Tutor profile: Gaurav G.

John is conducting an experiment in which he is growing bacteria. The population of a bacteria undergoes exponential growth. If there are 100 bacteria at noon on 15th June 2016. 200 bacteria at 2pm on the same date. John needs 1600 bacteria. At what time should be stop the experiment?

Approach 1. The bacteria population doubled from 100 to 200 in a span of 2 hours (noon to 2 pm). Thus the bacteria population is doubling every 2 hours. Therefore, after another 2 hours (i.e. at 4pm) the population will reach 400. Again after 2 hours (i.e. 6pm) the population will reach 800. Then after 2 hours (i.e. at 8pm), it will again double to reach 1600. Thus John should stop the experiment at 8pm Approach 2. The general formula for exponential growth is given by $$ y=y_{0}(e^{kt})$$. where t is the time, $$y_{0}$$ is the initial value which is equal to 100. At noon, t=0 thus at 2pm, $$t=2$$ , $$y=200$$, $$y_{0}=100$$, putting these values into the equation, we get $$200=100(e^{2k})$$ which equals $$2=e^{2k}$$ Taking log on both sides, we get $$\ln 2 = 2k$$ or $$(\ln 2)/2 = k$$ Plug the value of k from above into the general equation $$y=y_{0}(e^{t(1/2)(\ln 2})$$ which equals $$y=y_{0}(e^{(\ln 2) t/2})$$ $$y=y_{0}(2^{t/2})$$ (because $$e^{\ln x}=x$$ )...........(1) Now we want y to be 1600, and using $$y_{0}=100$$ substituting these values into equation 1, we get $$1600=100(2^{t/2})$$ $$16=(2^{t/2})$$ ....(2) we know $$2^4=16$$ therefore substituting that into equation 2, we get $$2^4=(2^{t/2})$$ As terms of LHS and RHS are equal and base is equal, there power is also equal (power can only be positive as time can never be negative) $$4=t/2$$ $$8=t$$ There, John should stop the experiment at 8pm

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