Psychguy98
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I was wondering why 8^0 would equal 1? Thanks!
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8^0
- Thread starter Psychguy98
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Psychguy98 said:I was wondering why 8^0 would equal 1? Thanks!
Suppose the variable, a, is any real number, except I will limit it to not being 0 for my demonstration.\(\displaystyle . . **\)
\(\displaystyle One \ interpretation \ is \ 8^0 \ (or \ any \ nonzero \ base \ raised \ to \ 0) \ = \ 8^{a - a} = \frac{8^a}{8^a} = 1.\)
\(\displaystyle This \ would \ be \ true \ because \ a \ number \ divided \ by \ the \ same \ number \\)
\(\displaystyle (except \ if \ that \ number \ equals \ 0) \ equals \ 1.\)
\(\displaystyle ** \ \ Otherwise, \ that \ would \ include \ \ 8^0, \ but \ we \ wouldn't \ know \ if \ \ 8^0 \ were \ equal \ \ to \ \ 0, \\)
\(\displaystyle or \ not, \ to \ even \ do \ an \ allowable \ division.\)
\(\displaystyle And \ we \ can't \ allow \ a \ situation \ where \ we \ have \ \ 0/0, \ in \ case \ it \ were \ the \ case \ that \ 8^0 \ could \ equal \ \ 0.\)
Psychguy98 said:I was wondering why 8^0 would equal 1? Thanks!
Here's one explanation I've seen:
One of the rules for exponents says that when you divide powers of the same base, you SUBTRACT the exponents.
b[sup:1ovn0kpx]m[/sup:1ovn0kpx] / b[sup:1ovn0kpx]n[/sup:1ovn0kpx] = b[sup:1ovn0kpx](m - n)[/sup:1ovn0kpx]
For example, 3[sup:1ovn0kpx]8[/sup:1ovn0kpx] / 3[sup:1ovn0kpx]6[/sup:1ovn0kpx] = 3[sup:1ovn0kpx](8 - 6)[/sup:1ovn0kpx] or 3[sup:1ovn0kpx]2[/sup:1ovn0kpx] or 9
And 5[sup:1ovn0kpx]2[/sup:1ovn0kpx] / 5[sup:1ovn0kpx]2[/sup:1ovn0kpx] = 5[sup:1ovn0kpx](2 - 2)[/sup:1ovn0kpx] or 5[sup:1ovn0kpx]0[/sup:1ovn0kpx]............but we know that 5[sup:1ovn0kpx]2[/sup:1ovn0kpx]/5[sup:1ovn0kpx]2[/sup:1ovn0kpx] = 25/25, or 1.
So, it MUST BE TRUE that 5[sup:1ovn0kpx]0[/sup:1ovn0kpx] = 1
And by similar thinking, we must agree that b[sup:1ovn0kpx]0[/sup:1ovn0kpx] = 1 for any NONZERO value of b, since (as long as b is not 0), b[sup:1ovn0kpx]n[/sup:1ovn0kpx] / b[sup:1ovn0kpx]n[/sup:1ovn0kpx] = 1 = b[sup:1ovn0kpx](n - n)[/sup:1ovn0kpx] = b[sup:1ovn0kpx]0[/sup:1ovn0kpx]