 Hello and welcome to the session. In this session we will discuss the following question that says, put that log of square root 27 plus log 8 plus log of square root 10, this one upon log of 120 is equal to 3 upon 2. Before we move on to the solution, let's discuss some laws of logarithm. 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. Then the second law is log of x1 to the power n and this to the base a is equal to n into log of x1 to the base a. Now here this a is any positive number such that it is not equal to 1 then x1 is also a positive number and x2 is also a positive number. And this n is a real number that is n belongs to key idea that we use for this question. Let's proceed with the solution now. We are supposed to prove that log of square root 27 plus log 8 log of square root 1000 and this whole upon log of 120 is equal to 3 upon 2. First of all consider log of square root 27 plus log 8 plus log of 2000 and this whole upon log of 120. This means that log of 27 this mean to the power 1 upon 2 plus log of 8 or 8 can be written as 2 cube plus log of 1000 to the power 1 upon 2 upon log 120. Now we have factorized 120 as 2 cube into 3 into 5. The denominator in place of log 120 we can write log of 2 cube into 3 into. Consider this key idea that is the second log of the key idea 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. Using this log we will rewrite the terms of this expression. So this is equal to 1 upon 2 into log 27 plus 3 into log 2 plus 1 upon 2 into log 1000 and this way we can log of 2 cube into 3 into. Now when we factorize 27 we get very 7 is equal to 3 cube so we can write this as 1 upon 2 into log of 3 cube and divide this term by 2 so we get 3 upon 2 into 2 log 2 plus 1 upon 2 into 10 cube. So it will be log of 10 cube and this way we get 1 log of. Now let's rewrite these terms and we can write this as 2 square into 3 into. Now using the second log stated in the key idea that is log of x1 to the power n to the base a equal to n into log x1 to the base a we can rewrite numerator 3 upon 2 into log 3 upon 2 into 2 log 2 3 upon 2 into log 3 written using this log of x1 to the base a equals log x2 to the base. log 2 square plus log 3 plus log taken 3 upon 2 common in the numerator we have 3 upon 2 into log 3 and the log will be the first term we get 2 into log 2 log 3 plus log and after the numerator are same that is log 3 plus 2 log 2 plus log 10 and 2 log 2 plus log 3 plus log 10 So we now have the root 27 plus log 8 plus square root n upon log of 120 is equal to 3 upon 2 and this is what we were supposed to do with the pure understanding solution of this question.