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Math Help - Laws of Exponents dealing with fractions

  1. #1
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    Laws of Exponents dealing with fractions

    I have a problem to solve but I am stuck trying to find out how to determine the correct answer. If it was the same variable I would be able to do it but since it's two variables (m and n) I'm not sure how to proceed. I am supposed to re-write it as a single exponent. Please forgive me if the equation format is not using the proper symbols.


    Code:
    (m/n)^5 (m/n)^-6 (m/n)^0
    -------(Fractional Line)-------
             (m/n)^20
    Thanks

    The answer should be: (n/m)^21
    Last edited by Joe222; March 22nd 2014 at 05:12 PM.
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  2. #2
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    Re: Laws of Exponents dealing with fractions

    An equation implies it is equal to something?
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  3. #3
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    Re: Laws of Exponents dealing with fractions

    Quote Originally Posted by fourierT View Post
    An equation implies it is equal to something?
    Updated the first post post to clarify
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    Re: Laws of Exponents dealing with fractions

    The mistake is calling it an equation, that is simply a fraction to solve, but anyway.

    The powers can be treated the same as if it was one variable seeing as every term in the brackets is the same.

    Attempt it and post your working, I'll help you to solve if you get stuck.
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  5. #5
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    Re: Laws of Exponents dealing with fractions

    Code:
    (m/n)^5 (m/n)^-6 (m/n)^0
    -------(Fractional Line)-------
             (m/n)^20
    =

    Code:
    (m^5/n^5) (m^-6/n^-6) (m/n)
    -------(Fractional Line)--------
    (m^20/n^20)
    And this is where I get stuck...

    If it were for example:

    Code:
    (m^5/m^5)
    I would know to just subtract the powers but since the variables are different, I don't know the proper way to procede.
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  6. #6
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    Re: Laws of Exponents dealing with fractions

    First of all, you don't need to take the power inside the brackets. Treat (m/n)^5 as you would say x^5 for the first few steps, it's only later that you may need to bring it inside (even then you don't need to but it will be easier for you to see)

    Also anything to the power of zero is 1.

    Anything to the power of one is itself
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  7. #7
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    Re: Laws of Exponents dealing with fractions

    So something like this then...

    Code:
    (m/n)^5 (m/n)^-6 (m/n)^0
    -------(Fractional Line)-------
    (m/n)^20
    =

    Code:
    (m/n)^5 (m/n)^-6 (1)
    -------(Fractional Line)-------
    (m/n)^20

    =

    Code:
    (m/n 5 - (-6)) (1)
    -------(Fractional Line)-------
    (m/n)^20
    =

    Code:
    (m/n)^11
    -------(Fractional Line)-------
    (m/n)^20
    =

    Code:
    (m/n)^11 - 20

    =

    Code:
    (m/n)^-9

    Not sure if I'm on the right track here...
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  8. #8
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    Re: Laws of Exponents dealing with fractions

    You are on the right track but you made a mistake, when you multiply terms the exponents are added not taken away.

    (m/n)^5 * (m/n)^-6 gives you 5-6 not 5--6 so you get

    (m/n)^-1

    The next part is correct, you are dividing by (m/n)^20 so you would subtract the 20, correcting your previous error you end up with (m/n)^-21

    Do you know how to proceed from here?
    Thanks from Joe222
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    Re: Laws of Exponents dealing with fractions

    I think the next step would be to invert from (m/n) to (n/m) so the exponent would become positive?
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    Re: Laws of Exponents dealing with fractions

    Do you know the steps, or did you just guess that because you have the answer?
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  11. #11
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    Re: Laws of Exponents dealing with fractions

    This is as easy as 1,2,3 if you are familiar with these exponential laws.
    • 1- When multiplying terms with the same base, we add the exponents
    • 2- When dividing terms with the same base, we subtract exponents

    Any way,

     \frac{ (\frac{m}{n})^5 \times (\frac{m}{n})^-^6 \times (\frac{m}{n})^0}{ (\frac{m}{n})^2^0 }

     =\frac{ (\frac{m}{n})^5^-^6^+^0}{\frac{m}{n}^2^1}

     =(\frac{m}{n})^-^1^-^2^0

     = (\frac{m}{n}) ^-^2^1 $
    Last edited by sakonpure6; March 22nd 2014 at 06:33 PM.
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  12. #12
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    Re: Laws of Exponents dealing with fractions

    .
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  13. #13
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    Re: Laws of Exponents dealing with fractions

    Quote Originally Posted by sakonpure6 View Post
    This is as easy as 1,2,3 if you are familiar with these exponential laws.
    • 1- When multiplying terms with the same base, we add the exponents
    • 2- When dividing terms with the same base, we subtract exponents

    Any way,

     \frac{ (\frac{n}{m})^5 \times (\frac{n}{m})^-^6 \times (\frac{n}{m})^0}{ (\frac{n}{m})^2^0 }

     =\frac{ (\frac{n}{m})^5^-^6^+^0}{\frac{n}{m}^2^1}

     =(\frac{n}{m})^-^1^-^2^0

     = (\frac{n}{m}) ^-^2^1 $
    This is wrong, the terms are (m/n) not (n/m).
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    Re: Laws of Exponents dealing with fractions

    Quote Originally Posted by fourierT View Post
    Do you know the steps, or did you just guess that because you have the answer?
    Well, it's a logical guess on my part because that's what I recall regarding negative exponents as a final answer need to be changed into positive exponents.


    Thank you for your help.
    Last edited by Joe222; March 22nd 2014 at 06:37 PM.
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  15. #15
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    Re: Laws of Exponents dealing with fractions

    @fourier Sorry, I fixed the response. But you should of seen the general idea.
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