# Really Basic Stuff: Exponents

1. Dec 4, 2006

### Swapnil

I always get confused when you get things like:
$${(-4)}^{\frac{3}{2}}$$ or $$8^{\frac{2}{3}}$$.

Which operation do you do first and why?
Do you take the square/squareroot first and then take the cube/cuberoot or what?

2. Dec 4, 2006

For $$(-4)^{\frac{3}{2}}$$ take the square root first and then cube it.

For $$(8)^{\frac{2}{3}}$$ take the cube root first and then square it.

Last edited: Dec 4, 2006
3. Dec 4, 2006

### Hurkyl

Staff Emeritus
To compute it directly, you do neither. You merely raise -4 to the 3/2 power, and raise 8 to the 2/3 power.

If you want to compute it in some other fashion, you have to invoke some extra knowledge. For example, if a > 0, you have a theorem1 that says

(ab)c = abc,

which would allow you to split this single exponent operation into two separate exponent operations. If you were able to do so, then it's clear what you would do first.

1: This theorem is for real arithmetic; things get messier when you are doing complex arithmetic

4. Dec 5, 2006

### Staff: Mentor

courtrigrad showed you the way we would do it -- perform the simplest operation first. Whichever makes the most sense if there is a simple operation to do in your head, do that first and deal with the rest second. So in the first one, taking the square root of the perfect square 4 is easiest (you have to deal with the imaginary part of sqrt(-4), but that's no big deal). In the second one, the cube root of 8 is easiest to do first in your head. In other examples, look for stuff you can do first for an even answer, then deal with the rest.

5. Dec 5, 2006

### Swapnil

But chosing which operation to do first actually gives you a different answer! Look at this:

$$(-4)^\frac{3}{2} = \Big[(-4)^\frac{1}{2}\Big]^3 = (2i)^3 = -8i$$

but then you can also do

$$(-4)^\frac{3}{2} = \Big[(-4)^3\Big]^\frac{1}{2} = (-64)^\frac{1}{2} = 8i$$

See what I mean?

Last edited: Dec 5, 2006
6. Dec 5, 2006

### d_leet

This is because the square root is not single valued, notice that both 2i and -2i are square roots of -4.

7. Dec 5, 2006

### Swapnil

But don't we always take the positive square root when we do these types of things. You know, taking only the principle square root...

8. Dec 5, 2006

### d_leet

Positive doesn't have much meaning when dealing with complex numbers, but yes I believe we usually do take the principle root.

9. Dec 5, 2006

### D H

Staff Emeritus
Two points:
1. you didn't read Hurkyl's note, repeated below with critical part in bold.
2. (-4)^(1/2) has two solutions, as does (-64)^1/2.

10. Dec 6, 2006

### Swapnil

I did read it. But the theorem doesn't help me when I have a negative base. And all he said about complex aritmetic was that things get messy. That didn't really helped me.

I see. But why this impartiality. Why is it that be consider both positive and negative complex roots but only consider positive real roots (i.e. the principle square root)?

Last edited: Dec 6, 2006
11. Dec 6, 2006

### Hurkyl

Staff Emeritus
The point of my post was "invoke other knowledge" -- use theorems and other facts you know about arithmetic in order to clarify the situation. (definitions are always a good start) The one theorem I put in my post was just an example of the process.

12. Dec 6, 2006

### Swapnil

I see, but I didn't meant any disrespect when I said that btw.

You know, now that I think about it why would it be messier for complex numbers? Doesn't that same theorem work for negative bases too?

Last edited: Dec 6, 2006
13. Dec 7, 2006

### Hurkyl

Staff Emeritus
My postings so far have been based on the assumption that you know the definition of the complex exponential,

$$a^b = \exp(b \log a) = e^{\Re (b \log a)} ( \cos (\Im (b \log a)) + i \sin (\Im (b \log a))$$

(and the principal value of $a^b$ is found by using the principal value of the compex logarithm)

Was that correct?

14. Dec 7, 2006

### Swapnil

What was what correct?

15. Dec 8, 2006

### Hurkyl

Staff Emeritus
My assumption that you knew the definition of complex exponentiation. I guess that's irrelevant now since I just told you. (Though you would need to know the complex logarithm to use it)

16. Dec 18, 2006

### Swapnil

Yes it was correct.

So if I want to evaluate $$(-4)^\frac{3}{2}$$ then

$$(-4)^\frac{3}{2} = \exp(\frac{3}{2}\log(-4)) = \exp( \Re (\frac{3}{2} \log(-4)) ) ( \cos (\Im (\frac{3}{2} \log(-4))) + i \sin (\Im (\frac{3}{2} \log(-4)))$$

$$= \exp( \Re (\frac{3}{2} (\log(4)+i\pi)) ) ( \cos (\Im (\frac{3}{2}(\log(4)+i\pi))) + i \sin (\Im (\frac{3}{2}(\log(4)+i\pi)))$$

$$= \exp(\frac{3}{2}\log(4)) ( \cos(\frac{3}{2}\pi) + i \sin (\frac{3}{2}\pi) )$$

$$= 4^\frac{3}{2}(-i)$$
$$= -8i$$

what happened to the solution $$+8i$$?

Last edited: Dec 18, 2006
17. Dec 18, 2006

### Hurkyl

Staff Emeritus
Well, it looks like you were taking the principal value of the logarithm. (You should capitalize Log when you do that) Thus, you got the principal value of the exponential.

18. Dec 19, 2006

### Swapnil

...or I can just use the fact that
$$4^\frac{3}{2} = \pm 8$$,
right?

19. Dec 19, 2006

### tehno

that's right.