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Question

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

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Solution

Let the number of bacteria at time t be y
Given that rate growth of bacteria is proportional to the number present
dydty
dydt=ky
dyy=kdt
Integrating both sides,
dyy=kdt
logy=kt+C ---- ( 1 )

Now according to question,
The bacteria court is 1,00,000. The number is increased by 10% in 2 hours, in how many hours will the count reach 2,00,000.
Putting t=0 and y=1,00,000 in ( 1 )
log1,00,000=k×0+C
C=log1,00,000
Putting value of C in ( 1 )
logy=kt+C
logy=kt+log1,00,000 ---- ( 2 )

Now,
Putting t=2 and y=1,10,000 in ( 2 )
log1,10,000=2k+log1,00,000
log1,10,000log1,00,000=2k
log(1,10,0001,00,000)=2k
12log(1110)=k
Putting value of k in ( 2 )
logy=kt+log1,00,000
logy=12log(1110)t+log1,00,000 ---- ( 3 )

Now, if bacterial =2,00,000 in ( 3 )
log2,00,000=12log(1110)t+log(1,00,000)
log2,00,000log1,00,000=12log(1110)t

log(2,00,0001,00,000)=12log(1110)t

log2=12log(1110)t

t=2log2log(1110)


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