Answer
(a) $k = 2.07944$
(b) $P(t) = (50)~e^{2.07944~t}$
(c) $P(6) = 13,107,078~cells$
(d) The rate of growth after 6 hours is $27,255,382~cells/h$
(e) $t = 4.76~h$
Work Step by Step
(a) We can find the relative growth rate $k$:
$P(t) = P(0)e^{kt}$
$P(1/3) = P(0)e^{k/3} = 2P(0)$
$e^{k/3} = 2$
$\frac{k}{3} = ln~2$
$k = 3~ln~2$
$k = 2.07944$
(b) $P(t) = P(0)e^{kt}$
$P(t) = (50)~e^{2.07944~t}$
(c) $P(t) = (50)~e^{2.07944~t}$
$P(6) = (50)~e^{(2.07944)(6)}$
$P(6) = 13,107,078~cells$
(d) We can find the rate of growth after 6 hours:
$\frac{dP}{dt} = k~P(6)$
$\frac{dP}{dt} = (2.07944)(13,107,078)$
$\frac{dP}{dt} = 27,255,382~cells/h$
(e) We can find the time $t$ when $P(t) = 1,000,000$:
$P(t) = (50)~e^{2.07944~t} = 1,000,000$
$e^{2.07944~t} = \frac{1,000,000}{50}$
$e^{2.07944~t} = 20,000$
$2.07944~t = ln(20,000)$
$t = \frac{ln(20,000)}{2.07944}$
$t = 4.76~h$
