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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Filo · Accepted Answer

Let the number of bacteria at time t be yGiven that rate growth of bacteria is proportional to the number presentdtdy​∝ydtdy​=kyydy​=kdtIntegrating both sides,∫ydy​=k∫dt⇒  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+CC=log1,00,000Putting 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+log1,00,000log1,10,000−log1,00,000=2klog(1,00,0001,10,000​)=2k21​log(1011​)=kPutting value of k in ( 2 )logy=kt+log1,00,000⇒  logy=21​log(1011​)t+log1,00,000          ---- ( 3 )Now, if bacterial =2,00,000 in ( 3 )log2,00,000=21​log(1011​)t+log(1,00,000)log2,00,000−log1,00,000=21​log(1011​)tlog(1,00,0002,00,000​)=21​log(1011​)tlog2=21​log(1011​)t⇒  t=log(1011​)2log2​

Question Text	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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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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