The concentration of a drug in the blood stream after t hours of being injected is given by C(0.5) =0.57; C(1.5) = 1.12 ; C(3) = 1.13; C(4) =1.22; C(6) = 0.92. Knowing that the concentration is modeled by $C(t) = Kte^{-at}$ determine C(5.5).
I tried to linearize $C(t) = Kte^{-at} = \frac{C}{t} = ke^{-at} = \ln \frac{C}{t} = \ln k -at$. And I set up the table:
$$ \begin{array}{c|lcr} n & \text{t} & \text{ln(c/t)} & \text{tln(c/t)} & \text{t}^2 \\ \hline 1 & 0.5 & 0.13103 & 0.06551 & 0.25 \\ 2 & 1.5 & -0.29214 & -0.43820 &2.25\\ 3 & 3 & -0.97639 & -2.92918&9\\ 4 & 4 & -1.18744 & -4.74977 &16\\ 5 & 6 & -1.875514 & -19.30249 &36\\ \sum&15 & -4.20009 & -19.30249 & 63.5 \end{array} $$
Using formulas a and b.
a= $\frac{n\sum t\ln(c/t) - \sum t \sum \ln(c/t)}{n \sum t^2 - (\sum t)^2} = -0.36228$
b=$\frac{\sum t\ln(c/t)- \sum \ln(c/t)\sum t^2}{(\sum t)^2-n \sum t^2 } = 0.24683$
so
$k = 2.71828^b = 1.27996$
$C(5.5) = K5.5e^{-a5,5} = 51.63157$
I also made variations in place of ln(c/t) I used ln c and even used only c. But none of the variations are matching the alternatives, so what am I doing wrong?
Thanks for any help.
Edited: The original answer did not keep the proposed functional form of the concentration.
As you mentioned, since $\log\frac{C(t)}{t}$ should be a polynomial of degree 1, we can use the least squares technique to fit the data $(t_i, \log \frac{C(t_i)}{t_i})$, yielding (as you also mention) $$ \log \frac{C(t)}{t} \approx 0.246831 - 0.362283 t $$
or
$$ C(t) \approx e^{0.246831} \, t\, e^{-0.362283 \, t} $$
This leads to $C(5.5)\approx 0.959852$. You just got a sign wrong in the exponent... If $a$ comes from the least squares fit, the exponential should be $e^{at}$. You can see the graphic below.