Results 1 to 2 of 2

Thread: Log question

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    61

    Log question

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by Detanon View Post
    1. From the LHS of the equation you know $\displaystyle x = 9^{\log_9(x)}$

    2. Plug in this term into the RHS:

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

    3. Consequently the given equation

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

    becomes

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

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

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

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

    Square both sides and solve the quadratic equation for x.

    Check afterwards if both solutions are valid.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum