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How does -2^(5/2) -2^(5/2) = -2^(7/2)?

  1. Jul 19, 2015 #1
    1. The problem statement, all variables and given/known data
    How does -2^(5/2) -2^(5/2) = -2^(7/2)?

    2. Relevant equations
    I've integrated a problem down to this and I know that the answer is -2^(7/2). Unfortunately, I've forgotten the algebraic steps required to get it into that form.

    3. The attempt at a solution
    I'm totally lost.
     
  2. jcsd
  3. Jul 19, 2015 #2
    [itex](-2)^{5/2}+(-2)^{5/2}=2(-2)^{5/2}[/itex]

    It might help to think about it like this. We write [itex]x=y^{a/b}[/itex] to mean roughly that [itex]x[/itex] is the number equal to [itex]y[/itex] multiplied together with itself "[itex]a/b[/itex] times". In this sense, how might you think we should write [itex]2(-2)^{5/2}[/itex]?
     
  4. Jul 20, 2015 #3
    Is this logic correct, (-)2^(1+5/2) = (-)2^(2/2)+(5/2) = -2^(7/2)? Can I do that with the negative sign? It doesn't seem like I can.
     
  5. Jul 20, 2015 #4
    If the negative sign is outside the power, you can pull it out all together until you're done simplifying. If the negative sign is being raised to a power as well, you must bring it along.

    So, if you have the negative signs all out front, and not being raised to a power, then yes, because [itex]-x=(-1)\cdot{x}[/itex]. So what you've done is precisely correct in that case, except for the notation. Writing (-) in any situation doesn't make any sense. We use [itex]-n[/itex] (for some number [itex]n[/itex]) to basically mean less than [itex]0[/itex] in the standard ordering of the real numbers (i.e., [itex]3\le\pi[/itex] etc.). In other words, I would write your answer as:

    -2^(5/2)-2^(5/2)=-(2^(5/2)+2^(5/2))=-(2^(1+5/2))=-(2^((2/2)+(5/2)))=-2^(7/2)

    Or in TeX

    [itex]-2^{(5/2)}-2^{(5/2)}=-(2^{(5/2)}+2^{(5/2)})=-(2^{(1+5/2)})=-(2^{(2/2)+(5/2)})=-2^{(7/2)}[/itex]
     
    Last edited: Jul 20, 2015
  6. Jul 20, 2015 #5

    Mentallic

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    Homework Helper

    NihilTico, what you wrote is different to what the OP has. [itex](-2)^n=(-1)^n2^n[/itex] while [itex]-2^n=-(2^n)[/itex]. The first is a positive number when n is an even integer, and negative when n is odd. It is also a complex number when n is neither of those. The second expression however is always a negative number.

    Yes, that is correct, but you don't need to surround the negative sign in brackets. I'd write it like this:

    [tex]-2^{5/2}-2^{5/2}[/tex]
    [tex]=2(-2^{5/2})[/tex]
    [tex]=-2^12^{5/2}[/tex]
    [tex]=-2^{1+5/2}[/tex]
    [tex]-2^{2/2+5/2}[/tex]
    [tex]=-2^{7/2}[/tex]

    But of course when you get more accustomed to the rules, you can skip a lot of these steps.
     
  7. Jul 20, 2015 #6
    Well, yes, I figured that when Rosebud replied with the same notation ;)
     
  8. Jul 20, 2015 #7
    Thank you both for your time and effort. I can finally go to sleep now, lol.
     
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