Answer
a) $k \approx 0.008357$ b) $P \approx 333361000$
Work Step by Step
a) Consider the exponential growth equation as follows: $P=P_0e^{kt}$
As we are given that $P=314419199$ and $P_0=314419198$; $t=12 sec= \dfrac{12}{(31536000)} $ year
Now, we have $314419199=314419198 e^{(\frac{12}{31536000})k}$
or, $k=[\dfrac{31536000}{12}]\ln [\dfrac{314419199}{314419198}] \approx 0.008357$
b) The equation of model will be formed as follows:
$P=(314419198) e^{(0.008357)t} $
or, $P=(314419198) e^{(0.008357)(7)} \approx 333361000$
