**Ex 9.4, 22**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.4, 22 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? Let the number of bacteria at time t be y Given that, 𝑑𝑦𝑑𝑡 ∝ y 𝑑𝑦𝑑𝑡 = ky 𝑑𝑦𝑦 = kdt Integrating both sides 𝑑𝑦𝑦=𝑘 𝑑𝑡 log y = kt + C According to Questions Now, Putting t = 0 and y = 1,00,000 in (1) log 1,00,000 = k × 0 + C ⇒ C = log 1,00,000 Putting value of C in (1) log y = kt + log 1,00,000 Now, Putting t = 2 and y = 1,00,000 in (2) log 1,10,000 = 2k + log 1,00,000 log 1,10,000 − log 1,00,000 = 2k log 1,10,0001,00,000 = 2k 12 log 1110 = k Putting value of k in (2) log y = 12 log 1110 t + log 1,00,000 Now, If Bacterial = 2,00,000, we have to find t Putting y = 2,00,000 in (3) log 2,00,000 = 12 log 1110 t + log (1,00,000) log 2,00,000 − log 1,00,000 = 12 log 1110 t log 2,00,0001,00,000 = 12 log 1110 t log 2 = 12 log 1110 t t = 𝟐 𝐥𝐨𝐠𝟐 𝐥𝐨𝐠 𝟏𝟏𝟏𝟎

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.