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Math Help - Quick Algebra question (logs and stuff)

  1. #1
    Junior Member reynardin's Avatar
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    Quick Algebra question (logs and stuff)

    I'm having trouble with this word problem:

    Assume $150 is deposited in a savings account. If the interest rate is 4.5% compounded continuously, after how many years will the investment be worth $450? (Round your answer to the nearest month.)

    I start off with
    y=ae^{rt}
    450=150e^{4.5t}
    3=e^{4.5t}
    \ln(3)=4.5t\ln(e)

    after this, my teacher says i did something wrong, but i can't figure out where i did it wrong.
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  2. #2
    Eater of Worlds
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    You used 4.5 instead of .045
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    I'm having trouble with this word problem:

    Assume $150 is deposited in a savings account. If the interest rate is 4.5% compounded continuously, after how many years will the investment be worth $450? (Round your answer to the nearest month.)

    I start off with
    y=ae^{rt}
    450=150e^{4.5t}
    3=e^{4.5t}
    \ln(3)=4.5t\ln(e)

    after this, my teacher says i did something wrong, but i can't figure out where i did it wrong.
    4.5\text{ }\text{percent}=\frac{4.5}{100}=.045
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  4. #4
    Junior Member reynardin's Avatar
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    aaaaaah!
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  5. #5
    Junior Member reynardin's Avatar
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    oh yay! mathstud28 again! =D
    I still got 2 more problems from yesterday that I don't get lol.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by reynardin View Post
    oh yay! mathstud28 again! =D
    I still got 2 more problems from yesterday that I don't get lol.
    Ok go!
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  7. #7
    Junior Member reynardin's Avatar
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    EDIT see next post
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  8. #8
    Junior Member reynardin's Avatar
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    ok;
    In 5 years, radioactivity reduces the msas of a 100-gram sample of an element to 60 grams. for this element, find the number of years it would take for the original sample to be reduces to half its original mass.

    im using the equation y=ae^{-kt} to start.
    60=100e^{-.5t}
    .6 = e^{-.5t}

    good so far? or not?
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  9. #9
    Eater of Worlds
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    Using the initial condition.

    y(0)=100, y(5)=60

    60=100e^{5k}

    Solving for k=\frac{-ln(\frac{5}{3})}{5}\approx{-.102165124753}

    So, now solve the following for t:

    50=100e^{\frac{-ln(5/3)}{5}t}
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  10. #10
    Junior Member reynardin's Avatar
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    eh im confused
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Using the initial condition.

    y(0)=100, y(5)=60

    60=100e^{5k}

    Solving for k=\frac{-ln(\frac{5}{3})}{5}\approx{-.102165124753}

    So, now solve the following for t:

    50=100e^{\frac{-ln(5/3)}{5}t}
    We know that when t=0 that y=100

    setting this up we get

    100=Ce^{k\cdot{0}}=Ce^0=C

    So C=100

    Next we have that t=5 y=60

    Setting this up we get

    60=100e^{k\cdot{50}}\Rightarrow{\frac{3}{5}=e^{5k}  }\Rightarrow{\ln\bigg(\frac{3}{5}\bigg)=5k}\Righta  rrow{k=\frac{\ln\bigg(\frac{3}{5}\bigg)}{5}}
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    We know that when t=0 that y=100

    setting this up we get

    100=Ce^{k\cdot{0}}=Ce^0=C

    So C=100

    Next we have that t=5 y=60

    Setting this up we get

    60=100e^{k\cdot{50}}\Rightarrow{\frac{3}{5}=e^{5k}  }\Rightarrow{\ln\bigg(\frac{3}{5}\bigg)=5k}\Righta  rrow{k=\frac{\ln\bigg(\frac{3}{5}\bigg)}{5}}
    O

    Just so our conflicting answers dont confuse you

    \frac{\ln(\frac{3}{5})}{5}=\frac{\ln\bigg(\bigg(\f  rac{5}{3}\bigg)^{-1}\bigg)}{5}=\frac{-1\ln(\frac{5}{3})}{5}
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  13. #13
    Junior Member reynardin's Avatar
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    oh, so both come out to the same thing?
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  14. #14
    Junior Member reynardin's Avatar
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    gah, back to the question:
    how does that answer it?

    "In 5 years, radioactivity reduces the msas of a 100-gram sample of an element to 60 grams. for this element, find the number of years it would take for the original sample to be reduces to half its original mass."

    the number that I get doesn't seem to fit, when i calculate the answer from what you guys showed me. (-.1021651248...)
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  15. #15
    Eater of Worlds
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    If you solve it correctly, you should get 6.78 years
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