Answer:
Step-by-step explanation:
Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...
[tex]P(t)=P_0\cdot e^{kt}[/tex]
We know from the given points that we can write the equation as ...
[tex]P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t[/tex]
Comparing this to the desired form, we see that ...
[tex]P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693[/tex]
So, the approximate equation for P is ...
[tex]P(t)=131.836\cdote^{0.032t}[/tex]
And the parameters of interest are ...
