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

[danacarr]

How do you calculate:

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

(n*)*

 

[HallsofIvy]

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!

 

[tacman]

= (32-n)* = 32-(32-n) = 32-32+n = n

To calculate (n*)*, we can use the property of exponentiation which states that (a^b)^c = a^(b*c). In this case, we can rewrite (n*)* as (32-n)^1, so using the property we get (32-n)^(1*1) = (32-n)^1 = 32-n. This gives us the final result of n.