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Solving Negative Exponents

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data
    8x^-3 = 64


    2. Relevant equations
    None.


    3. The attempt at a solution
    I tried doing all sorts of things, changing 8x^-3 to (1/8x)^3 or trying to get both sides to have the same base, but couldn't get it to work.

    The book I am using does not explain how to do so, I have already looked through the whole chapter that the problem is from.

    Primarily, I would like assistance in understanding the concepts behind this problem. With the current problem, it is relatively easy to figure out by educated guessing & checking, but when there are more difficult numbers and more challenging problems, guessing and checking won't work. Thanks for the assistance.
     
  2. jcsd
  3. Oct 23, 2007 #2

    berkeman

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    Staff: Mentor

    Hint: (x^-3)^3 = ?
     
  4. Oct 23, 2007 #3
    (x^-3)^3 = X^-9 but I'm not quite sure how that helps.

    Thanks for the response.
     
  5. Oct 23, 2007 #4

    Dick

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    Science Advisor
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    Berkeman meant (x^-3)^(-1/3).
     
  6. Oct 23, 2007 #5
    Got it. I didn't realize you could raise both sides to a power ((-1/3) in this case). Thanks for the help.
     
  7. Oct 23, 2007 #6

    berkeman

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    Staff: Mentor

    Whoops. Thanks Dick.
     
  8. Oct 24, 2007 #7
    Have you tried dividing both sides by 8?

    Then did you try expressing x^-3 as an expression with a positive exponent using the rule a^-b = 1/a^b
     
  9. Oct 24, 2007 #8

    CompuChip

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    You can always raise both sides to the same power.
    In fact this is exactly what you're doing when solving something like
    [tex]3 x^2 = 27[/tex]
    If you divide out the 3 you get
    [tex]x^2 = 9[/tex]
    and you would take the square root to get x = 3 (or - 3 of course). But you can also see it as raising both sides to the power 1/2:
    [tex](x^2)^{1/2} = x^{2 \times 1/2} = x \quad = \quad 9^{1/2} = \sqrt{9}[/tex]
    where the last equality is just a change of notation, so you see that [tex]\sqrt{x} = x^{1/2}[/tex].

    But the raising-both-sides-to-a-power-trick works even in the case of negative and fractional exponents.
     
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