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Exercises 1 Let's start by recalling how can we write a power with a rational exponent as a radical expression. an1=na These expressions are equivalent for any integer n greater than 1. Using this result, we can rewrite the exercise's expression. 8141=481 According to the definition of an nth root, the fourth root of 81 is the real number such that multiplied times it itself four times gives 81. Since 34=81, we can continue as shown below. 481=434=3 Therefore, we know that 8141=3. | |

Exercises 2 We are given the expressions shown below and asked to find the one that is different. We will label each of them in order to tell them apart. A) (327)2 B) 2732 C) 32 D)(227)3 Since these expressions involve rational exponents and radicals, it will be useful to review how can we relate them.Equivalence Beween Rational Exponents and Radicals Let a be a real number, and n and m are two positive integers. We can express a power with a rational exponent as a radical, as indicated below. anm=(na)m According to this property, we can rewrite expression B as expression A. 2732=(327)2 This allows us to evaluate B, 2732, by simplifying A, (327)2. We can start by simplifying the expression inside the parentheses by using the definition of an nth root. (327)2Rewrite 27 as 33(333)2nan=a{n}32Calculate power9 As we can see, expressions A and A are also equivalent to expression C, 32. Therefore, the expression that is different must be D, (227)3. This expression cannot be simplified any more, but we can approximate it using a calculator. (227)3Use a calculator140.296115…140.296 We just found that D (227)3≈140.296. As we can see, it is clearly different from the other expressions. Therefore, this is the odd one out. | |

Exercises 3 When rewriting a radical into rational exponent form, the exponent of the radicand is the numerator of the rational exponent, and the index of the radical is the denominator of the rational exponent. na=an1 Recall that when the index is not stated, it is 2. Therefore, 10=210. With this in mind, we can rewrite the given expression. 210 = 1021 | |

Exercises 4 When rewriting a radical into exponential form, the exponent of the radicand is the numerator of the rational exponent, and the index of the radical is the denominator of the rational exponent. na=an1 With this in mind, we can rewrite the given expression. 534=3451 | |

Exercises 5 To simplify the given expression, remember that the numerator of a rational exponent is the exponent of the expression, and the denominator is the index. anm=nam With this in mind, we can rewrite the given expression. 1531⇔3151⇔315 | |

Exercises 6 To simplify the given expression, remember that the numerator of a rational exponent is the exponent of the expression, and the denominator is the index. If the numerator is 1, we do not write it in the radical expression. an1=na With this in mind, we can rewrite the given expression. 14081=8140 | |

Exercises 7 For an integer n greater than 1, if bn=a then b is said to be an nth root of a, and is written as na. Let n be an integer greater than 1, and let a be a real number. Let's recall four rules used to calculate the real nth roots of a.If n is odd, then a has one real nth root, which is na = a1/n. If n is even and a>0, then a has two real nth roots, which are ±na = ±a1/n. If n is even and a=0, then a has one real nth root, which is n0 = 0. If n is even and a<0, then a has no real nth roots. We are given n=2, an even number, and a=36, a positive number. Therefore, by the above rules, 36 has two real squared roots. ±236=±6or±361/2=±6 Showing Our Work info Finding the Root We can find the 2nd root without a calculator by looking at the factors of 36. We want to decompose the number until we have a factor that shows up 2 times. 236Split into factors26⋅6a⋅a=a2262nan=a{n}±6 | |

Exercises 8 For an integer n greater than 1, if bn=a then b is said to be an nth root of a, and is written as na. Let n be an integer greater than 1, and let a be a real number. Let's recall four rules used to calculate the real nth roots of a.If n is odd, then a has one real nth root, which is na = a1/n. If n is even and a>0, then a has two real nth roots, which are ±na = ±a1/n. If n is even and a=0, then a has one real nth root, which is n0 = 0. If n is even and a<0, then a has no real nth roots. We are given n=4, an even number, and a=81, a positive number. Therefore, by the above rules, 81 has two real fourth roots. ±481=±3or±811/4=±3 Showing Our Work info Finding the Root We can find the 4th root without a calculator by looking at the factors of 81. We want to decompose the number until we have a factor that shows up 4 times. 481Split into factors49⋅9Split into factors43⋅3⋅3⋅3Write as a power434nan=a{n}±3 | |

Exercises 9 For an integer n greater than 1, if bn=a then b is said to be an nth root of a, and is written as na. Let n be an integer greater than 1, and let a be a real number. Let's recall four rules used to calculate the real nth roots of a.If n is odd, then a has one real nth root, which is na = a1/n. If n is even and a>0, then a has two real nth roots, which are ±na = ±a1/n. If n is even and a=0, then a has one real nth root, which is n0 = 0. If n is even and a<0, then a has no real nth roots. We are given n=3, an odd number, and a=1000, a positive number. Therefore, by the above rules, 1000 has one real cube root. 31000=10or10001/3=10 Showing Our Work info Finding the Root We can find the third root without a calculator by looking at the factors of 1000. We want to decompose the number until we have a factor that shows up 3 times. 31000Split into factors310⋅10⋅10a⋅a⋅a=a33103nan=a{n}10 | |

Exercises 10 For an integer n greater than 1, if bn=a then b is said to be an nth root of a, and is written as na. Let n be an integer greater than 1, and let a be a real number. Let's recall four rules used to calculate the real nth roots of a.If n is odd, then a has one real nth root, which is na = a1/n. If n is even and a>0, then a has two real nth roots, which are ±na = ±a1/n. If n is even and a=0, then a has one real nth root, which is n0 = 0. If n is even and a<0, then a has no real nth roots. We are given n=9, an odd number, and a=-512, a negative number. Therefore, by the above rules, -512 has one real ninth root. 9-512=-2or(-512)1/9=-2 Showing Our Work info Finding the Root We can find the ninth root without a calculator by looking at the factors of -512. We want to decompose the number until we have a factor that shows up 9 times. 9-512Split into factors9-32⋅16Split into factors9-2⋅16⋅8⋅2Split into factors9-2⋅(-2)⋅(-2)⋅(-2)⋅(-2)⋅(-2)⋅(-2)⋅(-2)⋅(-2)Write as a power9(-2)9nan=a{n}-2 | |

Exercises 11 When given the volume and asked to find the length of a side, it's best to remind yourself of the formula for the cube's volume. Vcube=s3 Let's substitute V=64 into the formula and solve for s. V=s3V=6464=s3 Solve for s 3LHS=3RHS364=3s3Calculate root4=sRearrange equation s=4 The side of a cube with volume 64 in3 is 4 in. | |

Exercises 12 When given the volume and asked to find the length of a side, it's best to remind yourself of the formula for the cube's volume. Vcube=s3 Let's substitute V=216 into the formula and solve for s. V=s3V=216216=s3 Solve for s 3LHS=3RHS3216=3s3Calculate root6=sRearrange equation s=6 The side of a cube with volume 216 in3 is 6 in. | |

Exercises 13 When we are given a number in a radical, we are typically being asked for the principal root. When the index is even, this is the positive root. Since the index of 4256 is even, let's find its principal root. 4256Split into factors416⋅16Split into factors44⋅4⋅4⋅4Write as a power444nan=a{n}4 The real root of 4256 is 4. | |

Exercises 14 A cube root is the number you can multiply by itself three times to get the value inside the radical. a⋅a⋅a=a3=-216 In this case, we need the number that can be multiplied three times to get -216. a=3-216Split into factorsa=3-6⋅(-6)⋅(-6)a⋅a⋅a=a3a=3(-6)3nan=a{n}a=-6 The cube root of 3-216 is -6. | |

Exercises 15 A cube root is the number you can multiply by itself three times to get the value inside the radical. a⋅a⋅a=a3 In this case, we need the number that can be multiplied three times to get -343. a=3-343Split into factorsa=3-7⋅(-7)⋅(-7)a⋅a⋅a=a3a=3(-7)3nan=a{n}a=-7 The cube root of 3-343 is -7. | |

Exercises 16 An nth root is the number you can multiply by itself n times to get the value inside the radical. a⋅a⋅a⋅a⋅a=a5 In this case, we need the number that can be multiplied five times to get 1024. a=-51024Split into factorsa=-564⋅16Split into factorsa=-54⋅4⋅4⋅4⋅4Write as a powera=-545nan=a{n}a=-4 The fifth root of -51024 is -4. | |

Exercises 17 We will rewrite the base as a perfect seventh to simplify this number. Because the denominator of the exponent is 7, this will allow us to simplify the rational exponent. Let's start! 12871Split into factors(2⋅2⋅2⋅2⋅2⋅2⋅2)71Write as a power(27)71(am)n=am⋅n27⋅717⋅7a=a21a1=a2 | |

Exercises 18 Let's examine the given expression. (-64)21⇔-64 Since any number multiplied by itself cannot produce a negative number, (-64)21 is not a real number. | |

Exercises 19 When rewriting a radical into rational exponent form, the exponent of the radicand is the numerator of the rational exponent, and the index of the radical is the denominator of the rational exponent. na=an1 and nam=anm With this in mind, we can rewrite the given expression. (58)4⇔854 | |

Exercises 20 When rewriting a radical into exponential form, the exponent of the radicand is the numerator of the rational exponent, and the index of the radical is the denominator of the rational exponent. (na)m=am/n With this in mind, we can rewrite the given expression. (5-21)6⇔(-21)6/5 | |

Exercises 21 To simplify the given expression, remember that the numerator of a rational exponent is the exponent of the expression, and the denominator is the index. anm=(na)m With this in mind, we can rewrite the given expression. (-4)72⇔(7-4)2 | |

Exercises 22 To rewrite the given expression, remember that the numerator of a rational exponent is the exponent of the expression, and the denominator is the index. anm=nam With this in mind, we can rewrite the given expression. 925=95 | |

Exercises 23 To evaluate the expression, we will rewrite the base as a perfect fifth. Because the denominator of the exponent is 5, this will allow us to simplify the rational exponent. Let's start! 3253Split into factors(2⋅2⋅2⋅2⋅2)53a⋅a=a2(25)53(am)n=am⋅n25⋅535⋅5a=a23Calculate power8 | |

Exercises 24 Since the denominator of the given rational exponent is 3, we will write the base as a power with exponent 3. This will allow us to simplify the rational exponent. 12532Split into factors(5⋅5⋅5)32a⋅a⋅a=a3(53)32(am)n=am⋅n53⋅323⋅3a=a52Calculate power25 | |

Exercises 25 Let's examine the given expression. (-36)23⇔(-36)3 Since any number multiplied by itself cannot produce a negative number, (-36)23 is not a real number. | |

Exercises 26 Since the denominator of the given rational exponent is 5, we will write the base as a power with exponent 5. This will allow us to simplify the rational exponent. (-243)52Split into factors((-3)⋅(-3)⋅(-3)⋅(-3)⋅(-3))52Write as a power((-3)5)52(am)n=am⋅n(-3)5⋅525⋅5a=a(-3)2Calculate power9 | |

Exercises 27 To evaluate the expression, we will rewrite the base as a perfect seventh. Because the denominator of the exponent is 7, this will allow us to simplify the rational exponent. Let's start! (-128)75Split into factors((-2)(-2)(-2)(-2)(-2)(-2)(-2))75Write as a power((-2)7)75(am)n=am⋅n(-2)7⋅757⋅7a=a(-2)5(-a)5=-a5-25Calculate power-32 | |

Exercises 28 We will rewrite the base as a perfect cube to simplify this number. Because the denominator of the exponent is 3, this will allow us to simplify the rational exponent. Let's start! 34334Split into factors(7⋅7⋅7)34a⋅a⋅a=a3(73)34(am)n=am⋅n73⋅343⋅3a=a74Calculate power2401 | |

Exercises 29 Let's look carefully at the equation. (32)4=23/4 The error is in the exponent. The root number, 3, of the radical sign goes in the denominator of the fractional exponent and the exponent, 4, belongs in the numerator. The mathematical rule is anm=nam. Done correctly, we get the following. (32)4=24/3 | |

Exercises 30 When describing errors, it's important to justify each step to see if you can identify the error. Let's look at the first step. (-81)3/4=[(-81)1/4]3 In this step we can see that the rule of (am)n=am⋅n was applied where m=1/4 and n=3. This step follows the rules for exponents, lets look at the next step. ==[(-81)1/4]3(-3)3 In this step, there is an error. There is no real solution when taking an even root of a negative number. Therefore, -3 is incorrect and should have been an imaginary number. (-81)3/4====[(-81)1/4]3(3i)3-27i3-27i | |

Exercises 31 To evaluate the expression, we will rewrite the base as a perfect cube. Because the denominator of the exponent is 3, this will allow us to simplify the rational exponent. Let's start! (10001)31Split into factors(10⋅10⋅101)31a⋅a⋅a=a3(1031)31(ba)m=bmam10331311a=110331Cancel out common factors101 | |

Exercises 32 Since the denominator of the given rational exponent is 6, we will write the base as a power with exponent 6. This will allow us to simplify the rational exponent. (641)61Split into factors(2⋅2⋅2⋅2⋅2⋅21)61Write as a power(261)611a=1(2616)61bmam=(ba)m((21)6)61(am)n=am⋅n(21)6⋅616⋅6a=a(21)1a1=a21 | |

Exercises 33 We want to simplify the given number. Notice that this expression has a negative exponent. When this is the case, the expression can be moved to the denominator and the exponent will become positive. a-nm=anm1⇒(27)-32=(27)321 We will split the base into perfect cube factors to simplify this fraction. Because the denominator of the exponent is 3, this will allow us to simplify the rational exponent. Let's start! (27)321Split into factors(3⋅3⋅3)321a⋅a⋅a=a3(33)321(am)n=am⋅n33⋅3213⋅3a=a321Calculate power91 | |

Exercises 34 We want to simplify the given number. Notice that this expression has a negative exponent. When this is the case, the expression can be moved to the denominator and the exponent will become positive. a-nm=anm1⇒(9)-25=(9)251 We will write the base as a perfect square to simplify the rational exponent. (9)251Write as a power(32)251(am)n=am⋅n(3)2⋅2512⋅2a=a351Calculate power2431 | |

Exercises 35 When asked to find the area, we need to identify the geometric figure to determine which formula to use. In this case, we are looking at a rectangle. ARectangle=ℓ⋅w For this rectangle ℓ=3729 and w=41/2. A=ℓ⋅wℓ=3729, w=41/2A=3729⋅41/2 Solve for A Calculate rootA=9⋅41/2Calculate powerA=9⋅2Multiply A=18 Therefore, the area of the bake sale sign is 18 ft2. | |

Exercises 36 When a number is written in power form, we tend to want to compute it. In this exercise, 275 is a very large number, let's leave it like it is and see what happens as we solve. First, let's recall the formula for a cube's volume. Vcube=s3 Now lets substitute V=275 and solve for s. V=s3V=275275=s3 Solve for s 3LHS=3RHS3275=3s3Simplify radical and power275/3=s3/3Calculate power243=sRearrange equation s=243 Therefore, if a cube has a volume of 275 mm3, then the length of its side is 243 mm. Alternative solution info Mental Math Another way to think about 275 is to rewrite 27 as a perfect cube, then take the cubed root. These steps can be done mentally, so they can eliminate the need for a calculator. 3s3=327527=333s3=3(33)5Rewrite (33)5 as (35)33s3=3(35)3Calculate roots=35Calculate powers=243 | |

Exercises 37 In this exercise, we are given a formula and a picture with values. We need to substitute the values from the picture into the formula and solve. Given the information, we can see that V=5, h=4, and they want us to use π=3.14. r=(πh3V)1/2Substitute valuesr=(3.14(4)3(5))1/2Multiplyr=(12.5615)1/2Calculate quotientr≈(1.194)1/2Calculate powerr≈1.093 From this we can say that the radius of the cup to the nearest inch is 1 in. | |

Exercises 38 In this exercise, we have a formula and a value for surface area. Let's substitute the values S=60 and estimate π as 3.14 into the formula and solve. V=6π1S3/2V=60, π=3.14V=63.141(60)3/2 Solve for V Calculate rootV≈6(1.77)1(60)3/2Calculate powerV≈63.141(464.76)MultiplyV≈10.631(464.76)Calculate quotientV≈(0.094)(464.76)Multiply V≈43.71 From this we can approximate the volume of the sphere to be 44 m3. | |

Exercises 39 Let's start by reviewing the definition of an nth root. If n is an integer greater than 1, and an=b, we say that a is an nth root of b. We denote this as shown below. nb=nan=a Now we can apply the Power of a Power Property to write nth roots as rational exponents. Let's consider the quantity (an1)n. (an1)n(am)n=am⋅nan1⋅nna⋅n=aa1a1=aa According to the definition of an nth root an1 is the nth root of a, because (an1)n=a. We can conclude that the expressions shown below are equivalent. an1=na Finally, we can raise both sides of the equation above to the mth power and simplify using the Power of a Power Property once again. an1=naLHSm=RHSm(an1)m=(na)m(am)n=am⋅nan1⋅m=(na)mb1⋅a=baanm=(na)m With this conclusion, we see that we can write a radical expression as a power with a rational exponent, as required. (na)m=anm | |

Exercises 40 We need to write an expression for the the side length of a square with area x in2. Let's start with the usual formula for area then substitute our value for area. Asquare=s2 Let's substitute A=x. A=s2A=xx=s2 Solve for s LHS=RHSx=s2Calculate root±x=sRearrange equations=±xRewrite x as x1/2 s=±x1/2 We are talking about the side length of a square and can ignore the negative solution. Therefore, s=x1/2, when the area is x in2. | |

Exercises 41 To calculate the annual inflation rate, we can use the formula. r=(PF)1/n−1 P is the initial value, in our exercise, P=800000. F is the new value, in this case, F=1100000. And n is the number of years, in our case, n=6. Let's substitute those values into our equation and solve. r=(PF)1/n−1Substitute valuesr=(8000001100000)1/6−1 Solve for r Calculate quotientr=(1.375)1/6−1Calculate powerr≈1.0545−1Subtract terms r≈0.0545 To convert that decimal to a percent, we need to multiply by 100 yielding r≈5.45%. We are asked to round that to the nearest tenth of a percent which gives us a 5.5% rate of inflation over the 6 years. | |

Exercises 42 To calculate the annual inflation rate, we can use the formula. r=(PF)1/n−1 P is the initial value, in our exercise, P=1.46. F is the new value, in this case, F=3.53. And n is the number of years, in our case, n=10. Let's substitute those values into our equation and solve. r=(PF)1/n−1Substitute valuesr=(1.463.53)1/10−1 Solve for r Calculate quotientr≈(2.4178)1/10−1Calculate powerr≈1.0923−1Subtract terms r≈0.0923 To convert that decimal to a percent, we need to multiply by 100 yielding r≈9.23%. We have been asked to round that to the nearest tenth of a percent which gives us a 9.2% rate of inflation over the 10 years. | |

Exercises 43 Consider numbers that always have the same answer for different exponents. In this case, 0n=0 and 1n=1. Let's look at why these work. x=x1/5x=00=?01/5Calculate power0=0 Zero to the power of any positive number is always zero meaning one solution is x=0. Similarly, one to the power of any number is always one. x=x1/5x=11=?11/51a=11=1 Therefore the second solution to this equation is x=1. Let's also consider that since x1/5=5x, we need to look at negative numbers as well. In this case, -1. x=x1/5x=-1-1=?-11/5an1=na-1=?5-1Calculate root-1=-1 We see that -1, is a third solution. To prove these three solutions are the only solutions it's helpful to look at the graphs. The extra with the graphs is provided below. Extra info Graph the Solutions One way to look for solutions on a graph is to see where each side of the equation crosses the other. The points of intersection are the solutions. Let's graph y=x1/5 and y=x.We can see the graphs intersect at our three solutions of x=-1, x=0, and x=1. | |

Exercises 44 One way to check if our friend is correct is to check our own values. We need to consider even and odd roots as well positive and negative arguments. Let's set up a table and compute different values for na and -na.nana-na 264264=8-264=-8 2-642-64=8i-2-64=-8i 364364=4-364=-4 3-643-64=-4-3-64=-(-4)=4 In the last case, where n is odd and a is negative, we get a negative answer for na and positive one for -na. Therefore, out friend is incorrect. | |

Exercises 45 To simplify the given expression, we will use the Properties of Exponents. For this exercise, we will begin by distributing the exponent to the base. Then we will try to simplify the exponent by canceling common factors. Let's do it! (y61)3⋅x(am)n=am⋅ny61⋅3⋅xb1⋅a=bay63⋅xSimplify powery21⋅xa21=ay⋅xa⋅b=a⋅bxya=a21(xy)21 | |

Exercises 46 To simplify the given expression, we will use the Properties of Exponents. For this exercise, we will begin by using the Product of Powers Property. Then we will use the Power of a Power Property. Let's do it! (y⋅y1/3)3/2Write as a power(y1⋅y1/3)3/2am⋅an=am+n(y1+1/3)3/2Rewrite 1 as 33(y3/3+1/3)3/2Add fractions(y4/3)3/2(am)n=am⋅ny4/3(3/2)Multiply fractionsy12/6Calculate quotienty2 Alternative solution info Alternative Solution There is another way of solving this exercise. We can start by distributing the exponent. (y⋅y1/3)3/2(a⋅b)m=am⋅bmy3/2(y1/3)3/2(am)n=am⋅ny3/2y1/3(3/2)Multiply fractionsy3/2y3/6ba=b/3a/3y3/2y1/2am⋅an=am+ny3/2+1/2Add fractionsy4/2Calculate quotienty2 | |

Exercises 47 To simplify the given expression, we will use the Properties of Exponents. For this exercise, we will begin by eliminating the nth roots. Then we will try to simplify the expression by factoring out common factors. Let's do it! x⋅3y6+y2⋅3x3Split into factorsx⋅3y2⋅3+y2⋅3x3am⋅n=(am)nx⋅3(y2)3+y2⋅3x3nan=a{3}x⋅y2+y2⋅xFactor out x⋅y2x⋅y2(1+1)Add terms2xy2 | |

Exercises 48 To simplify the given expression, we will use the Properties of Exponents. For this exercise, we will begin by using the Power of a Product Property. Then we use the Product of Powers Property. Let's do it! (x1/3⋅y1/2)9⋅y(a⋅b)m=am⋅bmx1/3(9)⋅y1/2(9)⋅yb1⋅a=bax9/3⋅y9/2⋅ya=a1/2x9/3⋅y9/2⋅y1/2am⋅an=am+nx9/3⋅y9/2+1/2Add fractionsx9/3⋅y10/2Calculate quotientx3⋅y5Multiplyx3y5 | |

Exercises 49 We are given the formula for the volume of a dodecahedron, as well as a picture of one with a volume of 20 ft3. Let's substitute V=20 into the given formula and solve for ℓ, the edge length. V≈7.66ℓ3V=2020≈7.66ℓ3 Solve for ℓ LHS/7.66=RHS/7.662.61≈ℓ3LHS1/3=RHS1/32.611/3≈(ℓ3)1/3Calculate power1.37≈ℓRearrange equation ℓ≈1.37 Therefore, the edge length of a dodecahedron with a volume of 20 ft3 is about 1.37 ft. | |

Exercises 50 There are many possible formulas we can choose from that contain a radical. For example, we can consider the formula for the period of a swinging pendulum. If the angle is small we can calculate the time that the pendulum takes to complete an oscillation using the expression shown below. T=2πgL In this equation T is the period of the motion, L is the length of the pendulum's string, and g is the Earth's gravity constant. Note that the radical sign doesn't have an index, therefore we know it is a square root. Recall that we can rewrite any nth root as a power. na=an1 With this in mind, we can rewrite the formula for the period of a pendulum. T=2πgL⇔T=2π(gL)21 | |

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