How to Simplify Exponential Expressions with Fractions?

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In summary, when evaluating expressions such as {(-4)}^{\frac{3}{2}} or 8^{\frac{2}{3}}, it is important to first simplify the expression by performing the simplest operation first, such as taking the square root or cube root. In cases where the base is a negative number, the principle value of the logarithm must be used to find the principle value of the exponent. In complex arithmetic, things can get more complicated due to the possibility of multiple solutions for the exponent. However, using known theorems and definitions can help clarify the situation and determine the correct solution.
  • #1
Swapnil
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I always get confused when you get things like:
[tex]{(-4)}^{\frac{3}{2}}[/tex] or [tex]8^{\frac{2}{3}}[/tex].

Which operation do you do first and why?
Do you take the square/squareroot first and then take the cube/cuberoot or what?
 
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  • #2
For [tex] (-4)^{\frac{3}{2}} [/tex] take the square root first and then cube it.


For [tex] (8)^{\frac{2}{3}} [/tex] take the cube root first and then square it.
 
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  • #3
Which operation do you do first and why?
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
Swapnil said:
I always get confused when you get things like:
[tex]{(-4)}^{\frac{3}{2}}[/tex] or [tex]8^{\frac{2}{3}}[/tex].

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

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

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

but then you can also do

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

See what I mean?
 
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  • #6
Swapnil said:
But chosing which operation to do first actually gives you a different answer! Look at this:

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

but then you can also do

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

See what I mean?

This is because the square root is not single valued, notice that both 2i and -2i are square roots of -4.
 
  • #7
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
Swapnil said:
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...

Positive doesn't have much meaning when dealing with complex numbers, but yes I believe we usually do take the principle root.
 
  • #9
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.

Hurkyl said:
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
 
  • #10
D H said:
[*] you didn't read Hurkyl's note, repeated below with critical part in bold.
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.

D H said:
[*] (-4)^(1/2) has two solutions, as does (-64)^1/2.
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)?
 
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  • #11
Swapnil said:
I did read it. But the theorem does 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.
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
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?
 
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  • #13
My postings so far have been based on the assumption that you know the definition of the complex exponential,

[tex]
a^b = \exp(b \log a) = e^{\Re (b \log a)}
( \cos (\Im (b \log a)) + i \sin (\Im (b \log a))
[/tex]

(and the principal value of [itex]a^b[/itex] is found by using the principal value of the compex logarithm)

Was that correct?
 
  • #14
Hurkyl said:
Was that correct?
What was what correct?
 
  • #15
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
Hurkyl said:
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)
Yes it was correct.

So if I want to evaluate [tex](-4)^\frac{3}{2}[/tex] then

[tex](-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)))[/tex]

[tex]= \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)))[/tex]

[tex]= \exp(\frac{3}{2}\log(4)) ( \cos(\frac{3}{2}\pi) + i \sin (\frac{3}{2}\pi) )[/tex]

[tex]= 4^\frac{3}{2}(-i) [/tex]
[tex]= -8i [/tex]

what happened to the solution [tex]+8i[/tex]?
 
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  • #17
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
...or I can just use the fact that
[tex]4^\frac{3}{2} = \pm 8[/tex],
right?
 
  • #19
that's right.
 

Related to How to Simplify Exponential Expressions with Fractions?

What is an exponent?

An exponent is a mathematical notation that represents the number of times a number is multiplied by itself.

What is the base number in an exponent?

The base number in an exponent is the number that is being multiplied by itself. It is the number that is raised to the power of the exponent.

What is the power of an exponent?

The power of an exponent is the small number written above and to the right of the base number. It represents the number of times the base number is multiplied by itself.

How do you read an exponent?

An exponent is read as "the base number raised to the power of the exponent." For example, 23 would be read as "2 raised to the power of 3."

What is the difference between a positive and negative exponent?

A positive exponent means the base number is being multiplied by itself a certain number of times. A negative exponent means the base number is being divided by itself a certain number of times. For example, 23 is equal to 2 x 2 x 2, while 2-3 is equal to 1 / (2 x 2 x 2).

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