Can someone please answer my question about radical exponents?

Question

This is the expression. ( / is what I am drawing as a radical sign)
____
3 / _
/64

That's the best I could do. 3 is the index in the a large radical sign over radical 64. I am trying to rationalize the denominator.

The answer is 6/64 (6 being the index of radical 64) = 2

I know you're supposed to multiply the indexes but my question is how did the 3 index become a 6. Why do you just out of the blue give the radical 64 an index of 2 when it didn't have one? Can anyone understand what I am asking?

Answer 1

OK, some basics first.
A radical sign with no number above it represents the square root. The square root of x is the same as x to the power (1/2).

A radical sign with a number (say n) above it represents the "nth root" (if n = 3 we call this the cube root). This is the same as x to the power (1/n). A radical sign with no number above it actually has an implicit 2 there - that is the default, anything else you have to specify.

So the cube root of the square root of 64 is (64^(1/2)) ^ (1/3) which by the laws of powers is 64^((1/2) * (1/3)) [this is why you multiply the indexes] = 64^(1/6) = 2. Or you can just say √64 = 8, so we have cube root of 8 = 2.

But to repeat, the answer to your main question is that a radical sign without a number above it is the same as one with a 2 above it. The 2 is understood to be there by default.

Answer 2

I'm not sure I understand your question, but here's what I think the problem is:

By convention, a radical symbol with no index specified is considered to be a square root. So you're taking the third root of the second root of 64. Multiplying the 3 and 2 together, you get the sixth root.

Answer 3

Are you saying you have the cube root of 64? Or the cube root of the square root of 64? I'm guessing the latter.

Remember that the nth root of a number x is just x^(1/n). So here I suppose they're doing:

(64^(1/2)) ^(1/3) =
64^(1/2 * 1/3) =
64^(1/6) =
2