides of the original equation by `3^(n+2)` to get:

`1 = n^(n+2)/3^(n+2) = (n/3)^(n+2)`

For any other real value of `n` than `n=3` or `n=-2`, we have `n/3 != 1` raised to a non-zero power. That cannot result in `1`.

I may think about possible complex solutions later.