Root Powers

Okay simplify 9^(1/2) how would you show your work?

You know 9^3 = 9*9*9 right?

So how would you do that "9*9*9" thing with a fractional power?

9^(1/2) = sqr(9) = 3

9^(1/3) would be the cubic root of 9

I'm just trying to give some ideas in case you need more than that:).

9^1 = 9
9^0 = 1

Therfore 9^0.5 must be somewhere in between that.

Also, x^y = sqr(x)^2y (since for each x you can substitute sqr(x) * sqr(x))

Therefore

9^0.5 = sqr(9)^1 = sqr(9) = 3^1 = 3

x^n for rational values of n can be done with powers and roots. For example.

x^(5/6) is the 6th root of (x^5).

For irrational values of n, it is more complicated, requiring natural logarithm and exponential functions.

ln(x^n) = n*ln(x)

x^n = e^[ ln(x^n) ]

Natural logarithms and the exponential function can be calculated using series which converge to the required values. The series are described in various mathematical texts and handbooks.

You guys didn't get the point. I know how to find the roots of numbers but how would it be simplified?

4^2 = 2*2
4^1/2 = ?*?*?*?*?
9^2 = 81
9^1/2 = ?*?*?*?*?

So what would you put in for the "?" on each one.

That is what I mean. What would you fill in for the "?"?

You can't, no such expansion exists.

Then how the hell do calculators do it? I guess they got about infinite memory to store each kind of root of every number. :rolleyes:

Actually, calculators (or so I've heard) GUESS they're answers, then improve upon that guess until they get it right ;)

Well, actually 2^0.5 = 2^0.25 * 2^0.25 = 2^0.125 * 2^0.125 * 2^0.125 * 2^0.125, etc:).

To calculate square roots, most modern calculators would use this formula:

X = 0.5 * (X + Y/X)

Where Y is the number you want the square root of, and X is the guess as to what it might be. Each time the guess is replaced with a more accurate value until the answer is as accurate as needed; then it stops:). Try it out:).

If you really wanted to you could use the binomial theorem, which works for all real powers, as well as natural powers, but becomes an infinite sum. I'm not sure if that's what you wanted though.