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Thread: sum of roots...

  1. #1
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    sum of roots...

    find Sum of Roots of the equation $\displaystyle 4x^{\lfloor log_{10}2x-3 \rfloor} = 1$

    Where $\displaystyle \lfloor x \rfloor$ = greatest integer function.

    like $\displaystyle \lfloor 2.3 \rfloor = 2$
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  2. #2
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    $\displaystyle 4x^{\lfloor \log_{10}2x-3 \rfloor} = 1$
    $\displaystyle \log_{10}(4x^{\lfloor \log_{10}2x-3 \rfloor}) = \log_{10}(1)$
    $\displaystyle \log_{10}4 + \log_{10}(x^{\lfloor \log_{10}2x-3 \rfloor})) = 0$
    $\displaystyle \log_{10}4 + \lfloor \log_{10}2x-3 \rfloor \log_{10}(x) = 0$
    $\displaystyle \log_{10}4 + \lfloor \log_{10}2x-3 \rfloor (\log_{10}2x - \log_{10}2) = 0$

    Let $\displaystyle y = \log_{10}2x$
    $\displaystyle \log_{10}4 + \lfloor y-3 \rfloor (y - \log_{10}2) = 0$

    First solve without greatest integer (to get close to the solution)
    $\displaystyle \log_{10}4 + (y-3) (y - \log_{10}2) = 0$
    $\displaystyle y^2 - (3 + \log_{10}2)y + 5 \log_{10}2 = 0$
    Solutions are around 2.75 and 0.55

    Now with this we guess values for $\displaystyle a = \lfloor y\rfloor$ (0,1,2,3)
    $\displaystyle \log_{10}4 + (a - 3) (y - \log_{10}2) = 0$
    $\displaystyle y = -\frac{\log_{10}4}{a - 3} + \log_{10}2$
    We see that $\displaystyle a = 0$ is the only one with a consistent solution.
    $\displaystyle y = \frac{\log_{10}4}{3} + \log_{10}2 = \log_{10}2^{\frac{5}{3}}$
    $\displaystyle y = \log_{10}2x$
    $\displaystyle x = \frac{10^y}{2} = 2^{\frac{2}{3}}$

    And this is the only solution for $\displaystyle x$
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