 Hello and welcome to the session. In this session we discussed the following question which says, prove that log of 16 upon 27 to the base 10 is equal to 4 into log 2 to the base 10 minus 3 into log 3 to the base 10. Before moving on to the solution, let's discuss some laws of logarithm to be used in this question. First we have log of m upon n to the base a is equal to log m to the base a minus log n to the base a, then log of m to the power n to the base a is equal to n into log of m to the base a. This is the key idea that we use for this question. Let's proceed with the solution now. We need to prove that log of 16 upon 27 to the base 10 is equal to 4 into log 2 to the base 10 minus 3 into log 3 to the base 10. Let us first consider the LHS, which is log of 16 upon 27 to the base 10. So this is further equal to log of. Now when we factorize 16, we get 16 is equal to 2 to the power of 4. So 16 could be written as 2 to the power of 4 and on factorizing 27, we get 27 equal to 3 to the power of 3. So we can write here 3 to the power of 3. So this is log of 2 to the power of 4 upon 3 to the power of 3 to the base 10. Now using this law, which says log of m upon n to the base a is equal to log m to the base a minus log n to the base a, we get this is equal to log of 2 to the power 4 to the base 10 minus log of 3 to the power 3 to the base 10. Now we use this law in which we have log of m to the power n to the base a is equal to n into log n to the base a. So using this we get this is equal to 4 into log 2 to the base 10 minus 3 into log 3 to the base 10. And this is equal to the RHS that is this. Thus we have LHS is equal to the RHS that is log of 16 upon 27 to the base 10 is equal to 4 into log 2 to the base 10 minus 3 into log 3 to the base 10. So hence proved this completes the session. Hope you have understood the solution of this question.