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Math Help - Another Logarithmic question

  1. #1
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    Another Logarithmic question

    Solve for x.

    a) 2^x+2-2^x =96

    2^x (2^2 +2) =96 nvm i did addition instead of subtraction i see my mistake... on this question.

    2^x(6)=96/6

    2^x=16

    2^4=16

    x=4 (the answer at the back of the book says its 5)


    c) 4^x+1 +4^x=160

    4^x (4^2+1)=160

    4^x(17)=160/17

    4^x=9.41

    (the answer is suppose to be 2.5 but i dont see where they are coming from.)
    Last edited by Skoz; December 16th 2009 at 03:56 PM.
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  2. #2
    Member kjchauhan's Avatar
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    do you mean

    2^x+2-2^x =96

    Is this correct?
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  3. #3
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    skeeter's Avatar
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    Quote Originally Posted by Skoz View Post
    Solve for x.

    a) 2^x+2-2^x =96

    2^x (2^2 +2) =96 nvm i did addition instead of subtraction i see my mistake... on this question.

    2^x(6)=96/6

    2^x=16

    2^4=16

    x=4 (the answer at the back of the book says its 5)


    c) 4^x+1 +4^x=160

    4^x (4^2+1)=160

    4^x(17)=160/17

    4^x=9.41

    (the answer is suppose to be 2.5 but i dont see where they are coming from.)
    please use parentheses to make your math expressions/equations clear, like so ...

    2^(x+2) - 2^x = 96

    4^(x+1) + 4^x = 160



    4^{x+1} + 4^x = 160

    4^x(4 + 1) = 160

    4^x = 32

    2^{2x} = 2^5

    2x = 5

    x = 2.5
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