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Math Help - 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)

    \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; December 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) \sqrt{\ln x}=\ln\sqrt{x}

    \ln x\geq 0\Rightarrow x\geq 1

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

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

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

    Now continue.

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

    Let \log_3x=t and solve the quadratic.

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

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

    Let 5^x=t and solve the quadratic.

    5) \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 . . .


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

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

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

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



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

    We have: . \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: . 4\ln x \:=\:(\ln x)^2 \quad\Rightarrow\quad (\ln x)^2 - 4\ln x \:=\:0

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

    Therefore: . \begin{array}{ccccccc}\ln x \:=\:0 & \Rightarrow & \boxed{x \:=\:1} \\<br />
\ln x - 4 \:=\:0 & \Rightarrow & \ln x \:=\:4 & \Rightarrow & \boxed{x \:=\:e^4} \end{array}



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

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

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

    Therefore: . \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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