im using the equation p(t) = L/1+Ce^-kt

the density was measured at days 0 and 10 and was found to be 2 and 194 respectively.when k = 0.74 i need to find L and C. can someone please show me how?

\(\displaystyle p(t) = \frac{L}{1 + Ce^{-kt}}\).

You are told that \(\displaystyle k = 0.74\) and you are given two \(\displaystyle (t, p)\) points.

So \(\displaystyle 2 = \frac{L}{1 + Ce^{-0.74(0)}}\) and \(\displaystyle 194 = \frac{L}{1 + Ce^{-0.74(10)}}\).

That means you have two equations in two unknowns:

\(\displaystyle 2 = \frac{L}{1 + C}\)

\(\displaystyle 194 = \frac{L}{1 + Ce^{-7.4}}\).

Dividing the second equation by the first yields:

\(\displaystyle \frac{194}{2} = \frac{\frac{L}{1 + Ce^{-7.4}}}{\frac{L}{1 + C}}\)

\(\displaystyle 97 = \frac{1 + C}{1 + Ce^{-7.4}}\)

\(\displaystyle 97(1 + Ce^{-7.4}) = 1 + C\)

\(\displaystyle 97 + 97Ce^{-7.4} = 1 + C\)

\(\displaystyle 97Ce^{-7.4} - C = 1 - 97\)

\(\displaystyle C(97e^{-7.4} - 1) = -96\)

\(\displaystyle C = -\frac{96}{97e^{-7.4} - 1}\)

\(\displaystyle C = \frac{96}{1 - 97e^{-7.4}}\).

Substituting back into the first equation:

\(\displaystyle 2 = \frac{L}{1 + C}\)

\(\displaystyle 2 = \frac{L}{1 + \frac{96}{1 - 97e^{-7.4}}}\)

\(\displaystyle 2 = \frac{L}{\frac{97 - 97e^{-7.4}}{1 - 97e^{-7.4}}}\)

\(\displaystyle 2 = \frac{L}{\frac{97(1 - e^{-7.4})}{1 - 97e^{-7.4}}}\)

\(\displaystyle L = \frac{194(1 - e^{-7.4})}{1 - 97e^{-7.4}}\).