 Hello and welcome to the session. Let us discuss the following question. Question says, in a bank principal increases continuously at the rate of r percent per year. Find the value of r if rupees 100 double itself in 10 years. Let us now start with the solution. First of all let us assume that principal is equal to rupees p. So here we can write let the principal be rupees p. And clearly we can see we have given rate r percent per year. So here we can write and the given rate percent is r percent per year. Now rate of change of p that is principal is given by dp upon dt and it is equal to r upon 100 multiplied by p. Below principal increases continuously at the rate of r percent per year. So rate of change of principal is equal to r percent of p that is r upon 100 multiplied by p. Now let us name this equation as equation 1. Now separating the variables in equation 1 we get dp upon p is equal to r upon 100 dt. Now integrating both the sides of this equation we get integral of dp upon p is equal to integral of r upon 100 dt. Now this further implies integral of dp upon p is equal to r upon 100 multiplied by integral of dt. Now using these two formulas of integration we get this integral is equal to log p where we will write is equal to sign. We will write this r upon 100 as it is and integral of dt is t plus c1 where c1 is the constant of integration. Now applying this law of logarithms on both the sides of this equation we get p is equal to e raised to the power rt upon 100 plus c1. Now this further implies p is equal to e raised to the power rt upon 100 multiplied by e raised to the power c1. Now substituting c for e raised to the power c1 we get p is equal to e raised to the power rt upon 100 multiplied by c. Now we are given in the question that rupees 100 double itself in 10 years. Now from this condition we get p is equal to rupees 100 when t is equal to 0 where t represents that time in years. Now let us name this equation as equation 2. Now substituting these values of p and t in this equation we get 100 is equal to e raised to the power 0 multiplied by c. We know e raised to the power 0 is equal to 1 only so we get 100 is equal to 1 multiplied by c. Now this further implies c is equal to 100. Now we will substitute this value of c in equation 2. Now substituting c is equal to 100 in equation 2 we get p is equal to 100 multiplied by e raised to the power rt upon 100. We also know that p is equal to rupees 200 after 10 years. Now let us name this equation as equation 3. Now substituting p is equal to rupees 200 and p is equal to 10 years. In equation 3 we get 200 is equal to 100 multiplied by e raised to the power r multiplied by 10 upon 100. Now this further implies 200 is equal to 100 multiplied by e raised to the power r upon 10. Now dividing both the sides of this equation by 100 we get 2 is equal to e raised to the power r upon 10. Now taking log on both the sides of this equation we get log 2 to the base e is equal to r upon 10. Now here we are given value of log 2 to the base e it is 0.6931. Now we will substitute 0.6931 for log 2 to the base e here and we get 0.6931 is equal to r upon 10. Now multiplying both the sides of this equation by 10 we get 6.931 is equal to r or we can simply write r is equal to 6.93 approximately. So this is the required value of r or we can say the required rate is equal to 6.93 percent. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.