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

danacarr · Sep 7, 2006

How do you calculate:

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

(n*)*

HallsofIvy · Sep 7, 2006

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 [itex]/frac{1}{32^n}[/itex]. Doing that twice,
[tex](n*)*= 32^{-\frac{1}{32^n}}[/tex]
which is 1 over the 32ⁿ root of 32.

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

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

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

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