 Hello and welcome to the session. In this session we will learn about loss of logarithms. Now let A be a positive number which is not equal to 1 that is A is greater than 0 and A is not equal to 1 and let x and y be any positive numbers that is x is greater than 0 and y is greater than 0 and n be any real number that is n belong to the set of real numbers. Then the first law of logarithm is log x1 to the base a is equal to log x to the base a log y to the base a. Now let us start with its proof. Now let log x to the base a is equal to m. Now let us give it equation number 1 log y to the base a is equal to m. Let this be equation number 2. Now we learn the definition of logarithm that is if three numbers are so related that a is equal to x is equal to m then in the log book form we can write log m to the base a is equal to x and here a x and n are all positive numbers. So from the first equation by definition it is equal to a raise to power n and from the second equation we have y is equal to a raise to power m. The supplies x into y is equal to a raise to power m into a raise to power m which is equal to now applying the product log of x1n this will be a raise to power m plus n. The supplies xy is equal to a raise to power m plus n. Now again by definition of logarithm we have log xy is equal to m plus n. Now m is equal to log x to the base a. So this implies log xy to the base a is equal to log x to the base a plus n which is log y to the base a. So we have proved the first law of logarithm assumption of this law that is log xyz and so on. Only the base a is equal to log x to the base a plus log y to the base a plus log x to the base a. So we have proved the first law of logarithm that is log of a product to any base is equal to sums of logs of the factors to the same base. Logarithm that is log over y is equal to log x to the base a minus log y to the base a. Now let us start with its truth m is equal to log y to the base a. By definition we have x is equal to a raise to power n and y is equal to a raise in case of the experience this will be equal to a raise to power m minus x over y is equal to a raise to power m minus n. Now again by the definition of logarithm minus n that is log y to the base a logarithm which is log of a fraction to the base a is equal to log of the numerator of that fraction to the base a minus log of the denominator of the fraction that is of this fraction to the base a. log x to the base a that is log x to into x 3 and so on log x to the base a plus log x to the base a plus log the whole minus log y to the base a and so on. The third law of logarithm and that is log x to the base a is equal to m let this be equation number so by definition of logarithm we have this is equal to a raise to power m. Now raising both sides to the power n this is equal to a raise to power m by a raise to power n. This fourth law of logarithm is equal to a raise to power now again by definition of logarithm we can write this equation as log is equal to m into n. And therefore this implies log x raise to power n to the base a is equal to that is m into log x to the base a here we have proved the third law of logarithm. And that is log is equal to 1 over n. Now let us start with its root log raise a raise to power n is equal to p so by definition this implies p is equal to x by using this definition of logarithm. Further this implies a raise to power n into p is equal to x which further implies. Now here again logarithm this will be is equal to n into p is equal to 1 over n into log x to the base a. We have taken p is equal to log x to the base a log x to the base a raise to power n is equal to 1 by n we have learnt about the laws of logarithms. And this completes our session hope you and have enjoyed the session.