• Jul 9th 2008, 06:46 PM
princess_anna57

16^(-1/2)
(27)^(1/3) - with questions that have two brackets, only the first number has brackets on paper.
(27)^(-1/3)
(-27)^(1/3)
(-27)^(-1/3)
(-2)^(-3)
• Jul 9th 2008, 06:59 PM
Reckoner
Quote:

Originally Posted by princess_anna57

16^(-1/2)
(27)^(1/3) - with questions that have two brackets, only the first number has brackets on paper.
(27)^(-1/3)
(-27)^(1/3)
(-27)^(-1/3)
(-2)^(-3)

$x^{m/n} = \sqrt[n]{x^m}$

$x^{-n} = \frac1{x^n}$
• Jul 9th 2008, 07:16 PM
princess_anna57
Thanks!
I got the first three, but I'm not sure about the last ones. What difference does it make if x is a negative? does it mean that it's undefined/does not exist?
• Jul 9th 2008, 07:23 PM
Reckoner
Quote:

Originally Posted by princess_anna57
I got the first three, but I'm not sure about the last ones. What difference does it make if x is a negative? does it mean that it's undefined/does not exist?

Well, if you try to take the square root (or any even root) of a negative number, you will not get a real answer (because the square of any real number is nonnegative). However, you can take cube roots of a negative number, and in fact you can take the $n^{\rm th}$ root of a negative number as long as $n$ is odd.

Thus, $\sqrt[3]{-27} = -3$, for example, because $(-3)^3 = (-3)(-3)(-3) = -27$.