Word problem to equation
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What is the equation for radioactive decay?
Thats the point, to create the equation, from that word problem.
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In that case, you should begin by solving the first order IVP:
[MATH]\d{A}{t}=-kA[/MATH] where \(0<k\) and \(A(0)=A_0\)
This tells is that the rate of change of the amount \(A\) with respect to time \(t\) is proportional to \(A\). Once you solve this IVP, then you may use the given information regarding the half-life to determine \(k\).
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Or, if you are not in a differential equations/calculus course, you can use:
[MATH]A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{8}}[/MATH]
Let \(A(t)=\dfrac{1}{20}A_0\):
[MATH]\frac{1}{20}A_0=A_0\left(\frac{1}{2}\right)^{\frac{t}{8}}[/MATH]
[MATH]\frac{1}{20}=\left(\frac{1}{2}\right)^{\frac{t}{8}}[/MATH]
Now, solve for \(t\).
[MATH]A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{8}}[/MATH]
Let \(A(t)=\dfrac{1}{20}A_0\):
[MATH]\frac{1}{20}A_0=A_0\left(\frac{1}{2}\right)^{\frac{t}{8}}[/MATH]
[MATH]\frac{1}{20}=\left(\frac{1}{2}\right)^{\frac{t}{8}}[/MATH]
Now, solve for \(t\).
Steven G
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Why do you think that this should not be a half life problem? Why should this problem be more simplistic than other half life problem?
If you do not need an exact answer, then:
8days = 1/2 = 50%
16 days = 1/4 = 25%
24 days 1/8 = 12.5%
32 days = 1/16 = 6.25%
64 days = 1/32 = 3.125%
I would say that it looks like day 35= 5%
We haven't studied those at all yet. This is pretty much "create a equation without logarithm" stuff, I think.
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What Have you not studied "at all yet"?
There is no reference to logarithms in the referenced response!
Dr.Peterson
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This discussion is a good example of why we ask for the context of a question, and also what you have tried. We need to know what methods you have learned, in order to give you appropriate help.
What, in your experience, does radioactive decay mean, in terms of an equation? Can you give an example of what you have been taught to do?
What, in your experience, does radioactive decay mean, in terms of an equation? Can you give an example of what you have been taught to do?
We havent been taught anything about it.You can replace radioactive with milk,stone or anything else, since it is not the point.It is confusing me also, so I asked.
Dr.Peterson
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Clearly you have learned something relevant, since you know that half-lives and logarithms are connected with this, and since it was assigned to you (I assume). Forget "radioactive"; what does "decay", or any other of the words in the problem, mean to you? Are you learning about exponential functions, or anything related? Since you haven't benefited from any suggestion you've been given, we need some idea of what you can do.
If you really have no background at all, then I guess Jomo's post #6 is the best you can do.
If you really have no background at all, then I guess Jomo's post #6 is the best you can do.
Start by setting up a notation.
[MATH]f(t) = \text {percentage of radioactive stuff left after t days.}[/MATH]
Next write down what you know.
[MATH]f(8t) = 100 * \left ( \dfrac{1}{2} \right)^t[/MATH]
What is the problem?
[MATH]\text {Find } t \text { such that } f(t) = 5.[/MATH]
Now you can find a crude approximation by arithmetic.
f(8) = 50. f(16) = 25. f(24) = 12.5. f(32) = 6.25, and f(40) = 3.125. So about 35 days.
Or you can get a crude approximation by sketching a graph.
Or you can do a little algebra if you have been taught logarithms.
[MATH]f(8t) = 5 \implies 5 = 100 * 0.5^t \implies log(5) = 2 + t * log(0.5) \implies t = \dfrac{log(5) - 2}{log(0.5)} \implies[/MATH]
[MATH]t \approx 4.32 \implies 8t \approx 34.56.[/MATH]
[MATH]f(t) = \text {percentage of radioactive stuff left after t days.}[/MATH]
Next write down what you know.
[MATH]f(8t) = 100 * \left ( \dfrac{1}{2} \right)^t[/MATH]
What is the problem?
[MATH]\text {Find } t \text { such that } f(t) = 5.[/MATH]
Now you can find a crude approximation by arithmetic.
f(8) = 50. f(16) = 25. f(24) = 12.5. f(32) = 6.25, and f(40) = 3.125. So about 35 days.
Or you can get a crude approximation by sketching a graph.
Or you can do a little algebra if you have been taught logarithms.
[MATH]f(8t) = 5 \implies 5 = 100 * 0.5^t \implies log(5) = 2 + t * log(0.5) \implies t = \dfrac{log(5) - 2}{log(0.5)} \implies[/MATH]
[MATH]t \approx 4.32 \implies 8t \approx 34.56.[/MATH]
Steven G
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Wow my guess of 35 was fairly good (I was too lazy to do the work you did. Thanks Jeff!)
We have been studying exponential functions, logarithms and converting numbers to same base numbers/potencys, but the task was to solve it using equation, so without use of logarithm I guess..?
Anyway, in JeffM answer, where does that -2/+2 come from? Otherwise I think that is the correct answer, thought logarithm was used. So maybe MarkFL way of solving might also be right, thought A0 doesn't mean anything to me, unless it is related to sequence of numbers...
Thank you all already btw.
Anyway, in JeffM answer, where does that -2/+2 come from? Otherwise I think that is the correct answer, thought logarithm was used. So maybe MarkFL way of solving might also be right, thought A0 doesn't mean anything to me, unless it is related to sequence of numbers...
Thank you all already btw.
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\(A_0\) represents the initial amount, at time \(t=0\). Continuing where I left off, we could write:
[MATH]20=2^{\frac{t}{8}}[/MATH]
This implies:
[MATH]t=8\log_2(20)\approx34.575424759[/MATH]
Dr.Peterson
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"Using equation" does not mean "not using logarithms". Surely you have solved exponential and logarithmic equations ...
We do ask that people state the context and the entire wording of their problems, though, in part to reveal any such restrictions.
Perhaps the primary practical purpose of logarithms today outside of calculus is to solve exponential equations. So you probably were meant to set up an exponential equation and then to solve it using logarithms. In general,
[MATH]0 < a,\ 0 < b,\ b \ne 1, d > 0, \text { and } c = d * a^x \implies[/MATH]
[MATH]x = \dfrac{log_b(c) - log_b(d)}{log_b(a)}.[/MATH]
For many (except in calculus courses) [MATH]log(a) \ MEANS \ log_{10}(a).[/MATH]
That is the way that most calculators are set up.
Now to answer your question about where the 2 came from, let's do it this way, which is a slightly closer to Mark's way.
Start by setting up a notation.
[MATH]f(t) = \text {percentage of radioactive stuff left after t days.}[/MATH]
Next write down what you know.
[MATH]f(t) = 100 * \left ( \dfrac{1}{2} \right)^{(t/8)}.[/MATH]
That is the equation, and it is an exponential equation.
We want to know t when f(t) = 5.
[MATH]\therefore 5 = 100 * \left ( \dfrac{1}{2} \right )^{(t/8)} \implies[/MATH]
[MATH]log_{10}(5) = log_{10}\left \{100 * \left ( \dfrac{1}{2} \right )^{(t/8)} \right \} = log_{10}(100) + log_{10}\left ( 0.5^{(t/8)} \right ) =[/MATH]
[MATH]log_{10}(10^2) + \dfrac{t}{8} * log_{10}(0.5) =2 * log_{10}(10) + \dfrac{t}{8} * log_{10}(0.5) =[/MATH]
[MATH]2 * 1 + \dfrac{t}{8} * log_{10}(0.5) = 2 + \dfrac{t}{8} * log_{10}(0.5) \implies[/MATH]
[MATH]log_{10}(5) = 2 + \dfrac{t}{8} * log_{10}(0.5) \implies \dfrac{t}{8} * log_{10}(0.5) = log_{10}(5) - 2 \implies[/MATH]
[MATH]t * log{10}(0.5) = 8\{log_{10}(0.5) - 2\} \implies t = \dfrac{8\{log_{10}(0.5) - 2\}}{log_{10}(0.5)} \approx 34.575424759.[/MATH]