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Simplify the following problem with exponents

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Simplify. Express final answers with positive exponents.

    ((64m^-36)(n^-15)(p^9)) ^ (2/3)

    2. Relevant equations
    [tex]\sqrt[3]{}[/tex] ((64m^-36)(n^-15)(p^9)) ^ 2


    3. The attempt at a solution

    ((4m^-36)(n^-15)(p^9))^2
    = (16m^-36)(n^-15)(p^9)

    = 1/(16m^36)(n^15)(p^-9)

    I can't seem to rid the negative exponents. It would be amazing if someone could find me a page which shows the laws and helps me with this kinda stuff. thanks.
     
  2. jcsd
  3. Dec 8, 2008 #2

    Mentallic

    User Avatar
    Homework Helper

    Re: Exponents

    ok you have the wrong idea of how the power outside the entire expression works.

    [tex](ab)^n=a^nb^n[/tex]

    So: [tex][(64m^{-36})(n^{-15})(p^9)]^{\frac{2}{3}}[/tex]
    becomes
    [tex](64m^{-36})^{\frac{2}{3}}(n^{-15})^{\frac{2}{3}}(p^9)^{\frac{2}{3}}[/tex]

    and remember the rule must be applied again for each part.
    It works like this:
    [tex](ab^x)^n=a^nb^{xn}[/tex]

    As for making all the exponents positive, you don't need to take all the terms into the denominator. For the 3 pronumerals in the question, you only need to take 2 pronumerals and you are allowed to do that.

    e.g. [tex]a^xb^{-x}c^{-y}=\frac{a^x}{b^xc^y}[/tex]
     
  4. Dec 9, 2008 #3
    Re: Exponents

    I guess I wrote it wrong, but it's [tex](64m^{-36}n^{-15}p^9)^{\frac{2}{3}}[/tex]
     
  5. Dec 9, 2008 #4
    Re: Exponents

    Also, would [tex]{(-2x^{-3}y)(-12x^{-4}y^{-2}) / {6xy^{-3}}[/tex]

    = [tex]1/4x^{-6}y^2[/tex]

    If so, how do I make the exponents positive??
     
  6. Dec 9, 2008 #5

    Mark44

    Staff: Mentor

    Re: Exponents

    No.
    In the numerator you have (-2)(-12) = 24, and in the denominator you have 6. You should get 24/6 = 4, not 1/4 for the numeric coefficient.

    The exponent on x is -3 + (-4) - 1 = -8.
    The exponent on y is 2.
    To make an exponent positive, replace the exponential factor by its reciprocal. For example, 2x^(-2) = 2*(1/x^2) = 2/(x^2).
     
  7. Dec 9, 2008 #6
    Re: Exponents

    I got p^18/16m^72n^30

    Is it right? :)
     
  8. Dec 9, 2008 #7

    Mark44

    Staff: Mentor

    Re: Exponents

    Not even close.
    You started with [tex](64m^{-36}n^{-15}p^9)^{\frac{2}{3}}[/tex]
    That's going to be [tex]64^{2/3}m^{-36 * 2/3}n^{-15*2/3}p^{9*2/3}[/tex]

    For now, simplify 64^(2/3), which is the same as the cube root of 64, squared, and get exponents on the variables that are integers.

    After you do that, any variables that have negative exponents can be put in the denominator with positive exponents.
     
  9. Dec 9, 2008 #8

    Mentallic

    User Avatar
    Homework Helper

    Re: Exponents

    This is exactly the same as what you posted in your first post. The extra brackets for each pronumeral make no difference.

    We have [tex](64m^{-36})^{\frac{2}{3}}(n^{-15})^{\frac{2}{3}}(p^9)^{\frac{2}{3}}[/tex]

    Again, use the rule that [tex](a^nb)^x=a^{nx}b^x[/tex]
    This must be used for each and every part. See if you can apply this idea to the problem above.
     
  10. Dec 12, 2008 #9
    Re: Exponents

    I got

    (p^6) / (16m^24)(n^10)
     
  11. Dec 12, 2008 #10
    Re: Exponents

    Here is the result, step by step:

    [tex][(64m^{-36})(n^{-15})(p^9)] ^ {2/3}[/tex]

    [tex]64^{2/3}*m^{-24}*n^{-10}*p^6[/tex]

    [tex]64=2^6[/tex]

    [tex]64^{2/3}=2^{6*2/3}=2^4=16[/tex]

    [tex]16*\frac{1}{m^{24}}*\frac{1}{n^{10}}*p^6[/tex]

    [tex]\frac{16*p^6}{m^{24}*n^{10}}[/tex]

    [tex](\frac{4*p^3}{m^{12}*n^{5}})^2[/tex]
     
    Last edited: Dec 12, 2008
  12. Dec 12, 2008 #11
    Re: Exponents

    Uhm, the last statement does not equal the one above it..
     
  13. Dec 12, 2008 #12
    Re: Exponents

    Sorry, I got problems with LaTeX code, now it is ok. It should be:

    [tex](\frac{4*p^3}{m^{12}*n^{5}})^2[/tex]
     
    Last edited: Dec 12, 2008
  14. Dec 12, 2008 #13

    Mark44

    Staff: Mentor

    Re: Exponents

    The last expression (not statement) DOES equal the one above.
     
  15. Dec 12, 2008 #14
    Re: Exponents

    Yes, now it does. Please care to read his edit.
     
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