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
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answer #1
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answered by Philo 7
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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
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answer #2
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answered by Mandél M 3
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check out this site.
also 0^0 does not equal 1.
2007-09-13 16:25:29
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answer #3
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answered by Anonymous
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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
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answer #4
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answered by The greatest and the best. 5
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