A scientist notes the bacteria count in a petrie dish is 50. Two hours later, he notes the count has increased to 80. If this rate of growth continues, how much more time will it take for the bacteria count to reach 100?

Formula for exponential growth is #p(t) = p_0 * e^(kt)#

Where #p(t) =# Number at time #t , p_0 =# Number at start.

#k = #rate of growth (when #>0#) or decay (when# <0) ,t# = time

#80 = 50 * e^(k*2) or e^(2k)=80/50=1.6# Taking log on both

sides we get #k*2= ln(1.6) or k = ln(1.6)/2= 0.235002#

#:.p(t)=100, p(0)=50 , k= 0.235002#

#:.100 = 50 * e^(0.235002*t) or e^(0.235002*t) =2 # Taking log

on both sides we get #0.235002*t= ln 2 #

#:. t= ln2/0.235002~~2.95# hour.

It will take #2.95# hour to reach #100# bacteria [Ans]

Let's have a go at doing this without (explicitly) taking logarithms, etc.:

Note that:

#sqrt(10) = 3+1/(6+1/(6+1/(6+...))) ~~ 3+1/6 = 19/6#

Denote the time in hours by #t#, with #t=0# denoting the time of the first observation.

Let #p(t)# denote the population at time #t#.

We are given:

#{ (p(0) = 50), (p(2) = 80) :}#

What is #p(1)# ?

If the rate of growth is constant then #p(1)# must be the geometric mean of #50# and #80#:

#p(1) = sqrt(p(0) * p(2)) = sqrt(50*80) = sqrt(4000) = 20sqrt(10)#

So every hour, the population increases by the factor:

#(p(1))/(p(0)) = (20sqrt(10))/50 = (2sqrt(10))/5#

We find:

#p(3) = 80 * (2sqrt(10))/5 = 32sqrt(10) ~~ (32*19)/6 = 101 1/3#

So there will be #1# bacterium too many after a further hour.

Let's linearly interpolate the last hour:

We want about #20/21# of the time, since we want #20# bacteria out of #21#, so approximately #19/20 = 0.95# hours.