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Math Help - Binomial expansion?

  1. #1
    Super Member fardeen_gen's Avatar
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    Binomial expansion?

    At what x the 6^{th} term in the expansion of the binomial:
    \left[\sqrt{\left(2^{\log (10 - 3^x)}\right)} + \sqrt [5]{\left(2^{(x - 2)\log 3}\right)}\right]
    is equal to 21 if it is known that the binomial coefficients of the 2^{nd},3^{rd} and the 4^{th} terms in the expansion represent respectively the first, third and fifth terms of an arithmetic progression?

    (A) x = 6
    (B) x = 2
    (C) x = -2
    (D) x = 11
    Last edited by fardeen_gen; May 3rd 2009 at 07:35 PM. Reason: 3^{rd} :)
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Maybe I'm mistaking... but if x=6, then 10-3^6<0 \Rightarrow \log(10-3^6) is not defined.
    Same thing for x=11.

    ------------------------
    If x=2
    then \log(10-3^x)=\log(1) \Rightarrow 2^{\log(10-3^x)}=1

    And 2^{(x-2)\log(3)}=1

    So the expression is 2.
    ------------------------
    Therefore, x=-2 ??
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  3. #3
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Maybe I'm mistaking... but if x=6, then 10-3^6<0 \Rightarrow \log(10-3^6) is not defined.
    Same thing for x=11.

    ------------------------
    If x=2
    then \log(10-3^x)=\log(1) \Rightarrow 2^{\log(10-3^x)}=1

    And 2^{(x-2)\log(3)}=1

    So the expression is 2.
    ------------------------
    Therefore, x=-2 ??
    Ahh! Back to basics Nice.
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