Question 1099143
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The logistic function is<br>
{{{R(t) = L/(1+Me^(-kt))}}}<br>
The numerator "L" is the limiting value of the function, as t goes to infinity.  This is because as t gets very large, the expression {{{e^(-kt)}}} goes to 0, making the function value L/1 = L.<br>
So in your problem, L is 200, and the function is<br>
{{{R(t) = 200/(1+Me^(-kt))}}}<br>
The function value at t=0 is the starting value of the function.  The function evaluated at t=0 is {{{200/(1+M)}}}, because when t=0 {{{e^(-kt) = 1}}}.<br>
In your problem, where the initial number of people who have heard the rumor is 1, we have {{{200/(1+M) = 1}}} which means M is 199.<br>
So the function in this problem is<br>
{{{R(t) = 200/(1+199e^(-kt))}}}<br>
To find the value of k to finish the problem, we use the given information that 10 students have heard the rumor after 1 day.  That is, R(1) = 10.<br>
{{{R(1) = 10 = 200/(1+199e^(-k))}}}
{{{1+199e^(-k) = 200/10 = 20}}}
{{{199e^(-k) = 19}}}
{{{e^(-k) = 19/199}}}<br>
Now in the denominator of the logistic function, notice that {{{e^(-kt) = (e^(-k))^t)}}}<br>
Since we have found that the value of {{{e^(-k)}}} is {{{19/199}}}; we can now write the completed function for this problem as<br>
{{{R(t) = 200/(1+199(19/199)^t)}}}<br>
With the logistic function in this form, it is easy to see that the given three data points are satisfied:
(1) {{{R(0) = 200/(1+199(1)) = 200/200 = 1}}}
(2) {{{R(1) = 200/(1+199(19/199)) = 200/(1+19) = 200/20 = 10}}}
(3) For t "large", {{{R(t) = 200/(1+199*0) = 200/1 = 200}}}<br>
Now I note that the way you have asked the question, we are supposed to find the values of L, M, and k.  Finding the values of L and M was easy.<br>
But we didn't find the value of k in the work above.  However, that is easy:
{{{e^(-k) = 19/199}}}
{{{-k = ln(19/199)}}}
{{{k = -ln(19/199)}}}<br>
We could write the logistic function using {{{e^(-kt)}}}; but it would look ugly:
{{{R(t) = 200/(1+199(e^((-ln19/199)*t)))}}}