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Math Help - Rational Exponents

  1. #1
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    Smile Rational Exponents

    i dont understand what to do when is it like this

    (9+16)^1/2

    my answer was 7? is that correct???


    and what if it is like this

    9^1/2 + 16^1/2

    i also got 7???

    i have more questions...
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  2. #2
    Senior Member Stroodle's Avatar
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    Hi.

    (9+16)^{\frac{1}{2}}=25^{\frac{1}{2}}=\sqrt{25}=5

    9^{\frac{1}{2}}+16^{\frac{1}{2}}=\sqrt{9}+\sqrt{16  }=3+4=7
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  3. #3
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    9+16=25. So 25^{1/2}=\sqrt{25}=5.

    Your answer to the second question was correct.

    EDIT: Too slow...
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  4. #4
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    hi, thank you :]

    my other questions are...

    27^2/3

    and i got (27^1/3)^2 = 3^2 = 9?

    andd the next one is

    x^5/8 divided by X ^1/6

    i dont know how to start that one...

    and my other question is...

    x^2/3(x^1/4 -x)

    i dont knowww :[
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  5. #5
    Senior Member Stroodle's Avatar
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    When dividing you subtract the indices and when multiplying you add the indices (first the indices need to have a common denominator):

    x^{\frac{5}{8}}\div x^{\frac{1}{6}}= x^{\frac{15}{24}}\div x^{\frac{4}{24}} =x^{\frac{11}{24}}

    For the second one you need to add the indices:

    x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)= x^{\frac{8}{12}}\times x^{\frac{3}{12}}-x^{\frac{2}{3}}\times x^{\frac{3}{3}}=x^{\frac{11}{12}}-x^{\frac{5}{3}}

    Does that make sense?
    Last edited by Stroodle; July 16th 2009 at 09:11 PM.
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  6. #6
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    Your first answer is correct.

    For the second part, find common denominators among the fractional exponents:

    \frac{x^{5/8}}{x^{1/6}}=\frac{x^{15/24}}{x^{4/24}}=x^{11/24}.

    Similarly for the third question, which I'll let you try yourself.
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  7. #7
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    For the second one you need to add the indices:

    x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)= x^{\frac{8}{12}}\times x^{\frac{3}{12}}-x^{\frac{2}{3}}\times x^{\frac{3}{3}}=x^{\frac{11}{12}}-x^{\frac{5}{3}}

    Does that make sense?[/quote]



    im a little confused how do you get x^11/12?
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  8. #8
    Senior Member Stroodle's Avatar
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    x^{\frac{2}{3}}(x^{\frac{1}{4}}-x) is the same as x^{\frac{8}{12}}(x^{\frac{3}{12}}-x)

    So as you're multiplying x^{\frac{8}{12}} by x^{\frac{3}{12}} the indices added together =\frac{11}{12}

    Similarly, as x^{\frac{2}{3}}(x^{\frac{1}{4}}-x) is the same as x^{\frac{2}{3}}(x^{\frac{1}{4}}-x^{\frac{3}{3}}) you get x^{\frac{5}{3}} as the second term.
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  9. #9
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    ohhhh okayyyy thanks so much


    i have another simliar question


    (x^1/3 + x^-1/3)^2

    how would you do that?
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  10. #10
    Senior Member Stroodle's Avatar
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    (x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2

    =\left( x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\right)\l  eft( x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\right)

    =(x^{\frac{1}{3}})^2+2(x^{\frac{1}{3}})(x^{-\frac{1}{3}})+(x^{-\frac{1}{3}})^2

    =x^{\frac{2}{3}}+2+\frac{1}{x^{\frac{2}{3}}}
    Last edited by Stroodle; July 17th 2009 at 02:41 AM. Reason: fixed error
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  11. #11
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    so the x's cant be simplified anymore after that???



    and i have another question, wanted to see if my answer was correct...

    cube root 125

    and the answer is 5?
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  12. #12
    Senior Member Stroodle's Avatar
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    Nope. That's fully simplified.

    Your answer is correct. \sqrt[3]{125}=5 as 5\times 5\times 5=125

    You should check out the exponent section in this site Practical Algebra Lessons , it explains them really well.
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  13. #13
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    Quote Originally Posted by Stroodle View Post
    (x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2

    =x^{\frac{2}{3}}+x^{-\frac{2}{3}}

    =x^{\frac{2}{3}}+\frac{1}{x^{\frac{2}{3}}}
    Whoa, that's not right. You have to FOIL it:
    (x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2
    \begin{aligned}<br />
&= (x^{\frac{1}{3}})^2 + 2(x^{\frac{1}{3}})(x^{-\frac{1}{3}}) + (x^{-\frac{1}{3}})^2 \\<br />
&= x^{\frac{2}{3}} + 2 + x^{-\frac{2}{3}}<br />
\end{aligned}


    01
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  14. #14
    Senior Member Stroodle's Avatar
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    Whoa, yeah.

    Sorry illeatyourxface. Silly mistake.
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