Question 757191
The mass of the colony increases by a factor of 1.5 each day,
so if {{{t}}}= time elapsed, in days,
and {{{m}}}= mass of bacteria, in grams,
we could write {{{m(t)=20*1.5^t}}}
However, it is traditional (and more convenient when calculating derivatives) to use
{{{highlight(m(t)=20*e^(t*ln(1.5)))}}} <-- answer to part a) of the problem
 
b) The rate of change of the mass of the colony would be the derivative of {{{m(t)}}}
{{{dm/dt=20*ln(1.5)*e^(t*ln(1.5))}}}
Fot {{{t=0}}} that rate is {{{20*ln(1.5)}}}= approximately 8.1 g/day
 
c) During which day is the mass of the colony increasing by 60 g/day?
we need to find the value of {{{t}}} that makes the rate, {{{dm/dt=20*ln(1.5)*e^(t*ln(1.5))}}}, equal to 60 g/day
{{{20*ln(1.5)*e^(t*ln(1.5))=60}}} --> {{{e^(t*ln(1.5))=60/20/ln(1.5)}}} --> {{{t*ln(1.5)=ln(3/ln(1.5))}}} --> {{{t=ln(3/ln(1.5))/ln(1.5)}}} --> t = approx. {{{4.94}}} days.
I would call that time (and any time {{{4<=t<5}}} between 4 and 5) day {{{highlight(5)}}}.
 
d) When will the mass of the colony be 60 Kg?
60Kg is a lot. It's 60,000 grams.
{{{m(t)=20*e^(t*ln(1.5))=60000}}} --> {{{e^(t*ln(1.5))=60000/20}}} --> {{{t*ln(1.5)=ln(3000)}}} --> {{{t=ln(3000)/ln(1.5)}}} --> {{{t=about19.75}}}, so we approximate to {{{t=highlight(20days)}}}.
So, in the ideal world of math, it would take 20 days to get 60 kg of bacteria.
 
In real life, a colony would not keep growing at the same rate for that long. Even with a lot of attention, "feeding", and "transplanting", grow would not stay constant.