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Math Help - Logarithms (advanced)

  1. #1
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    Logarithms (advanced)

    When it comes to math i usually have an indication as to what I am supposed to do but this task was completely impossible for me.

    M(t)=53e-t/223

    t=amount of years. I am supposed to calculate when the function has decreased by 1/3 of it's original size using logarithmical laws. This time I can't show you my steps because I don't even know where to begin!


    I also have this problem that i find equally difficult.

    (ex-5)/(5e-x-4)=2

    I don't even know where to begin and it's driving me absolutely insane lol, would a kind soul please help me ?
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  2. #2
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    Re: Logarithms (advanced)

    $M(t)=53e^{-t/223}$

    let the original size be at $t=0$

    $53e^{0} = 53=M_0$

    we want to find $t$ such that

    $M(t)=\dfrac {M_0}{3} = M_0 e^{-t/223}$

    $\dfrac 1 3 = e^{-t/223}$

    $\ln(1/3) = -t/223$

    $t = -223 \ln(1/3) \approx 245$

    For the second problem

    $\dfrac {e^x-5}{5e^{-x}-4}=2$

    $e^x-5 = 10e^{-x}-8$

    $e^x =10e^{-x}-3$

    Now a trick. Multiply everything by $e^x$ and get all the terms on one side of the equation.

    $e^{2x} + 3e^x -10=0$

    Now let $u=e^x$ and we transform this to a quadratic equation in $u$.

    $u^2 + 3u - 10=0$

    $(u+5)(u-2)=0$

    $u=5 \vee u=-2$

    $x=\ln(u)$

    $\ln(-5)$ is not defined for real numbers so we ignore -5. Thus

    $x=\ln(2)$
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    Re: Logarithms (advanced)

    Thank you so, so much for the help! but i tried the i tried the calculation of the first one and it's apperantly incorrect it says that the awnser is supposed to be x=(a)ln(b) where a and b are whole numbers and where a can't asume the value of 1
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    Re: Logarithms (advanced)

    Quote Originally Posted by Foxtrot View Post
    Thank you so, so much for the help! but i tried the i tried the calculation of the first one and it's apperantly incorrect it says that the awnser is supposed to be x=(a)ln(b) where a and b are whole numbers and where a can't asume the value of 1
    You're correct. I screwed up.

    $M(t)=53e^{-t/223}$

    let the original size be at $t=0$

    $53e^{0} = 53=M_0$

    we want to find $t$ such that

    $M(t)=\dfrac {M_0}{3} = M_0 e^{-t/223}$

    $\dfrac 1 3 = e^{-t/223}$

    Apply $\ln(1/x)=-\ln(x)$

    $\ln(3) = t/223$

    $t = 223 \ln(3)$


    The correct answer is $223\ln(3)$
    Thanks from Foxtrot
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    Re: Logarithms (advanced)

    Quote Originally Posted by romsek View Post
    Quote Originally Posted by Foxtrot View Post
    Thank you so, so much for the help! but i tried the i tried the calculation of the first one and it's apperantly incorrect it says that the awnser is supposed to be x=(a)ln(b) where a and b are whole numbers and where a can't asume the value of 1
    You're correct. I screwed up.
    Maybe it's the beer talking, but as far as I'm concerned, there was no screw up on the psart of Sir romsek. If there was a screw up, I'd say it was on Foxtrot's part for not including that condition on his/her first post in the first place.
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