Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Write an equivalent expression using radical notation and if possible simplify. The thing we need to understand is this bottom is our index. So when we say Y to the one 7th power, we're saying the index is 7 and Y to the first. Well, anything to the first power is really itself. So the 7th root of Y is equivalent to Y to the one 7th, 8 to the one third. Our index is going to be 3 and 8 to the first would be inside, but 8 is really 2 ^3 2 * 2 * 2. So the cube root of 2 ^3, the cube root and the cube cube will cancel, leaving us just 264 to the one sixth. The index is 664 to the 1st on the inside. Anything to the first is itself. 64 is 2 to the 6th, so the 6th root and 6th power are going to cancel, leaving us 2 AB to the one fourth. The index is 4, so the 4th root of AB. Looking at a few more B to the three halves power, the index is 2 and we have b ^3 left. Now that 2 is understood to be just a plain square root, so you could write it without the two in it. 4 to the seven halves sqrt 4 to the seven. Now we know sqrt 4, and then we could take that to the 7th power. So sqrt 4 is 2, 2 to the 7th, 2 * 2 is 4816326412881 to the three halves power. That bottom number is our index, so we're saying 81 ^3. Well we could do sqrt 81 first and then cube it. Sqrt 81 is 9/9 cubed is 7/29 125 A all to the 2/3 power. So the index is 325 A squared. Now we actually know the cube root of 125, and once we find that we're going to square it. We do not know the cube root of the a ^2 cube root of 125 is 5 S 5 ^2. The cube root of a ^2 we're going to have to leave. So 25 cube roots of a ^2 would be our final answer for that one. In simplified form, this next 1 sqrt 9 Y to the 6th, and then we can cube that. So sqrt 9 is 3 sqrt y to the 6th. Now we could we could think of this as undoing what we had been doing a minute ago. We could think of that as 6 / 2, the exponent. Well, 6 / 2 is really just three, so now we'd have 3 Y cubed. All cubed 3 ^3 is 27 powers to powers we multiply, so we get 27 Y to the 9th. As a final answer there, we want to write these as equivalent expressions using exponential notation. So our index was three, so the three goes in the denominator, so 1/3. Our index here is an understood 2, so 11 to the one half. Our index here once again is an understood 2, but we have A5 in the power to start with. So now it's going to be A to the five halves, N to the seven sixths, this next one, XY to the one fifth. Or we could think of that as X to the one fifth, Y to the 1/5, this next X to the fourth, Y to the fifth, Z cubed, all to the one 7th. So a final answer of X to the 4 sevenths, Y to the 5 sevenths, Z to the three sevenths, This next 17 XY is all going to be to the one third, and then that's all going to be to the fourth. Well, powers to powers, we multiply. So we have 7 XY to the Four Thirds or 7 to the Four Thirds, X to the Four Thirds, and Y to the Four Thirds. This next one three a / C to the 2/5 because the five is the index and the index goes in the denominator. We're going to write these next type of problems with positive exponents. So instead of negative 1/5, it's 1 / y to the one fifth three X ^2 y to the -3 fourths. That's one over three X ^2 y to the 3/4, which is 1/3 to the 3/4 powers to powers we multiply. So 2 * 3/4 would give us three halves for the X exponent and Y to the 3/4. This next one, the negative in the exponent, is going to tell us to flip the whole thing and change the negative exponent to a positive. 7 / 1 is just 7, so 7 to the 3/4 power negative in the exponent, and the denominator tells us to take the reciprocal and make it a positive, so A to the 3/5. You could think of this one if you wanted to do it in steps as 1 / A to the 3/5 and then 1 / 1 / A to the 3/5. Remember, we're going to skip, flip, and multiply. So we get A to the 3/5. This next one we want to get rid of the negative exponents. The negative exponents only on the X. So we're only moving the X to the denominator, everything else stays where it is. So our final answer 5 why did the 4 fifths Z / X to the 2/3 23? The only thing with a negative exponent is the X, so it's going to move up to the top. Everything else is going to stay where it's at. SO2X to the one third Z / 5. Remember I'm due I'm going to put the variables in alphabetical order X then Z8 to the 1 / 8 to the negative 2/3. That's going to be 8 to the positive 2/3. But we know the cube root of 8. The cube root of 8, we said a few moments ago, is just two. So then we have 2 ^2 or 4. When we have power, a base to a power times of the same base to a power, we add the exponents. So 2/3 + 1/2. We'd have to get a common denominator of 6th, so we'd have 4 sixths plus three sixths or seven sixths. It says we don't want to use exponents in any of our answers. If we had wanted to give this in radical notation, it would have been the 6th root of 5 to the 7th. This next one, the bases are the same, but we're subtracting or we're dividing, so we subtract the exponents. So 7 thirteenths minus -3 thirteenths. That's going to give us 10 thirteenths. If we wanted it in radical notation, the index would be thirteen, 9 to the 10th on the inside powers to powers we multiply, so we'd have 5 fourths times 3/10. If we wanted to reduce this five and that 10 would go to 1 / 2 or 1/2, so we'd get 5 to the three eighths, 1 * 3 in the top, 4 * 2 in the bottom. If we wanted to write it in radical notation, it D be the 8th root of 5 ^3 powers to powers. Once again, we're going to multiply, so we get -3 halves times 2 ninths. The twos are going to cancel, and the three and the 9 are each reducible by three. So we're going to get A to the -1 third, but we don't want any negatives. So we're going to take the reciprocal 1 / A to the one third powers to powers again. Now we have to do that 1/4 to both the terms. So we get negative 1/3 * 1/4 and Y to the 2/5 * 1/4. So we'd get X to the negative 112th Y. The two and the four are going to reduce to the 110th. The directions say no negative exponents, so we're going to take that X to the bottom. The next one we're going to use rational exponents to simplify. So T to the four, sixth four 6th reduces to 2/3. So the cube root of t ^2. Now what I show students is that that six and four are both evenly divisible by two. If you thought about 6 / 2, you'd get 3, and 4 / 2 you'd get 2. So you could actually get to the final answer without putting it into exponential. Here we'd have A to the 20 / 4. Well, 20 / 4 is 5. If you did the shortcut. 4 and 20 are both divisible by what, four? 4 / 4 is 1, so we don't have an radical anymore. 20 / 4 is 5 here. We'd have 6 and 18 are each divisible by 6, so 6 / 6 is 1. The radical would go away. 18 / 6 is 3. If you don't see that, we can always write it in exponential 18. The index is 618 / 6 is 37 and 14, each divisible by 7. So we're going to get XY squared because we have XY all to the 14th over seven. Well, 14 / 7 is 2, so X ^2 y ^2. This next one we'd have 5X to the 2 fourths. 2 fourths is 1/2. We don't want it in exponential form. We want to give it the answer back in radical form, so we change it back to sqrt 5 X. If we looked at the four and the two, they each had a 2IN common, so 4 / 2 is 2. Understood 2 here 2 / 2 is 1. So that power goes to the first power. This next one. This inside part is X to the half. This outside index is to the one fourth powers. To powers we multiply and put it back into radical form. Now when we have an index to an index, the shortcut method multiply those two numbers four times the understood 2 and we get the 8. Here 3 * 6 is going to give us a final of the 18th root of M If we don't see it, the cube root of M to the one 6th, and then to get rid of the cube root, we put that as the one third power powers to powers, we multiply, and then we put it back into radical notation. This next one, four and 12 each have a four in common. 4 into 4 is one, so the radical notation goes away. 4 into 12 is 3, so our final answer is going to be XY cubed or X ^3, y ^3. We can rewrite that if we don't see it as 12 / 4 and the 12 / 4 would reduce to the three and then take each of them separately. This one we want to find the domain, so if we start by rewriting it, we get sqrt X + 5 / sqrt X + 7. Well, we know that this inside of a square root always has to be positive or zero. So that sqrt X + 5 had to be greater than or equal to 0. Now the X + 7, because it's in the denominator, has to be greater than 0. We need both of these to be true. So I need X greater than or equal to -5, X greater than -7. If we thought about looking at this on a number line, we'd have X greater than or equal to -5 and would have X greater than -7. We need both of those pieces to be true, so we need where both of them are correct. So our final answer would be -5 to Infinity. If I had said -6 negative 6 wouldn't have worked in the top. It would have worked in the bottom, but not the top. If I say -5 negative 5 works in both on top, top and bottom. Anything to the right of -5 works in both the top and the bottom. Thank you and have a wonderful day.