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Math Help - Exponent Problem

  1. #1
    Junior Member BobBali's Avatar
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    Exponent Problem

    Hi All & Good-day,

    The problem states, Show that:

    (\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }

    * Above Denimonator should be 2 power 2x (latex trouble)

    (2^{-x + 4})(2^{-x + 4})

    2^{-2x + 4}(2^{-x}) + 4(2^{-x}) + 16

    2^{-2x} + 2^{-x} (4+4) + 16

    2^{-2x} + 2^{-x} (2^2 + 2^2) + 2^4

    2^{-2x} + 2^{-x+2} + 2^{-x+2} + 2^4


    After this i am at a loss... But i do know that if i take the above first term's
    reciprocal > \frac{1}{2^2x }

    I now have the denominator in the answer, but how to distinguish the 2nd and 3rd terms in the answer...?
    Last edited by BobBali; November 21st 2011 at 12:14 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Exponent Problem

    Quote Originally Posted by BobBali View Post
    Hi All & Good-day,

    The problem states, Show that:

    (\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }

    * Above Denimonator should be 2 power 2x (latex trouble)

    (2^{-x + 4})(2^{-x + 4})

    2^{-2x + 4}(2^{-x}) + 4(2^{-x}) + 16

    2^{-2x} + 2^{-x} (4+4) + 16

    2^{-2x} + 2^{-x} (2^2 + 2^2) + 2^4

    2^{-2x} + 2^{-x+2} + 2^{-x+2} + 2^4


    After this i am at a loss... But i do know that if i take the above first term's
    reciprocal > \frac{1}{2^2x }

    I now have the denominator in the answer, but how to distinguish the 2nd and 3rd terms in the answer...?
    for the first one is not clear in the left hand side there is not 2^x how in the right side there is ??
    for the second one

    (2^{-x+4} ) (2^{-x+4}) = 2^{-x+4-x+4} = 2^{-2x+8} = 2^{-2x} \cdot 2^8
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  3. #3
    MHF Contributor

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    Re: Exponent Problem

    Quote Originally Posted by BobBali View Post
    Hi All & Good-day,

    The problem states, Show that:

    (\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }

    * Above Denimonator should be 2 power 2x (latex trouble)
    This is obviously NOT true. If x= 1, the left side is 6^2= 36 while the left side is \frac{14}{4}= \frac{7}{2}

    To put more than one character in an exponent use ^{ }. For example 2^{2x} requiires 2^{2x}.
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  4. #4
    Junior Member BobBali's Avatar
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    Re: Exponent Problem

    Sorry the RHS of the equation is supposed to look like this>
    1 + 2^{x+3} + 2^{2x+4} \div 2^{2x}
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  5. #5
    MHF Contributor Amer's Avatar
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    Re: Exponent Problem

    Quote Originally Posted by BobBali View Post
    Sorry the RHS of the equation is supposed to look like this>
    1 + 2^{x+3} + 2^{2x+4} \div 2^{2x}
    what about the left hand side, use brackets
    Last edited by Amer; November 21st 2011 at 08:17 PM.
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  6. #6
    Junior Member BobBali's Avatar
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    Re: Exponent Problem

    The LHS is as was shown originally>  (2^{-x} + 4) (2^{-x} + 4)
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  7. #7
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    Re: Exponent Problem

    Hello, BobBali!

    \text{Show that: }\:\left(2^{-x} + 4\right)^2 \:=\: \frac{1 + 2^{x+3} + 2^{2x+4}}{2^{2x} }

    \left(2^{-x} + 4\right)^2 \;=\;\left(\frac{1}{2^x} + 2^2\right)^2 \;=\;\left(\frac{1 + 2^2\!\cdot\!2^x}{2^x}\right)^2 \;=\;\left(\frac{1+2^{x+2}}{2^x}\right)^2

    . . . . . . . . =\;\frac{1 + 2\!\cdot\!2^{x+2} + 2^{2(x+2)}}{(2^x)^2} \;=\;\frac{1 + 2^{x+3} + 2^{2x+4}}{2^{2x}}

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