SOLUTION: x^0=1 how?

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Question 937500: x^0=1
how?
Found 2 solutions by adrian1baldovino, josgarithmetic:
Answer by adrian1baldovino(1) About Me  (Show Source):
You can put this solution on YOUR website!
Ex.1: 5^3 / 5^2 = 125 / 25 = 5
5^3 / 5^2 = 5 ^(3-2) = 5^1 = 5
Ex.2: 5^3 / 5^3 = 125 / 125 = 1
5^3 / 5^3 = 5 ^(3-3) = 5^0 =
... and the result must be 1. So 5^0 =1.
The rule is: x^b / x^c = x^(b-c).
In order to generalize this rule for the case b=c, it must be defined that x^0 = 1 (x is any number different from 0). In mathematics, usefulness and consistency are very important. This convention allows us to extend definitions of power that would otherwise require treating 0 as a special case.
This method can also explain the definition: x^(-b) = 1 / x^b.
Ex. 5^2 / 5^4 = 25 / 625 = 1 / 5^2
5^2 / 5^4 = 5 ^(2-4) = 5^(-2)

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Rules of exponents is how.

x%5E4%2F%28x%5E1%29=%28x%5E3%29%28x%2Fx%29=x%5E3%2A1=x%5E3
but also
x%5E4%2Fx%5E1=x%5E%284-1%29=x%5E3

x%5E2%2Fx%5E1=x%2Ax%2Fx=x%5E1=x
but also
x%5E2%2Fx=x%5E%282-1%29=x%5E1=x

Any real number , especially rational number r,
x%5Er%2Fx%5Er=1, already easily understood from rules for fractions.

x%5Er%2Fx%5Er=x%5E%28r-r%29=x%5E0
As already understood, x%5Er%2Fx%5Er=1=x%5E0