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Thread: Exponential and Logarithmic Equations

  1. #1
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    Exponential and Logarithmic Equations

    how do i solve

    (e^x + e^-x)/ (e^x - e^-x)

    ln(log(x)) = 0

    sqrt(ln x) = ln sqrt(x)

    (log3X)^2 - log3X^2 = 3

    ln x^2 = (lnx)^2

    5^(2x) - 3(5^x) +2 = 0

    ln x ^(lnx) = 4
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  2. #2
    Super Member bigwave's Avatar
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    Cool (e^x + e^-x)/ (e^x - e^-x)

    $\displaystyle \frac{(e^x + e^{-x})}{ (e^x - e^{-x})}$

    this one is just an expression did you mean to set to zero

    otherwise all one can is just come up with an alternate form
    Last edited by bigwave; Dec 13th 2009 at 11:19 PM. Reason: latex
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  3. #3
    MHF Contributor red_dog's Avatar
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    1) $\displaystyle \sqrt{\ln x}=\ln\sqrt{x}$

    $\displaystyle \ln x\geq 0\Rightarrow x\geq 1$

    The equation can be written as $\displaystyle \sqrt{\ln x}=\frac{1}{2}\ln x$

    Square both members: $\displaystyle \ln x=\frac{1}{4}\ln^2x$

    $\displaystyle \ln x\left(\frac{1}{4}\ln x-1\right)=0$

    Now continue.

    2) $\displaystyle \log_3^2x-\log_3x^2=3\Rightarrow \log_3^2x-2\log_3x-3=0$

    Let $\displaystyle \log_3x=t$ and solve the quadratic.

    3) $\displaystyle \ln x^2=\ln^2x\Rightarrow\ln^2x-2\ln x=0\Rightarrow\ln x(\ln x-2)=0$

    4) $\displaystyle 5^{2x}-3\cdot 5^x+2=0$

    Let $\displaystyle 5^x=t$ and solve the quadratic.

    5) $\displaystyle \ln x^{\ln x}=4\Rightarrow\ln^2x=4\Rightarrow \ln x=\pm 2$
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  4. #4
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    Hello, cheezeandbunnies!

    Here's some help . . .


    $\displaystyle 1)\;\;\frac{e^x + e^{-x}}{e^x - e^{-x}}$ . . . . not an equation

    $\displaystyle 2)\;\;\ln(\log x) \:=\: 0 $

    We have: .$\displaystyle \ln(\log x) \:=\:0 \quad\Rightarrow\quad \log x \:=\:e^0 \:=\:1$

    Therefore: .$\displaystyle \log x \:=\:1 \quad\Rightarrow\quad x \:=\:10^1 \:=\:\boxed{10}$



    $\displaystyle 3)\;\;\sqrt{\ln x} \:=\: \ln(\sqrt{x})$

    We have: .$\displaystyle \sqrt{\ln x} \:=\:\ln\left(x^{\frac{1}{2}}\right) \quad\Rightarrow\quad \sqrt{\ln x} \:=\:\tfrac{1}{2}\ln x \quad\Rightarrow\quad 2\sqrt{\ln x} \:=\:\ln x$

    Square both sides: .$\displaystyle 4\ln x \:=\:(\ln x)^2 \quad\Rightarrow\quad (\ln x)^2 - 4\ln x \:=\:0$

    Factor: .$\displaystyle \ln x(\ln x - 4) \:=\:0$

    Therefore: .$\displaystyle \begin{array}{ccccccc}\ln x \:=\:0 & \Rightarrow & \boxed{x \:=\:1} \\
    \ln x - 4 \:=\:0 & \Rightarrow & \ln x \:=\:4 & \Rightarrow & \boxed{x \:=\:e^4} \end{array}$



    $\displaystyle 4)\;\;(\log_3x)^2 - \log_3x^2 \:=\: 3$

    We have: .$\displaystyle (\log_3x)^2 - 2\log_3x -3 \:=\:0$

    Factor: .$\displaystyle (\log_3x + 1)(\log_3x - 3) \:=\:0$

    Therefore: .$\displaystyle \begin{array}{ccccccc}\log_3x+1\:=\:0 & \Rightarrow & \log_3x \:=\:-1 & \Leftarrow & \text{no real solution} \\ \\[-4mm] \log_3x-3\:=\:0 & \Rightarrow & \log_3x\:=\:3 & \Rightarrow & x \:=\:3^3 \:=\:\boxed{27} \end{array}$

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