 Hey guys, so having seen the theory part of Rational exponents of real numbers now it is time to solve a few problems So I am going to solve some of them and then later on you can solve these problems from any Standard textbook. Okay. Now, let us see what is the first problem. The first question says simplify 64 by 25 to the power minus 3 by 2 so how to solve this problem So let us say solution. We are now trying to do for a and this is 64 to the power and sorry 64 by 25 whole to the power minus 3 by 2 Okay, so by now laws of exponents. We know that this particular thing can be written as What can I write about this? So this is nothing but 64 by 25 whole To the power half and then whole again to the power minus 3. So which law did I apply? I applied this Let's say a to the power m by n is nothing but a to the power 1 by n Then whole to the power m, right? This is laws of rational exponents, right now To anything to the power half is nothing but the square root, isn't it? So I can write now as it has 64 by 25 Square root and whole to the power minus 3 correct now root 64 by root 25 So this can be written as root of 64 divided by root of 25 right and this whole to the power Minus 3 and what is a rule? So a to the power a by b To the power m is equal to a to the power m by b to the power m Is it it? now hence root of 64 is clearly 8 and Square root of 25 is 5 and now it is 8 by 5 to the power minus 3 right Which will be nothing but 1 upon 8 by 5 to the power 3 by which rule a to the power minus n is equal to 1 by a to the power in Correct. So hence now what can we do? We can now write this as 1 upon 8 cubed divided by 5 cubed right which is equal to 5 cubed by 8 cubed and hence it is 125 by 512 This is a simplified form right so this is what is the solution to the first question now Let us see the second question so the second question is 25 to the power minus 1 upon 3 into third root of 16 so what to do so if you see 25 to the power minus 1 by 3 can be written as 1 by 25 to the power 1 by 3 is it it and this is 16 to the power 1 by 3 which rules are we using so first rule is a to the power Minus n is equal to 1 upon a to the power n. This is for the first step and Let's say a to the power 1 by n can be written as nth root of a Isn't it so these two things I am using now. What is this? So if you see this is nothing but 16 by 25 This is 1 by 3. This is 1 by 3, right? So hence this can be written as 16 to the 16 by 25 Whole to the power 1 by 3 by which rule a to the power Sorry a by b to the power n will be equal to a to the power n P to the power n and vice versa. So in this case we had this and we went to this Okay, now what so let us write the factors of 16. So if you if you see 16 can be written as 2 into 2 into 2 into 2 to the power 4 and 25 likewise is 5 into 5 Isn't it so this is And the whole to the power 1 by 3 Correct. So what can we say now so you can either write this as 2 to the power 3 into 2 whole to the power 1 by 3 Divide by 25 to the power 1 by 3 whichever way, right? So this can then be written as 2 to the power 3 Whole to the power 1 by 3 into 2 to the power 1 by 3 divided by 25 to the power 1 by 3 Right. Why why is why am I writing like that? So a Times b whole to the power n is equal to a to the power n into b to the power n So in this case, I have considered 2 to the power 3 as 1 entity a correct. So hence you got this So what is the final value then so 2 to the power 3 into 1 by 3 it will become into 2 to the power 1 by 3 divided by 25 to the power 1 by 3 so which is equal to 2 to the power 1 into 2 upon 25 To the power 1 by 3 so it is right This is the final value so 2 into 2 by 25 to the power 1 by 3 now you can go ahead one more step by you know doing some Manipulation by let's say because this can be further done as 2 by 5 square isn't it this is 25 is 5 square to the power 1 by 3 Now what I can do is I can write this as 2 into 2 into 5 I multiply and divide by 5 so this will become 5 to the power q to the power 1 by 3 isn't it see I Multiplied and divide the numerator and denominator by 5 each so 5 square became 5 cube in the bottom right So hence what will this be now? So this is nothing but 2 into 10 to the power 1 by 3 Divide by 5 to the power 3 To the power 1 by 3 isn't it So this will finally become 2 into 10 to the power 1 by 3 divided by 5 why because 5 to the power 3 whole to the power 1 by 3 will be equal to 5 to the power 3 into 1 by 3 Which is 5 to the power 1 hence 5 Okay, so hence the final result will be 2 by 5 Into 10 to the power 1 by 3 So this could be another way of expressing this So this is question number 2 now the third question is Fifth root of 32 to the power minus 3. So hence again, what what would I do? I will write this as 32 to the power minus 3 upon 5 right, which is equal to 32 to the power 1 by 5 Whole to the power minus 3 Right now why because a to the power first is nth root of a is nothing but a to the power 1 by n So in this case fifth root it was y root 1 by 5 of 32 to the power minus 3 and then what a to the power p by q is simply a to the power 1 by q to the power p Right, so hence, what will it become? It is now Equal to 32 to the power 1 by 5 can be written as 2 to the power 5 whole to the power 1 by 5 And this is minus 3, which will equal to 2 to the power 5 into 1 by 5 whole to the power minus 3 Which will be equal to 2 to the power 1 whole to the power minus 3 Which is equal to 2 to the power minus 1 Sorry minus 3 And hence it is 1 upon 2 to the power 3 is equal to 1 upon 8 Okay, what all rules did I apply? I applied a to the power m to the power n is a to the power m times n for this particular step Right, and then a to the power negative n is equal to 1 upon a to the power n. That's for this to this step Okay, and then simply I simplified and then found the final answer. So this should be Uh, this is how you should be Uh, you know taking up some more problems like these and Solving them. Is it okay? So let's meet with another set of problems in the next video