Answer
(a) $P(t) = 100~e^{1.435~t}$
(b) 7407
(c) $10,629~cells/h$
(d) 3.2 hours
Work Step by Step
(a) We can find the value of $k$:
$P(t) = P(0)e^{kt}$
$P(1) = 100~e^{k} = 420$
$e^k = 4.2$
$k = ln~4.2$
$k = 1.435$
Then:
$P(t) = 100~e^{1.435~t}$
(b) We can find the number of bacteria after 3 hours:
$P(t) = 100~e^{1.435~t}$
$P(3) = 100~e^{(1.435)(3)}$
$P(3) = 7407$
(c) We can find the rate of growth after 3 hours:
$\frac{dP}{dt} = k~P(3)$
$\frac{dP}{dt} = (1.435)(7407)$
$\frac{dP}{dt} = 10,629~cells/h$
(d) We can find the time $t$ when $P(t) = 10,000$:
$P(t) = (100)~e^{1.435~t} = 10,000$
$e^{1.435~t} = 100$
$1.435~t = ln(100)$
$t = \frac{ln(100)}{1.435}$
$t = 3.2~h$
