What is zero to the zero power...?

Question

If x to the zero power is 1, and zero to the power of x is 0, then what is zero to the power of zero?

Answer 1

Actually, it is possible to have exponents of 0 and lower, though negative exponents are more frequently used in connection with powers of 10 (for example, 10^-3=.001).

You answered your own question in the first clause. No matter what the value of X is, as long as it's 0 or positive (and this may be true for negative numbers too; I'm not sure), x^0 will always equal 1.

But 0^x doesn't always equal 0. If x=0, then we have 0^0=1.

It's x/0 that is undefined. If x<>0, then 0/x=0 because x*0=0. But no matter what the value of x, you couldn't be able to calculate x/0 because what could you multiply by 0 to get x? By this definition, technically 0/0 should equal 0 because 0*0=0, but it's undefined by the accepted rules of mathematics.

Answer 2

0^0 is undefined. However, if some expression equals 0^0, like x^x when x = 0, then it is actually an example of what is called an *indeterminate* form, which in some ways is even worse than undefined. Indeterminate means that it *could* be undefined, but it could also have a defined value. To find out the status of an indeterminate form, you need to use L'hopital's Rule, which says that if you have some expression that ends up being 0/0 or inf/inf, you take the derivative of the top and bottom, and the result will be equal. If you still have 0/0 or inf/inf, you do it again, and again....

for lim x->0 x^x, first take the natural log, which equals ln(x^x) = x*lnx. @ x=0, this ends up being 0 * -inf, which in this case is still indeterminate, so to make this in a form where we can use L'hopital's Rule, change the lnx to 1//1/lnx (they are equal). Now, we have x on the top, and -1/lnx on the bottom. Since both approach 0 as x approaches zero, we use L'hopital's Rule, and obtain 1 in the numerator, and 1/(ln(x))^2. This number is 0 when x approaches 0, so now we know that lim x->0 x*lnx = 0. However, we took the natural log of our original expression before we began, and need to undo that: lim x->0 e^(x // -1/lnx) = e^0 = 1.

You can also confirm this with a non-rigorous limit: on a calculator, find .5^.5, then .1^.1, then .05^.05, and so on. Notice how the answers get closer and closer to 1? That serves as empirical evidence that while 0^0 may be undefined, the limit of x^x as x approaches 0 is actually equal to 1, which as close as you can get to a value of 0^0.

Answer 3

Zero to the zero power is equal to 0/-0, which is not possible so the product is undefined.

Answer 4

is 0

Answer 5

It's called an indeterminate form. See

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Answer 6

0 to the 0 power is undefined.
You can prove this by taking the logarithm...

Answer 7

Undefined

Answer 8

The answer is zero, but this can not be. An exponet has to be 2 or greater.

Answer 9

Zero.
Thanks