 Hi and welcome to the session. Let us discuss the following question. Question says, in a culture the bacteria count is 1 lakh. The number is increased by 10% in 2 hours. In how many hours will the count reach 2 lakhs if the rate of growth of bacteria is proportional to the number present? Let us now start with the solution. Now we are given that initial bacteria count is equal to 1 lakh. So here we can write initial bacteria count is equal to 1 lakh. We are also given that rate of growth of bacteria in every 2 hours is equal to 10% and final bacteria count is 2 lakhs. Now we have to find in how many hours bacteria count reach 2 lakhs and we are also given that rate of growth of bacteria is proportional to number present. Let us assume that number of bacteria present is equal to V. Now according to the question dB upon dt is equal to K multiplied by 10 upon 100 B. Here dB upon dt represents the rate of growth of bacteria. We know growth of bacteria in 2 hours is equal to 10% of B and we can say it is 10 upon 100 multiplied by B and we are given that dB upon dt is directly proportional to 10 upon 100 multiplied by B. Now we can write dB upon dt is equal to K multiplied by 10 upon 100 multiplied by B where K is the constant of proportionality. Integrating the variables in this equation we get dB upon B is equal to 0.1 K dt. Now integrating both the sides of this equation we get integral of dB upon B is equal to 0.1 K multiplied by integral of dt. Now using this formula of integration we get this integral is equal to log B and here this integral is equal to t. We have used this formula to find this integral. So here we can write 0.1 K t plus log C where log C represents constant of integration. Now this further implies log B minus log C is equal to 0.1 K t. Now this further implies log B upon C is equal to 0.1 K t. Now applying this law of logarithms on both the sides of this equation we get B upon C is equal to e raised to the power 0.1 K t. Now multiplying both the sides of this equation by C we get B is equal to C multiplied by e raised to the power 0.1 K t. Let us name this equation as equation 1. Now we know initial bacteria count is equal to 1 lakh. Now after 2 hours bacteria count is equal to 1 lakh plus 10 percent of 1 lakh. Now we know rate of increase of bacteria count is 10 percent in 2 hours. Now initial count of the bacteria is 1 lakh and in next 2 hours they will increase by 10 percent of 1 lakh. So this is further equal to 1 lakh plus 10 upon 100 multiplied by 1 lakh. Now simplifying we get bacteria count is equal to 1 lakh 10,000. Now we know bacteria count represents B or we can say bacteria present is equal to 1 lakh 10,000 when time is equal to 2 hours. Now we know time is represented by t. So we can write when t is equal to 2 B is equal to 1 lakh 10,000. Now substituting B is equal to 1 lakh 10,000 and t is equal to 2. In this equation we get 1 lakh 10,000 is equal to C multiplied by e raised to the power 0.2 K. Let us name this equation as 2. We also know that initial count of bacteria is equal to 1 lakh. So we can write when t is equal to 0 B is equal to 1 lakh. Now substituting t is equal to 0 and B is equal to 1 lakh. In equation 1 we get 1 lakh is equal to C multiplied by e raised to the power 0. Now this implies 1 lakh is equal to C multiplied by 1. We know e raised to the power 0 is equal to 1. Now this further implies C is equal to 1 lakh. Now we will substitute this value of C in equation 2. Now equation 2 becomes 1 lakh 10,000 is equal to 1 lakh multiplied by e raised to the power 0.2 K. Now dividing both the sides of this equation by 1 lakh we get 11 upon 10 is equal to e raised to the power 0.2 K. Now taking log on both the sides we get log 11 upon 10 is equal to 0.2 K. Now this implies log of 11 upon 10 is equal to 1 upon 5 multiplied by K. We know 0.2 is equal to 1 upon 5. Now this further implies 5 log 11 upon 10 is equal to K. Multiplying both the sides of this equation by 5 we get this equation. Now we can simply write K is equal to 5 multiplied by log of 11 upon 10. Now we will substitute values of C and K in equation 1. Now equation 1 becomes V is equal to 1 lakh multiplied by e raised to the power 0.5 log of 11 upon 10 multiplied by T. Now we have to find the R's in which bacteria count reaches 2 lakhs. So we can write when B is equal to 2 lakhs this equation becomes 2 lakhs is equal to 1 lakh multiplied by e raised to the power 1 upon 10. 1 upon 2 log of 11 upon 10 multiplied by T. We know 0.5 is equal to 1 upon 2 so here we can write 1 upon 2. Now dividing both the sides by 1 lakh we get 2 is equal to e raised to the power 1 upon 2 log of 11 upon 10 multiplied by T. Now taking log on both the sides we get log 2 is equal to 1 upon 2 log of 11 upon 10 multiplied by T. Now multiplying both the sides of this equation by 2 we get 2 log 2 is equal to log of 11 upon 10 multiplied by T. Dividing both the sides by log of 11 upon 10 we get 2 log 2 upon log of 11 upon 10 is equal to T or we can write T is equal to 2 log 2 upon log of 11 upon 10. We know T represents the time in R's so here we can write R's this is the required value of T. So we get number of R's in which bacteria count reaches 2 lakhs is equal to 2 log 2 upon log of 11 upon 10. This is our required answer this completes the session hope you understood the solution take care and have a nice day.