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Exercises 1 When we are asked for the root of a number, we are typically being asked for the principal root. When the index is even, this is the positive root. Since the index of -416​ is even, let's find its principal root. -416​Split into factors-42⋅2⋅2⋅2​a⋅a⋅a⋅a=a4-424​a2​=a-2 The real root of -416​ is -2.
Exercises 2 We will rewrite the base as a perfect sixth to simplify this number. Because the denominator of the exponent is 6, this will allow us to simplify the rational exponent. Let's start! 72961​Split into factors(3⋅3⋅3⋅3⋅3⋅3)61​a⋅a⋅a⋅a⋅a⋅a=a6(36)61​(am)n=am⋅n36⋅61​6a​⋅6=a31a1=a3
Exercises 3 We will rewrite the base as a perfect fifth to simplify this number. Because the denominator of the exponent is 5, this will allow us to simplify the rational exponent. Let's start! (-32)57​Split into factors((-2)(-2)(-2)(-2)(-2))57​Write as a power((-2)5)57​(am)n=am⋅n(-2)5⋅57​5⋅5a​=a(-2)7Calculate power-128
Exercises 4 To simplify the given expression, we will use the Properties of Exponents. For this exercise, we will begin by multiplying powers with the same bases. Let's do it! z-2⋅z4am⋅an=am+nz-2+4Add termsz2
Exercises 5 To simplify the given exponential expression, we will use the Zero Property. a0=1, for every nonzero number a​ Let's try to simplify the given expression using this property. a0b-8b-5​a0=11⋅b-8b-5​Identity Property of Multiplicationb-8b-5​anam​=am−nb-5−(-8)-(-a)=ab-5+8Add termsb3
Exercises 6 To simplify the given expression, we will use the Properties of Exponents. For this exercise, we will begin by eliminating the negative sign in the exponent. Then we will distribute the exponent to the numerator and denominator. Let's do it! (52c4​)-3\FracToNegPow(2c45​)3(ba​)m=bmam​(2c4)353​(a⋅b)m=am⋅bm23(c4)353​(am)n=am⋅n23c1253​Calculate power8c12125​
Exercises 7 To write an exponential function to model the given situation, let's first recall the general form of an exponential equation. y=abx​ In this formula, a is the initial value and b=1+r, where r is the rate of change. If the function represents growth then r>0, and if it represents decay then r<0.Writing the Equation To write the equation, we first need to define the variables. Let y be the annual salary, and let x be the number of years after the initial value. In this case, the initial value is an annual salary of \$42500. Since the salary increases by 3% each year, we have that r=0.03. y=42500(1+0.03)x⇕y=42500(1.03)x​ To graph the function we will make a table of values. Since time is always greater than or equal to 0, we will assign nonnegative values for x.x42500(1.03)xy=42500(1.03)x 042500(1.03)042500 142500(1.03)143775 342500(1.03)3≈46440 542500(1.03)5≈49269 942500(1.03)9≈55452 Let's now plot and connect the obtained points. Since both variables are nonnegative, we will only draw in the first quadrant.
Exercises 8 Compound interest is the interest earned on the principal and on previously earned interest. Let's recall the formula that gives the balance y of an account earning compound interest. y=P(1+nr​)nt​ In this formula, P is the principal or initial amount, r is the annual interest rate written in decimal form, t is the time in years, and n is the number of times the interest is compounded in one year. Let's pay close attention to the given exercise.You deposit \$500 in an account that earns 6.5% annual interest compounded yearly.We can immediately identify P as 500. Also, the annual interest rate, written as a decimal number, is 0.065. Finally, since the interest is compounded yearly, we have that n=1. Let's substitute these values into the formula and simplify. y=P(1+nr​)ntSubstitute valuesy=500(1+10.065​)1tCalculate quotienty=500(1+0.065)1tIdentity Property of Multiplicationy=500(1+0.065)tAdd termsy=500(1.065)t To graph the function we will make a table of values. Since time is always greater than or equal to 0, we will assign nonnegative values for t.t500(1.065)ty=500(1.065)t 0500(1.065)0500 1500(1.065)1532.5 3500(1.065)3≈604 5500(1.065)5≈685 10500(1.065)10≈939 Let's now plot and connect the obtained points. Since both variables are nonnegative, we will only draw in the first quadrant.
Exercises 9 Let's analyze the given table and calculate the differences between consecutive terms in the sequence.Since each term is larger than the previous one by 14, the given sequence is arithmetic and its common difference is d=14. First, we will write a recursive rule for an arithmetic sequence. To do this, we have to give the value of the first term, a1​, and write an​=an−1​+d. In our case, a1​=-6 and d=14. Recursive rule: a1​=-6,an​=an−1​+14​ Now, let's use the explicit rule of arithmetic sequences. an​=a1​+(n−1)d​ Now, we will substitute known values into the explicit rule and find it. an​=a1​+(n−1)da1​=-6, d=14an​=-6+(n−1)(14) Simplify right-hand side Distribute 14an​=-6+n(14)−1(14)Multiplyan​=-6+14n−14Subtract terms an​=14n−20
Exercises 10 Let's analyze the given table and calculate the differences between consecutive terms in the sequence.Since each term is smaller than the previous one by different values, the given sequence is not arithmetic.Our sequence has the property that each term is multiplied by the same number, 41​, to get the next term. Therefore it is a geometric sequence and its common ratio is r=41​. First, we will write a recursive rule for a geometric sequence. To do this, we have to give the value of the first term, a1​, and write an​=r⋅an−1​. In our case, a1​=400 and r=41​. Recursive rule: a1​=400,an​=41​⋅an−1​​ Next, let's use the explicit rule of geometric sequences. an​=a1​⋅rn−1​ Now, we will substitute known values into the explicit rule and find it. an​=a1​⋅rn−1a1​=400, r=41​an​=400⋅(41​)n−1
Exercises 11 To solve the given exponential equation, we will start by rewriting the terms so that they have a common base. 2x=1281​Write as a power2x=271​a1​=a-12x=2-7 Now, we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. 2x=2-7⇔x=-7​
Exercises 12 To solve the given exponential equation, we will start by rewriting the terms so that they have a common base. 256x+2=163x−1Write as a power(162)x+2=163x−1(am)n=am⋅n162x+4=163x−1 Now, we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. 162x+4=163x−1⇔2x+4=3x−1​ Finally, we will solve the equation 2x+4=3x−1. 2x+4=3x−1LHS−3x=RHS−3x-x+4=-1LHS−4=RHS−4-x=-5LHS⋅(-1)=RHS⋅(-1)x=5
Exercises 13 We want to draw a graph of the given exponential function. f(x)=2(6)x​ Because the base of the function is greater than 1, we know that this is an exponential growth function. To do so, we will start by making a table of values.x2(6)xy=2(6)x -22(6)-20.055… -12(6)-10.333… 02(6)02 12(6)112 The ordered pairs (-2,0.055), (-1,0.333), (0,2), and (1,12) all lie on the function. We will plot and connect these points with a smooth curve.Comparison to the Graph of g(x) Now we are asked to compare the graph of f(x) to the graph of g(x)=6x. Let's draw them!Notice that the graph of f(x) is a vertical stretch of the function g(x) by a factor of 2. Furthermore, the domain of all exponential functions is all real numbers, and from the graph of f(x) we can see that the range of f(x) is y>0.
Exercises 14 Let's analyze the given equation. 5b5a​=5-3​ We are asked to complete the statement a<​b with the symbol <, >, or =. Let's use the formula anam​=am−n on the left-hand side of the equation. 5b5a​=5-3⇓5a−b=5-3​ Now, we have two equivalent expressions with the same base, 5. If both sides of the equation are equal, the exponents must also be equal. 5a−b=5-3⇓a−b=-3​ Since -3 is negative, the expression a−b is also negative. a−b<0LHS+b=RHS+ba<b Therefore, we should complete the statement a<​b with the symbol <.
Exercises 15 Let's analyze the given equation. 9a⋅9-b=1​ We are asked to complete the statement a<​b with the symbol <, >, or =. Let's use the formula am⋅an=am+n on the left-hand side of the equation and use the fact that 90=1 on the right-hand side. 9a⋅9-b=1⇓9a−b=90​ Now we have two equivalent expressions with the same base, 9. If both sides of the equation are equal, the exponents must also be equal. 9a−b=90⇓a−b=0​ Next, let's move b to the right side of the equation. a−b=0LHS+b=RHS+ba=b Therefore, we should complete the statement a<​b with the symbol =.
Exercises 16 We are given the first two terms of the sequence: a1​=3 and a2​=-12. We know that a3​ is the third term when the sequence is arithmetic, and b3​ is the third term when the sequence is geometric. Let's recall the basic facts about these two types of sequences.A sequence is arithmetic if the difference between consecutive terms is constant. This difference is usually called the common difference, and is denoted by d. A sequence is geometric if the ratio between consecutive terms is constant. This ratio is usually called the common ratio, and is denoted by r.Finding a3​ Since a1​=3, a2​=-12, and a3​ form an arithmetic sequence, let's find its common difference d by finding the difference between the first and the second term. d=a2​−a1​a2​=-12, a1​=3d=-12−3Subtract termsd=-15 Each subsequent term in the arithmetic sequence can be obtained by adding d to the previous one. Knowing this, we can find a3​. a3​=a2​+da2​=-12, d=-15a3​=-12+(-15)Add termsa3​=-27Finding b3​ Since a1​=3, a2​=-12, and b3​ form a geometric sequence, let's find its common ratio d by finding the ratio between the first and the second term. r=a1​a2​​a2​=-12, a1​=3r=3-12​Calculate quotientr=-4 Each subsequent term in the geometric sequence can be obtained by multiplying the previous one by r. Knowing this, we can find b3​. b3​=a2​⋅ra2​=-12, r=-4b3​=-12⋅(-4)Multiplyb3​=48Finding a3​−b3​ Finally, we only have to substitute the appropriate values and calculate a3​−b3​. a3​−b3​a3​=-27, b3​=48-27−48Subtract terms-75
Exercises 17
Exercises 18