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Negative exponents and calculation rules

  1. Feb 21, 2006 #1
    I just want to know if a negative exponent is as just the same as saying one over another number.
    For example:
    5^1/3 = 5^-3

    Another thought
    would base numbers only affect base numbers and exponents only affect exponents?
     
    Last edited: Feb 21, 2006
  2. jcsd
  3. Feb 21, 2006 #2
    No, that is not right. To put it in your word: a negative exponent is the same as saying one over the whole number. In other words:

    [tex]x^{-a}=\frac{1}{x^a}[/tex]
     
  4. Feb 21, 2006 #3
    oops i meant to say

     
  5. Feb 21, 2006 #4
    .

    Anyway, it it still not correct. See my previous post.
     
  6. Feb 22, 2006 #5

    HallsofIvy

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    That would be saying [itex]a^{-x}= \frac{a^1}{something}[/itex], wouldn't it?? That, of course, is wrong. Again
    [tex]a^{-x}= \frac{1}{a^x}[/tex]
     
  7. Feb 22, 2006 #6

    arildno

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    Eeh? Come again?
    This is just incomprehensible.
     
  8. Feb 22, 2006 #7
    Just figured it out in the calculator,
     
  9. Feb 23, 2006 #8

    HallsofIvy

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    Figured out what? Are you saying you now know how to simplify something like [itex]\frac{a^6}{a^{-4}}[/itex] or are you just saying you'll let your calculator do it for you?
     
  10. Feb 23, 2006 #9
    No i just put 2^2 gave me an answer 2^1/2 gave me an answer 2^-2 gave me an answer and none of the answers were the same.
     
  11. Feb 24, 2006 #10
    Specifically, [tex]5^{1/3}=\sqrt[3]{5}[/tex] and [tex]5^{-3}=\frac{1}{5^3}[/tex]

    Not sure what you mean by your second question. Do you have an example in mind?

    -Dan
     
  12. Mar 1, 2006 #11
    It's not a question i just wanted to know the question previous to that dilemna had any way to relate to the dilemna. For even lamens terms:
    1st part of first question.
    2 over 3 is exactly the same as 4 over 6 only its simplified.
    would 5^-3 be the same as 5^1/3? (Yes it does)
    In number 2*2 = 4, 2^2*2^2 = 2^4, you know what i mean... example of a problem.
    3(2x*3) = 6x*9 notice 3 outside of the brackets affect the numbers inside of it? If the 2x was 2x^2 would it be then after 6x^6*9.

    Sorry, my english. I am trying to improve on it, please correct my grammar if you can. (It was all a misunderstanding, lol)
     
  13. Mar 1, 2006 #12

    chroot

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    No. 5^-3 = 0.008.

    5^(1/3) = 1.70998

    The reason they're different is because the exponents are different. If -3 does not equal 1/3 (it does not) then 5^-3 cannot equal 5^(1/3).

    - Warren
     
  14. Mar 1, 2006 #13
    No

    3(2x*3) = 6x*3 you essentiall multiplied by 9. and lets say you have just x2 for a second if you multiply that by 3 you get 3x2 not 3x6 The exponent is unaffected because you don't know for sure if x = 3 or what x equals
     
  15. Mar 2, 2006 #14

    VietDao29

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    No, you should know that:
    [tex]\sqrt[n] {a ^ m} = a ^ {\frac{m}{n}}[/tex].
    And:
    [tex]a ^ {-m} = \frac{1}{a ^ m}[/tex].
    This is due to:
    [tex]\frac{a ^ m}{a ^ n} = a ^ {m - n}[/tex].
    So:
    [tex]a ^ {-m} = a ^ {0 - m} = \frac{a ^ 0}{a ^ m} = \frac{1}{a ^ m}[/tex]
    Since a0 = 1 for all a <> 0.
    Now, back to your problem:
    [tex]5 ^ {\frac{1}{3}} = \sqrt[3] {5} \approx 1.71[/tex], whereas:
    [tex]5 ^ {-3} = \frac{1}{5 ^ 3} = \frac{1}{125} = 0.008[/tex].
    And of course you know that:
    0.008 <> 1.71, right? :)
     
  16. Mar 2, 2006 #15
    correct me if i'm wrong
    Question=(2x+2)(3x+3)
    -Foiled
    =6x^2+6x+6x+6
    =6x^2+12x+6
    so....
    3(2x+3) = 6x+9 (so side note to that) 3(2x*3)= 6x*3
    so....
    2^1/3 = 3v--2 and 2^-3 = 1/2^3 = 1/8
    so....
    can anyone give me some good ways of remembering this stuff? Or atleast tips?
     
  17. Mar 2, 2006 #16

    chroot

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    Everything you've posted looks correct, runicle. I'd advise that you use the notation sqrt(2) instead of "v--2" to represent the square root, or use the latex features built into the site.

    How to remember this stuff? Most of it becomes second nature once you being using it a bit. Which of your "operations" are you having trouble remembering?

    - Warren
     
  18. Mar 2, 2006 #17
    Do 30 problems and I would think that you would know every single thing without thinking about it anymore.
     
  19. Mar 3, 2006 #18
    I am still a little fuzzy with what happens when you add, multiply exponents and what and what not can you add or multiply with. Like as an example 2x^2 + 2x can't be added... Do you catch my drift? Along with what you can and cannot do when doing certain tasks. Is there a very good website that can tell you right away what expressions or equations would bring you to know common tasks?
     
  20. Mar 4, 2006 #19

    VietDao29

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    My suggestion is that you should go over your textbook again thoroughly, try to understand the concept, then try your hands on some problems, and remember the formulae.
    [tex]a ^ x \times a ^ y = a ^ {x + y}[/tex]
    [tex]\frac{a ^ x}{a ^ y} = a ^ {x - y}[/tex]
    ---------------
    Now of course, you cannot "add" 2x2 + 2x to get 4x2 or 4x. Just think like this:
    Writing 3x2 means that you have three x2's (it's like you have 3 apples), 5x2 means that you have five x2's. If you add them together, you'll have 8 x2's, right?
    3x2 + 5x2 = 8x2.
    Now 3x2 + 2x cannot be added since x is not the same as x2, you cannot add 3 apples, and 2 orranges, right? However, it can be factored like this:
    3x2 + 2x = x(3x + 2).
    Can you get this? :)
     
  21. Mar 4, 2006 #20

    HallsofIvy

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    Yes, that's correct.
    I wish you wouldn't use different symbols for the same thing!
    Does (2x*3) mean the same as 2x^3? If so then both of those are correct.
    It took me a while to figure that out! the v-- thing is a root!
    Yes, [itex]2^{\frac{1}{3}}= ^3\sqrt{2}[/itex]. Click on that to see the LaTex code I used.

    Yes, that also is true.
    The same way you get to Carnegie Hall- practice, practice, practice! Do lots of homework problems. If you teach assigns half the exercises on a page- do all of them!
     
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