 Hello and welcome to the session. In this session, we discussed the following question which says if a to the power 4 plus b to the power 4 is equal to 14 a square b square show that log of a square plus b square is equal to log a plus log b plus 2 log 2 and a and b are two positive numbers. Before we move on to the solution, let's discuss some laws of logarithms to be used in this question. First we have log of x1 x2 to the base a is equal to log of x1 to the base a plus log of x2 to the base a. Now the second law is log of x1 to the power n to the base a is equal to n into log x1 to the base a. Where we have this a is a positive number and it is not equal to 1 then x1 is a positive number and x2 is also a positive number and this n is a real number that is n belongs to r. This is the key idea that we use for this question. Let's proceed with the solution now. In the question we are given that a to the power 4 plus b to the power 4 is equal to 14 a square b square that is we have a to the power 4 plus b to the power 4 is equal to 14 a square b square and we are supposed to show that log of a square plus b square is equal to log a plus log b plus 2 log 2 we have but a and b are two positive numbers. Now a to the power 4 plus b to the power 4 equal to 14 a square b square can be written as a square b whole square plus b square b whole square is equal to 14 a square b square. Now square b square to both sides we get a square b whole square plus b square b whole square plus 2 a square b square is equal to 14 a square b square plus 2 a square b square. This requires an identity which is a plus b b whole square is equal to a square plus 2 a b plus b square. So using this identity we can rewrite the left hand side so we get so we get this as a square plus b square the whole square is equal to 14 a square b square plus 2 a square b square is 16 a square b square. Now further taking log on both sides we get log of a square plus b square the whole square is equal to log of 16 a square b square. Now for the left hand side we would use the second law of logarithm which says log of x1 to the power n to the base a is equal to n into log x1 to the base a. So we get this as 2 into log of a square plus b square is equal to 2 plus since right hand side is the log of the product so we would use the first theorem in which we have the log of the product to any base is equal to some of the logs of the factors to the same base. So using this that is this law we would get the right hand side as log of 16 plus log a square plus log b square. So further we have 2 into log of a square plus b square is equal to log 16 and here 16 can be written as 2 to the power of 4 log of 2 to the power 4 plus log a square plus log b square. Now for these three terms on the right hand side we would use the second law in which we have log of x1 to the power n to the base a is equal to n into log x1 to the base a. So we get further 2 into log of a square plus b square is equal to 4 into log 2 plus 2 into log a 2 into log b. Now further divided we get log of a square plus b square is equal to 2 log 2 plus log a plus log b and are given a and b are two positive numbers. And this is what we were supposed to prove that is log of a square plus b square is equal to log a plus log b plus 2 log 2 and a and b are two positive numbers. And that's proved. This completes the session. Hope you have understood the solution of this question.