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Thread: very hard question

  1. #1
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    very hard question

    hi this is m question:

    2X$\displaystyle 81^\frac{1}{4}$
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    hi this is m question:

    2x$\displaystyle 81^\frac{1}{4}$
    hint : $\displaystyle 3^4=81$
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  3. #3
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    dunno, how would i do it?
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  4. #4
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    Quote Originally Posted by andyboy179 View Post
    dunno, how would i do it?
    $\displaystyle
    2\times (3^4)^{\frac{1}{4}}
    $
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  5. #5
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    i don't understand!!!!!!!!!!!

    this is a question which i have done:
    3X$\displaystyle 125^\frac{1}{3}$ = cube root 125= 5x3=15

    i need to set it out like this for the question i posted, do you know how to work it out like this?
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  6. #6
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    Quote Originally Posted by andyboy179 View Post
    i don't understand!!!!!!!!!!!

    this is a question which i have done:
    3X$\displaystyle 125^\frac{1}{3}$ = cube root 125= 5x3=15

    i need to set it out like this for the question i posted, do you know how to work it out like this?
    $\displaystyle 5\times 5\times 5=5^3=125$ , therefore , $\displaystyle \sqrt[3]{125}=5$

    so in your case now ,

    $\displaystyle 3\times 3\times 3\times 3=3^4=81 $, therefore , $\displaystyle \sqrt[4]{81}=3$

    Then proceed , $\displaystyle 2\times 3 =6$
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    $\displaystyle 5\times 5\times 5=5^3=125$ , therefore , $\displaystyle \sqrt[3]{125}=5$

    so in your case now ,

    $\displaystyle 3\times 3\times 3\times 3=3^4=81 $, therefore , $\displaystyle \sqrt[4]{81}=3$

    Then proceed , $\displaystyle 2\times 3 =6$

    thankyou very much for all the question you have helped me on
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  8. #8
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    Quote Originally Posted by andyboy179 View Post
    thankyou very much for all the question you have helped me on
    you are welcome .
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