Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
∴dtdy∝y⇒dtdy=ky(where k is a constant)
⇒ydy=kdtIntegrating both sides,we get:
∫ydy=k∫dt⇒logy=kt+C...(1)Let
y0 be the number of bacteria at t=0.
⇒logy0=CSubstituting the value of C in equation(1),we get:
logy=kt+logy0⇒logy−logy0=kt⇒log(y0y)=kt⇒kt=log(y0y)...(2)Also,it is given that the number of bacteria increased by
10% in 2hours.
⇒y=100y0110⇒y0y=1011...(3)Substituting this value in equation(2),we get:
k.2=log(1011)⇒k=21log(1011)Therefore,equation(2) becomes:
21log(1011).t=log(y0y)⇒t=log(1011)2log(y0y)...(4)Now,let the time when the number of bacteria increases from 100000 to 200000 be
t1.
⇒y=2y0 at t=t1From equation(4),we get:
t1=log(1011)2log(y0y)=log(1011)2log2Hence,in
log(1011)2log2 hours the number of bacteria increases from 100000 to 200000.