A cell of some bacteria divides into two cells every 10 minutes.The initial population is 3 bacteria. (a) Find the size of the population after t hours (function of t) (b) Find the size of the population after 7 hours. # Preview | Preview (c) When will the population reach 21? t42 Preview

Answer:(a) [tex]P_{t}=3(2)^{6t}[/tex](b) [tex]3(2)^{42}[/tex](c) 28.07 minutesStep-by-step explanation:A cell of some bacteria divides itself into 2 cells in every 10 minutes and initial population of the bacteria was 3.That means sequence formed will be 3, 6, 12, 24............We can easily say that this sequence is a geometric sequence having common ratio (r) = [tex]\frac{T_{2}}{T_{1}}=\frac{6}{3}[/tex]r = 2Now we know the explicit formula of a geometric sequence is given by [tex]P_{t}=P_{0}(r)^{\frac{60t}{10}}=P_{0}(r)^{6t}[/tex]Where a = Initial population = 3 bacteriar = common ratio = 2and t = time in hoursSo explicit formula will be [tex]P_{t}=3(2)^{6t}[/tex](a) Now we have to calculate the size of population after t hours [tex]P_{t}=3(2)^{6t}[/tex](b) We have to find the size of population after 7 hours or 420 minutes[tex]P_{t}=3(2)^{6\times7}[/tex]= [tex]3(2)^{42}[/tex]After 7 hours bacteria population will be [tex]3(2)^{42}[/tex](c) Time to reach population as 21By the explicit formula [tex]21=3(2)^{6t}[/tex][tex]2^{6t}=\frac{21}{3}=7[/tex]Now we take log on both the sides of the equation [tex]log(2^{6t})=log(7)[/tex]6t log2 = log 76t(0.301) = 0.845t(1.806) = 0.845t = [tex]\frac{0.845}{1.806}=0.468[/tex] hoursOr t = 0.468×60 = 28.07 minutesTherefore, after 28.07 minutes bacterial population will be 21