Answer
(a) The relative growth rate is $104\%$
(b) $P(0) = 50$
(c) $P(t) = 50~e^{1.03972~t}$
(d) 5382 cells
(e) $5596~cells/h$
(f) 6.6 hours
Work Step by Step
(a) We can find the value of $k$:
$P(6) = P(0)e^{6k} = 25,600$
$P(2) = P(0)e^{2k} = 400$
$\frac{P(6)}{P(2)} = e^{6k-2k} = \frac{25,600}{400}$
$e^{4k} = 64$
$4k = ln(64)$
$k = \frac{ln(64)}{4}$
$k = 1.03972$
The relative growth rate is $104\%$
(b) We can find the initial size:
$P(t) = P(0)~e^{kt}$
$P(2) = P(0)~e^{2k} = 400$
$P(0) = \frac{400}{e^{(2)( 1.03972)}}$
$P(0) = 50$
(c) $P(t) = P(0)~e^{kt}$
$P(t) = 50~e^{1.03972~t}$
(d) We can find the number of bacteria cells after 4.5 hours:
$P(t) = 50~e^{1.03972~t}$
$P(4.5) = 50~e^{(1.03972)(4.5)}$
$P(4.5) = 5382$
(e) We can find the rate of growth after 4.5 hours:
$\frac{dP}{dt} = k~P(4.5)$
$\frac{dP}{dt} = (1.03972)(5382)$
$\frac{dP}{dt} = 5596~cells/h$
(f) We can find the time $t$ when $P(t) = 50,000$:
$P(t) = (50)~e^{1.03972~t} = 50,000$
$e^{1.03972~t} = 1000$
$1.03972~t = ln(1000)$
$t = \frac{ln(1000)}{1.03972}$
$t = 6.6~h$
