 Hi and welcome to the session. Let us discuss the following question, question says, The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009? Let us now start with the solution. First of all, let us assume that P be the population of the village. Now we are given that the population increases at the rate proportional to the number of inhabitants present at any time. So we can write Dp upon Dt is directly proportional to P. Here Dp upon Dt represents rate of increase of population and P shows number of inhabitants present at any time. Now we can write Dp upon Dt is equal to Kp where K is the constant of proportionality. We know we can replace this proportionality sign by equality sign by putting a constant here and this constant is constant of proportionality. Now let us name this equation as equation 1. Now separating the variables in equation 1, we get Dp upon P is equal to Kdt. Now integrating both the sides of this equation, we get integral of Dp upon P is equal to integral of Kdt. Now using this formula of integration, we can find this integral. So this integral is equal to log P to the base E and using this formula of integration, we can find this integral. This integral is equal to Kt and here we will write plus C where C is the constant of integration. Now we are given that the population of the village was 20,000 in 1999. Now if we consider T is equal to 0 in 1999 where T represents the time then P is equal to 20,000 where P represents the population. Now let us name this equation as equation 2. Now substituting T is equal to 0 and P is equal to 20,000. In equation 2, we get log 20,000 to the base E is equal to K multiplied by 0 plus C. Now this further implies log 20,000 to the base E is equal to C or we can simply write C is equal to log of 20,000 to the base E. Now we are also given that population of the village was 25,000 in the year 2004. Now if T is equal to 0 in 1999 then in 2004, T is equal to 5. Now substituting T is equal to 5 and P is equal to 25,000. In equation 2, we get log 25,000 to the base E is equal to 5K plus C. Now substituting this value of C in this equation we get log 25,000 to the base E is equal to 5K plus log of 20,000 to the base E. Now subtracting this term from both the sides of this equation we get log of 25,000 to the base E minus log of 20,000 to the base E is equal to 5K. Now applying this law of logarithms in reference side of this equation we get log of 25,000 upon 20,000 to the base E is equal to 5K. Now we will cancel common factor 5000 from numerator and denominator both and we get log of 5 upon 4 to the base E is equal to 5K. Now dividing both the sides of this equation by 5 we get 1 upon 5 log of 5 upon 4 to the base E is equal to K or we can simply write K is equal to 1 upon 5 multiplied by log 5 upon 4 to the base E. Now applying this law of logarithms in right hand side of this equation we get K is equal to log of 5 upon 4 raised to the power 1 upon 5 to the base E. Now we will substitute corresponding values of K and C in equation 2. Now equation 2 becomes log P to the base E is equal to T multiplied by log of 5 upon 4 raised to the power 1 upon 5 to the base E plus log 20,000 to the base E. Now applying this law of logarithms in this term this term becomes log 5 upon 4 raised to the power T upon 5 to the base E and we can write this expression as log of P to the base E is equal to log of 5 upon 4 raised to the power T upon 5 to the base E plus log 20,000 to the base E. Now we have to find the population of the village in the year 2009. Let us now name this equation as equation 3. Now we will substitute T is equal to 10 in this equation. We know if we are taking T is equal to 0 for the year 1999 then T is equal to 10 for the year 2009. So we will substitute T is equal to 10 in this equation and find the value of P. So we can write substituting T is equal to 10 in equation 3 we get log P to the base E is equal to log of 5 upon 4 raised to the power 10 upon 5 to the base E plus log of 20,000 to the base E. Now we know 10 upon 5 is equal to 2. So here we will get square of 5 upon 4. Now applying this law of logarithms in right hand side of this equation we get log of 20,000 multiplied by 25 upon 16 is equal to log P to the base E. We know square of 5 upon 4 is 25 upon 16. Now we will cancel common factor 16 from numerator and denominator both and we get log P to the base E is equal to log of 31250 to the base E. Multiplying these two terms we get 31250. Now this further implies P is equal to 31250. So we get population in the year 2009 will be 31250. This is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.