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Math Help - [SOLVED] need logarithm help

  1. #1
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    [SOLVED] need logarithm help

    hi i need help on solving these questions with all the log rules i'm confusing myself..

    1). 7^(x-2) = 1/2^(x-2)

    2). 10^(x^2+2x-8) = 1

    3). (1/4)^(x+2) * 32^(x-1) = 64^(x-2)

    i know i have to solve for x but i keep getting the wrong answers when i plug it into my graph and find the intersections for the answer.

    thanks!
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  2. #2
    Super Member Anonymous1's Avatar
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    For (1) do you mean 7^{x-2}= \frac{1}{2^{x-2}}?

    Then, (x-2)\log(7) = (2-x)\log(2)

    => (x-2)=(2-x)\frac{\log(2)}{\log(7)}

    Let \frac{\log(2)}{\log(7)} = \alpha

    and solve to find x= 2

    For (2) take the log of both sides:

    \log_{10} (10^{(x^2+2x-8)}) = log_{10}(1)

    => x^2+2x-8 = \log_{10}(1)

    => x^2+2x-8-\log_{10}(1) = 0

    Now use the quadratic equation to find your solutions. Where,

    a= 1
    b= 2
    c= -(8+\log_{10}(1))

    For (3) again take the log of both sides.

    \log[(1/4)^{x+2}\times 32^{x-1}] = \log[64^{x-2}]

    => (x+2)\log(1/4)+(x-1)\log(32) = (x-2)\log(64)

    ...
    Last edited by Anonymous1; March 17th 2010 at 06:48 PM.
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  3. #3
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    wow thanks a lot! it really helped, i was really off with some of them
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  4. #4
    Super Member Anonymous1's Avatar
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    Logarithms with start to get intuitive, just keep practicing.

    Good luck,
    Anonymous
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