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Math Help - Logarithms

  1. #1
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    Logarithms

    The concentration of alcohol C% in John's blood when he consumes alcohol could be represented by C=0.3te^{-0.912t} where t is the number of hours after he consumes alcohol.

    Find the value of t when the concentration of alcohol in John's blood is a maximum.

    Attempt:

    100=0.3te^{-0.912t}

    \frac{100}{0.3}=te^{-0.912t}

    \frac{100}{0.3}=lnt+lne^{-0.912t}

    \frac{100}{0.3}=lnt-0.912t

    stucked
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  2. #2
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    Quote Originally Posted by Punch View Post
    The concentration of alcohol C% in John's blood when he consumes alcohol could be represented by C=0.3te^{-0.912t} where t is the number of hours after he consumes alcohol.

    Find the value of t when the concentration of alcohol in John's blood is a maximum.

    Attempt:

    100=0.3te^{-0.912t}

    \frac{100}{0.3}=te^{-0.912t}

    \frac{100}{0.3}=lnt+lne^{-0.912t}

    \frac{100}{0.3}=lnt-0.912t

    stucked
    Are you familiar with basic optimization problems using differentiation?

    You want to find a value of t that satisfies these two equations:  \frac{dC}{dt} = 0 and  \frac{d^2C}{dC^2} < 0

    Having found that value of t substitute it into your original equation.
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  3. #3
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    Quote Originally Posted by Punch View Post
    The concentration of alcohol C% in John's blood when he consumes alcohol could be represented by C=0.3te^{-0.912t} where t is the number of hours after he consumes alcohol.

    Find the value of t when the concentration of alcohol in John's blood is a maximum.

    Attempt:

    100=0.3te^{-0.912t}

    \frac{100}{0.3}=te^{-0.912t}

    \color{red}{ln\frac{100}{0.3}} =lnt+lne^{-0.912t}
    But I think he would be dead long before 100%.
    Last edited by Anonymous1; May 2nd 2010 at 09:37 PM.
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  4. #4
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    Quote Originally Posted by Anonymous1 View Post
    But I think he would be dead long before 100%.
    Yes that would be correct, but 100 is not the max.
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  5. #5
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    Quote Originally Posted by Gusbob View Post
    Are you familiar with basic optimization problems using differentiation?

    You want to find a value of t that satisfies these two equations:  \frac{dC}{dt} = 0 and  \frac{d^2C}{dC^2} < 0

    Having found that value of t substitute it into your original equation.
    I have tried that but I got stucked.

    <br />
C=0.3te^{-0.912t}<br />

    \frac{dC}{dt} = 0.3t(e^{-0.912t})(-0.912)+e^{-0.912t}(0.3)

    = -0.912te^{-0.912t}+e^{-0.912t}

    then, 0.912te^{-0.912t} = e^{-0.912t}

    but this seems so wrong
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  6. #6
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    Quote Originally Posted by Punch View Post
    I have tried that but I got stucked.

    <br />
C=0.3te^{-0.912t}<br />

    \frac{dC}{dt} = 0.3t(e^{-0.912t})(-0.912)+e^{-0.912t}(0.3)

    = -0.912te^{-0.912t}+e^{-0.912t}

    then, 0.912te^{-0.912t} = e^{-0.912t}

    but this seems so wrong
    Your differentiation is correct. I got

     \frac{dC}{dt} = 0.3 ( e^{-0.921t} - 0.921t e^{-0.921t})

    You've done well so far, you just need to continue. Exponentials should cancel out -> divide both sides by  e^{-0.912t} . Just because the number isn't pretty doesn't mean it's wrong.
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  7. #7
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    I got it!! 0.912t=1

    t=1.1 ^_^ thanks
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