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2007-09-13 16:19:14 · 4 answers · asked by pink_kitties_and_rhinoplasty 2 in Science & Mathematics Mathematics

4 answers

look at the pattern:
10² = 100, 10^1 = 10, 10^0 = ???
each time we drop the exponent by 1, we divide the previous value by 10, so the last one should be 10/10 = 1. This works with any nonzero base.

Also, if the rule for multiplying is to add exponents, (x^a)(x^b) = x^(a+b), then if (x^a)(x^b) = x^a, then a+b = a, so b must be 0, and multiplying x^a by x^b left x^a unchanged, so x^b must be 1. So x^0 = 1 for any x <> 0.

2007-09-13 16:31:13 · answer #1 · answered by Philo 7 · 1 0

We know from the laws of indices that a^n/a^p = a^(n-p)

Where a, n and p are real numbers

But consider and instant where n = p

we would get a^n/a^p = 1

but a^n/a^p = a^(n-p) = a^0

so a^0 = 1

which is true for any number

2007-09-13 16:28:28 · answer #2 · answered by Mandél M 3 · 1 0

check out this site.
also 0^0 does not equal 1.

2007-09-13 16:25:29 · answer #3 · answered by Anonymous · 1 0

alrighty...
if you multiply something by itself one time, what do you get?
if you multiply something by itself -1 time what do you get?
interpolate.

2007-09-13 16:25:50 · answer #4 · answered by The greatest and the best. 5 · 1 2

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