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Math Help - Log question

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    Log question

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  2. #2
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    Quote Originally Posted by Detanon View Post
    1. From the LHS of the equation you know x = 9^{\log_9(x)}

    2. Plug in this term into the RHS:

    \frac12 \log_3\left({9^{\log_9(x)}}\right) = \frac12 \log_3(9) \cdot \log_9(x) = \frac12 \cdot 2 \cdot \log_9(x) = \log_9(x)

    3. Consequently the given equation

    \log_3(2x+1) = \log_9(5x^2-x+5)

    becomes

    \log_3(2x+1) = \frac12 \log_3(5x^2-x+5)

    \log_3(2x+1) =  \log_3(\sqrt{5x^2-x+5})

    4. Now use both sides as exponents to the base 3:

    2x+1 = \sqrt{5x^2-x+5}

    Square both sides and solve the quadratic equation for x.

    Check afterwards if both solutions are valid.
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