P(0)=Ce^(k*0)=C
P(1)=Ce^(k*1)=Ce^k
P(1)/P(0)=2=(ce^k)/C=e^k
Take natural log
ln(2)=ln(e^k)=k
k=ln(2)=0.6931 approx.
The size of a bacteria population is given by P=C*e^(kt) Where C is the initial size of the population, k is the growth rate constant and t is time in minutes.
a) If the bacteria in the population double their number every minute, what is the value of k?
I need help please!
1 answer
1 answer