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Math Help - Logarithmic equation

  1. #1
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    Logarithmic equation

    Hi!
    Today at math class we were given some examples to solve, all were easy to solve, except for this one:

    I tried to rewrite 3^log(3,x) as simply x and (3^log(2,x))^2 as 3^2log(2,x) or 3^log(2,x^2), but couldn't see anything to do further.
    I asked the teacher and she didn't seem to know either :P
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  2. #2
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    Quote Originally Posted by regdude View Post
    Hi!
    Today at math class we were given some examples to solve, all were easy to solve, except for this one:

    I tried to rewrite 3^log(3,x) as simply x and (3^log(2,x))^2 as 3^2log(2,x) or 3^log(2,x^2), but couldn't see anything to do further.
    I asked the teacher and she didn't seem to know either :P
    That must be a typo and the equation should read:

    \left(3^{\log_2 x} \right)^2 - 4 \cdot 3^{\log_2 x} + 3 = 0

    Substitute y = 3^{\log_2 x}

    and solve for y.

    Consequently re-substitute and solve for x.

    I've got x = 2~\vee~x=1

    If there isn't a typo you can only use numerical methods (for instance Newton) to get an approximate solution.
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  3. #3
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    Well not a typo in my notebook, the book has exactly how I had it.
    I tried to use Microsoft Math to solve it for me, aprox. answer was 1.
    The answer in book was x=2 and x=1.
    So to get to the answer the second logarithm base must be 2.
    Thanks for clearing that up
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