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Thread: Logarithmic inequality

  1. #1
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    Logarithmic inequality

    $\displaystyle log _{ \frac{1}{3} } x> log _{x}3-2,5 $

    Most important for me is the result as I made this exercise and my result is different as given in the book.
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  2. #2
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    Hello, achacy!

    I have a somewhat intricate solution.
    . . Maybe someone can do better . . .


    $\displaystyle \log _{\frac{1}{3}}x \;>\; \log _{x}3 - 2.5$ .[1]

    I converted the terms to the same base.


    $\displaystyle \text{Let: }\log_{\frac{1}{3}}(x) \:=\:P \quad\Rightarrow\quad \left(\frac{1}{3}\right)^P =\:x$

    . . Take logs: .$\displaystyle \ln\left(\frac{1}{3}\right)^P =\:\ln(x) \quad\Rightarrow\quad P\ln\left(\frac{1}{3}\right) \:=\:\ln(x) $

    . . $\displaystyle P\ln\left(3^{-1}\right) \:=\:\ln(x) \quad\Rightarrow\quad-P\ln(3) \:=\:\ln(x) \quad\Rightarrow\quad P \:=\:-\frac{\ln(x)}{\ln(3)}$

    . . Hence: .$\displaystyle \log_{\frac{1}{3}}(x) \;=\;-\frac{\ln(x)}{\ln(3)} $ .[2]

    From the Base-Change Formula: .$\displaystyle \log_x(3) \:=\:\frac{\ln(3)}{\ln(x)}$ .[3]


    Substitute [2] and [3] into [1]: .$\displaystyle -\frac{\ln(x)}{\ln(3)} \;> \;\frac{\ln(3)}{\ln(x)} - 2.5$

    We have: .$\displaystyle -\frac{\ln(x)}{\ln(3)} - \frac{\ln(3)}{\ln(x)} + 2.5 \;>\;0 \quad\Rightarrow\quad \frac{\ln(x)}{\ln(3)} + \frac{\ln(3)}{\ln(x)} - 2.5 \;<\;0$


    Let $\displaystyle u \:=\:\frac{\ln(x)}{\ln(3)} \quad\Rightarrow\quad u + \frac{1}{u} - 2.5 \;<\;0$

    Muliply by $\displaystyle u\!:\;\;u^2 + 1 - 2.5u \;<\;0$ . . . assuming $\displaystyle u > 0\;\;(x > 1)$

    Multiply by 2: .$\displaystyle 2u^2 - 5u + 2 \;<\;0\quad\Rightarrow\quad (u-2)(2u-1) \;<\;0$


    There are two cases:

    . . $\displaystyle \begin{array}{ccc}u-2\:>\:0 & \Rightarrow & u \:>\:2 \\ 2u-1 \:<\:0 & \Rightarrow & u \:<\:\frac{1}{2}\end{array}\quad\hdots \text{ impossible}$

    . . $\displaystyle \begin{array}{ccc}u-2 \:<\:0 & \Rightarrow & u \:<\:2 \\ 2u-1 \:>\:0 & \Rightarrow & u \:>\:\frac{1}{2}\end{array}\quad \hdots\;\tfrac{1}{2}\:<\:u\:<\:2$


    Then we have: . $\displaystyle \tfrac{1}{2}\;<\:\frac{\ln(x)}{\ln(3)} \;<\;2$

    Multiply by $\displaystyle \ln(3)\!:\;\;\tfrac{1}{2}\ln(3) \;<\;\ln(x) \;<\;2\ln(3) \quad\Rightarrow\quad \ln\left(3^{\frac{1}{2}}\right) \;<\;\ln(x) \;<\;\ln\left(3^2\right) $

    Therefore: .$\displaystyle \boxed{\sqrt{3}\;<\;x\;<\;9} $

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  3. #3
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    put y = lnx/ln3 ===> -y > 1/y - 5/2


    2y^2 -5y + 2 < 0 ===> 1/2< y < 2 ===> rotsq( 3) < x < 9

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