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Math Help - Need Help for Half Life Problems using Logs

  1. #1
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    Need Help for Half Life Problems using Logs

    Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.

    The equation that I got was:

    N = No (1/2) ^ t

    N is final amount, No is the initial ammount and t is time.

    I was given the data:

    Time (min) 10 20 30 40
    Amount (g) .83 .68 .56 .46

    and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.


    How do I do this? I have no clue...
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  2. #2
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    Do you have Excel?. You can do all sorts of regressions with it. Also, many calculators do regressions. Do you have one?. If so, what is it. Maybe I can step you through it.

    Anyway, using Excel I got an equation of

    y=1.0091e^{-.0196x}

    This has an R^2 of 1, so it is as good as it gets.

    Enter in x=10, do you get .83?.
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  3. #3
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    hi thanks for replying, I have a TI-83+.
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  4. #4
    Eater of Worlds
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    I would have to re-familiarize myself. I suppose the 83 does exponential regressions. I am not sure.
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  5. #5
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    it does, i did it before, I think.
    I just forgot how to do it.

    I think it's storing the data into 2 list and then compare them or something...

    but then, how do I set the equations up?

    I mean that N = No (1/2) ^ t thing , hate math final, lol.

    Maybe you can show me how you did it in excel?
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  6. #6
    Eater of Worlds
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    You can find the half life by using the equation k=\frac{-1}{T}ln(2)

    We know k=-.0196

    So, -.0196=\frac{-1}{T}ln(2)

    t=35.36

    We could also do it using the equation.

    Half of 1.0091 is .50455. Because 1.0091 is the amount at t=0

    So, .50455=1.0091e^{-.0196t}

    t=35.36

    Same as before.

    To find a regression on Excel, I make the graph, then add a trendline.

    Click on Insert, then Chart, then Scatter

    Highlight your data which is entered in columns A and B and proceed.

    It is pretty much self-explanatory. Then onece you have the graph, click on

    Chart, Add Trendline, Exponential, Click on Options and check the box that says to display the equation on the chart.

    There you have it.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Billwaa View Post
    Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.

    The equation that I got was:

    N = No (1/2) ^ t

    N is final amount, No is the initial ammount and t is time.

    I was given the data:

    Time (min) 10 20 30 40
    Amount (g) .83 .68 .56 .46

    and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.


    How do I do this? I have no clue...
    N = N_0 \left ( \frac{1}{2} \right ) ^{kt}

    ln(N) = ln(N_0) - [k~ln(2)]t

    Put t on the horizontal axis, ln(N) on the vertical axis, and do a linear regression. The slope will be k~ln(2) which you can then solve for k, and the intercept will be ln(N_0).

    -Dan
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  8. #8
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    thanks guys, this clear it up a little bit. ^_^
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  9. #9
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Billwaa View Post
    hi thanks for replying, I have a TI-83+.
    To do this on the 83+, input the data into two lists:

    L1: {10,20,30,40}
    L2: {.86,.68,.56,.46}

    Now, hit [STAT] and scroll over to [CALC] Select option [0]:ExpReg.

    This is an exponential regression. The solution will have the form of y=a\times b^x.

    At the home screen, you should now see:

    ExpReg

    Now hit [2nd][1][,][2nd][2]

    You should now have the following on the screen:

    \text{ExpReg } L_1\text{,}L_2

    Hit [ENTER]

    You'll get the regression equation:

    This is what the screen should say:

    ExpReg
    \begin{aligned}<br />
y&=a*b^x \\<br />
a&=1.00914514 \\<br />
b&=.9805442185<br />
\end{aligned}<br />

    Note that TheEmtpySet's answer was y=1.0091e^{-.0196x}

    We can rewrite the answer that the calculator gave us to get his solution.

    Note that b^x= e^{x\ln(b)}.

    Thus, .9805442185^x=e^{x\ln(.9805442185)}=e^{-.0196475365x}

    Rounding the regression equation, we get:

    \color{red}\boxed{y=1.0091e^{-.0196x}}

    Hope this makes sense!
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