Exponents with fractions and - sign

[ttttrigg3r]

what does this simplify to: (-3^3/2)^2/3

I use the exponent rules to multiply the fractions to get (-3¹) which comes out to -3, but the answer key says the real answer is 3. Is there something important when doing fractional exponent that I'm missing out? Thanks in advance.

 

[HallsofIvy]

\left(-3^{3/2}\right)^{2/3}=((-3^{3/2})^2)^{1/3}
Does that clarify it?

Another way to look at it is that (-3^{3/2})^{2/3}=(-1)^{2/3}(3^{(3/2)(2/3)}). (-1)^{2/3}= 1, not -1.

 

[Ray Vickson]

what does this simplify to: (-3^3/2)^2/3

I use the exponent rules to multiply the fractions to get (-3¹) which comes out to -3, but the answer key says the real answer is 3. Is there something important when doing fractional exponent that I'm missing out? Thanks in advance.

Use of the "rules of exponents" is NOT recommended when you have fractional powers of negative quantities. If you interpret your quantity as A = [(-3)^(3/2)]^(2/3), using the laws of exponents gives A = (-3)^1 = -3. However, the actual value (using general definitions in terms of functions in the complex plane) is A = (3/2) - I*(3/2)*sqrt(3), where I = sqrt(-1). On the other hand, if you meant to write B = [-(3^(3/2)]^(2/3), the laws of exponents gives B = (-1)^(2/3)*3 = (-1)^2*3 = 3. However, the actual value is B = -(3/2) + I*(3/2)*sqrt(3).

RGV

 

[Mark44]

Use of the "rules of exponents" is NOT recommended when you have fractional powers of negative quantities. If you interpret your quantity as A = [(-3)^(3/2)]^(2/3), using the laws of exponents gives A = (-3)^1 = -3.

I don't believe this is a correct interpretation of the problem. The OP shows it as -3^3/2, not (-3)^3/2.

However, the actual value (using general definitions in terms of functions in the complex plane) is A = (3/2) - I*(3/2)*sqrt(3), where I = sqrt(-1). On the other hand, if you meant to write B = [-(3^(3/2)]^(2/3), the laws of exponents gives B = (-1)^(2/3)*3 = (-1)^2*3 = 3. However, the actual value is B = -(3/2) + I*(3/2)*sqrt(3).

This is probably overkill in the context in which this problem was posed (as a prealgebra problem). I believe HallsOfIvy's approach is the right way to go, although it appears that he omitted a factor of 3 towards the end of his calculation.

Corrected, it would be (−3^3/2)^2/3=(−1)^2/3(3^(3/2)(2/3)) = 1 * 3 = 3

 

[ttttrigg3r]

(−33/2)2/3=((−33/2)2)1/3 That I got but (−33/2)2/3=(−1)2/3(3(3/2)(2/3)). (−1)2/3 . 3 = 3 this i do not get. Why is there a multiplication factor there?

side note. how do you copy and paste while keeping super and subscripts?

 

[Mark44]

(−33/2)2/3=((−33/2)2)1/3 That I got but (−33/2)2/3=(−1)2/3(3(3/2)(2/3)). (−1)2/3 . 3 = 3 this i do not get. Why is there a multiplication factor there?

Because -3 = -1 * 3. That's really all that's going on for that part.

side note. how do you copy and paste while keeping super and subscripts?

You can't. I had to reproduce the superscripts from what I had copied.

 

[ttttrigg3r]

(−3^3/2)^2/3=(−1)^2/3(3^(3/2)(2/3)). (−1)^2/3 . 3

looking at that equation, I understand that -3 is the same thing as -1*3 and so that is where (−1)^2/3(3^(3/2)(2/3)) came from but the additional (−1)^2/3 . 3 after that I don't get. where did that last part come from? is the first . supposed to be a multiplication sign or a period? if it is a period it would make sense.

sidenote:
is (x^-1/3)/(y^-1/3)=(y^1/3)/(x^1/3)

 

[Mark44]

(−3^3/2)^2/3=(−1)^2/3(3^(3/2)(2/3)). (−1)^2/3 . 3

looking at that equation, I understand that -3 is the same thing as -1*3 and so that is where (−1)^2/3(3^(3/2)(2/3)) came from but the additional (−1)^2/3 . 3 after that I don't get.

You're right - it shouldn't be there. I copied from HallsOfIvy's post and didn't notice that extra stuff. I corrected it in my earlier post. The expression in HTML is quite complicated, with all those SUP and /SUP tags, so I can see that it would be easy to forget to put something in, which is what I believe happened. Sorry for adding to the confusion.

where did that last part come from? is the first . supposed to be a multiplication sign or a period? if it is a period it would make sense.

The period is meant to indicate multiplication. I changed it to * to make that clearer.

sidenote:
is (x^-1/3)/(y^-1/3)=(y^1/3)/(x^1/3)

Yes.

 

[ttttrigg3r]

Ah then it is perfectly clear. Thanks for the help.