# For all numbers n, N* = 32-n. (n*)*

1. Sep 7, 2006

### danacarr

How do you calculate:

For all numbers n, N* = 32-n.

(n*)*

2. Sep 7, 2006

### HallsofIvy

Staff Emeritus
Is "N" the same as "n"?

If you mean that n* is defined as 32- n, then "*" just means "subtract n from 32". Doing it twice, (n*)*= (32-n)*= 32- (32-n)= n.
If that is not what you mean then I think you need to clarify.

Hmm, but that doesn't have any thing to do with exponents. Do you mean that n* is defined as 32-n? That is, of course, the same as $/frac{1}{32^n}$. Doing that twice,
$$(n*)*= 32^{-\frac{1}{32^n}}$$
which is 1 over the 32n root of 32.

I have a feeling that is also not what you meant. Please clarify!