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Math Help - Logs

  1. #1
    Member M670's Avatar
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    Logs

    OK so i have \log_2(2x)+\log_2(x-3)=3
    I think I should rewrite it as follows \log_2(2x(x-3))=3 then what?
    and I need help with e^-2x+3\bullet e^-x=10
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Logs

    Quote Originally Posted by M670 View Post
    OK so i have \log_2(2x)+\log_2(x-3)=3
    I think I should rewrite it as follows \log_2(2x(x-3))=3 then what?
    So far so good. Now, since 2^{\log_2(a)} = a what's your next step?

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Logs

    Quote Originally Posted by M670 View Post
    and I need help with e^-2x+3\bullet e^-x=10
    First the LaTeX code. Instead of using \bullet I'd use \cdot. It looks a bit nicer.

    e^{-2x} + 3 \cdot e^{-x} = 10

    For the sake of clarity, I'd multiply both sides by e^{2x}. This step is not vital and you can easily alter the substitution I'm going to use.

    So the equation is going to be:
    1 + 3 e^x = 10e^{2x}

    10e^{2x} - 3e^x - 1 = 0

    Now let y = e^x. This gives
    10y^2 - 3y - 1 = 0

    Solve this then back substitute to get x.

    -Dan
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  4. #4
    Member M670's Avatar
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    Re: Logs

    Quote Originally Posted by topsquark View Post
    So far so good. Now, since 2^{\log_2(a)} = a what's your next step?

    -Dan
    I have seen in the past that you can use this 2^{\log_2(a)} = a to get this (2x\cdot(x-3))=3^2 which I believe would give me  2x^2-6x-9=0 then I solve it using the quadratic formula. But I don't know the reason why I am able to square it and then drop the log off?
    Last edited by M670; December 9th 2012 at 06:57 AM.
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  5. #5
    Member M670's Avatar
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    Re: Logs

    Quote Originally Posted by M670 View Post
    I have seen in the past that you can use this 2^{\log_2(a)} = a to get this (2x\cdot(x-3))=3^2 which I believe would give me  2x^2-6x-9=0 then I solve it using the quadratic formula. But I don't know the reason why I am able to square it and then drop the log off?
    So I believe my answer is x=  \frac{6+\sqrt108}{4}
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