Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Radical expressions Write an equivalent expression using radical notation and if possible simplify so the top number or the numerator is the exponent and the bottom number is the index. So Y to the one seventh is just the 7th root of Y to the 1st. Well, Y did the first. We know is itself, or just the 7th root of Y8 to the one third means the cube root of 8, and we know that eight we could think of as 2 * 4 or 2 * 2 * 2. So the cube root of 8 could also be thought of as the cube root of 2 ^3. And then the cube root in the cube would cancel, leaving us a final answer of just two, the 6th root of 64. So we're going to do the same thing. But lots of us might not know the 6th root of 64 off top of our heads. So we could think about 64 is 8 * 8, eight being 2 * 4 and 2 * 4, and this four being 2 * 2. So now when we count, we can see that there's really 6 twos. So we're going to have the 6th root of 2 to the 6th, and the 6th root and the 6th power are going to cancel, leaving us 2. If we have more than one thing, we just do each one separately. So AB to the one fourth is just the 4th root of A and of BB to the three halves. This one's tricky because we have sqrt b ^3, but the index is smaller than the exponent. So two goes into three one full time with one leftover. Now I have to use the absolute value because I don't know what B is. B has to come out as a positive number. So B to the three halves. There's one hole here with one leftover. I'm going to let you do a few of these. I'm going to do this 125 A to the 2/3. So this is saying the cube root of 125 A, and then whenever we get that, we're going to square it. So the cube root of 125 we know is actually 5, and the cube root of a we don't know. So we're going to leave it inside. And then when we square it, 5 ^2 is 25. And the cube root of a ^2 is the same thing as the cube root of a with the square on the inside. So I'm going to have you finish up those few that I didn't do this next section right? An equivalent expression using exponential notation. So if there is no exponent, we understand it's a one, and that index always goes in the denominator. So the cube rate of 19 is 19 to the one third. This next 1A to the five. The understood number, if there isn't 1 written, is a 2 because it's a square root. So A to the five halves. I'm going to jump down here to the fifth root of XY. That's really XY to the one fifth, but we're going to write each one separately. So X to the one fifth, Y to the 1/5. This last one, we don't have the three A being to an index, so we're going to just change the C and the C is going to be to the 2/5. It wasn't really the last one. I'm going to do this one also. So this next one and this grouping, we're going to get 2 to the 111th, X to the 6th, 11th and Y to the 111th. And then we're going to take all of those to the 9th power. So we're going to end up with two to the 9 elevenths, X to the 54 elevenths, and Y to the 9 elevenths as our final. The next type of grouping says write an equivalent expression with positive exponents, and if possible, simplify. Well, if it's got a negative exponent, it's going to move. So if it's in the numerator, it's going to move down to the denominator, and if it's in the denominator, it's going to move up to the numerator. Now remember that if there's nothing in the denominator, it's an understood 1. So for this first one, we're just going to have 1 / y to the one fifth because we move that Y to the one fifth down. And of course it's always one times anything in the numerator. This next one we're going to have this negative tells us it's going to go down and it's all going to go down. So it's going to be three X ^2 y to the 3/4, but we can actually multiply each of those out. So we're going to get 1 / 3 to the 3/4. X powers to powers 2 * 3/4. We multiply them and simplify SO3 halves and Y to the 3/4. This next one. If it's a fraction, the numerator goes to the denominator. The denominator goes to the numerator, so it's going to turn into 7 / 1 to the 3/4 or just 7 to the 3/4. This 8 and the -3 fifths just comes up to the top and we get A to the 3/5. This last one. The only thing with the negative exponent is the X, so everything else stays where it is. SO5Y to the 4/5 over Z or times Z / X to the 2/3. I'm going to leave that last one for you to do, and these next three I'm going to let you do also. So the next grouping says use the law of exponents. To simplify, do not use negative answers, exponents and any answer. So if the bases are the same, we add the exponents SO5 to the 2/3 + 1/2. Remember, we have to get a common denominator here. So 4 sixths plus three sixths, so 5 to the seven sixths. This next one, when we're dividing, we subtract the exponents, so 7 thirteenths minus -3 thirteenths or 10 thirteenths. Now a different way to do the same problem would have been to take that negative up to the numerator. So we would have had 9 to the seven thirteenths times 9 to the three thirteenths. And then when we add those, we can see 7 + 3 is 10. I do this one down here next. So X to the negative, 1/3 to the 1/4 powers to powers we multiply. So we get X to the -112 Y to the 220th which is really 110th and then that negative exponents going to come down. So we get Y to the 1/10 / X to the 112th, leaving some for you to practice. Also, the next grouping use rational exponents. To simplify, do not use fractional exponents in the final answer. So what they're really wanting us to do is they're wanting us to say T to the four sixths, reduce 4 sixths to 2/3, and then put it back into the radical. But there's an easy shortcut that we can do instead. When we look here, if we look at the index of six and the exponent of four, is there a number that boasts 6 and 4 evenly divisible by and? The answer is yes, 2. So 6 / 2 is that 3, 4 / 2 is that 2, And this will work every time, so let's practice that. 4 and 20 are both divisible by four. 4 / 4 is one. So the radicals going to actually go away a divide or 20 / 4 is five. Now, because it was an even index, we have to have absolute values around it. 7 and 14, they're both divisible by seven. 7 / 7 is one, so the radicals going to go away. 14 / 7 is 2, so we're going to get XY squared or X ^2, y ^2. We don't need the absolute value here because the index started out as being odd. Here. Four and two are both divisible by two, so 4 / 2 is 2, 2 / 2 is 1, so that one just simplifies to sqrt 5 X. Now radicals of radicals. Actually the shortcut here is we just multiply the indices. So this one's going to be the square root. No, sorry, the 8th root of X. And the reason we could think of this as the 4th root of X to the half and then X to the half to the 1/4 powers to powers, we multiply and then we put it back into the radical form. So the 8th root of X, the last ones. I'm going to have you practice the last type of problem. We want to find the domain of F. So remember that this is going to say that's sqrt X + 5, and the negative tells us it's sqrt X + 7. So in this problem, the inside of a radical always has to be positive or zero if it's an even index. But in the denominator it can never be 0. So this one's just going to be X + 7 greater than 0X greater than or equal to -5 X greater than -7. They both have to be true. So we have a hollow dot at -7 going to the right using a different color just so we can visualize it a little bit better. So where are they both true? They're both true from -5 to Infinity. Thank you and have a wonderful day.