#set document(title: "3.4 Rational Exponents", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.4#h(0.6em)Rational Exponents === Powers of the Form #math.equation(block: false, alt: "a to the power m / n")[$a^(m / n)$] In the last section, we considered powers of the form #math.equation(block: false, alt: "a to the power 1 / n")[$a^(1 / n)$], such as #math.equation(block: false, alt: "x to the power 1 / 3")[$x^(1 / 3)$] and #math.equation(block: false, alt: "x to the power minus 1 / 4")[$x^(− 1 / 4)$], and saw that #math.equation(block: false, alt: "a to the power 1 / n")[$a^(1 / n)$] is equivalent to the root #math.equation(block: false, alt: "the nth root of a")[$root(n, a)$]. What about other fractional exponents? What meaning can we attach to a power of the form #math.equation(block: false, alt: "a to the power m / n")[$a^(m / n)$]? Consider the power #math.equation(block: false, alt: "x to the power 3 / 2")[$x^(3 / 2)$]. Notice that the exponent #math.equation(block: false, alt: "the fraction 3 over 2 equals 3 open parenthesis the fraction 1 over 2 close parenthesis")[$display(frac(3, 2)) = 3 ( display(frac(1, 2)) )$], and thus by the third law of exponents, we can write #math.equation(block: true, alt: "open parenthesis x to the power 1 / 2 close parenthesis cubed equals x to the power open parenthesis 1 / 2 close parenthesis cubed equals x to the power 3 / 2")[$( x^(1 / 2) )^(3) = x^(( 1 / 2 )^(3)) = x^(3 / 2)$] In other words, we can compute #math.equation(block: false, alt: "x to the power 3 / 2")[$x^(3 / 2)$] by first taking the square root of #math.equation(block: false, alt: "x")[$x$] and then cubing the result. For example, #math.equation(block: true, alt: "100 to the power 3 / 2, equals open parenthesis 100 to the power 1 / 2 close parenthesis cubed, Take the square root of 100.; equals 10 cubed equals 1000, Cube the result.")[$100^(3 / 2) & = ( 100^(1 / 2) )^(3) & & "Take the square root of 100." \ & = 10^(3) = 1000 & & "Cube the result."$] We will define fractional powers only when the base is a positive number. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Rational Exponents] #math.equation(block: true, alt: "a to the power m / n equals open parenthesis a to the power 1 / n close parenthesis to the power m equals open parenthesis a to the power m close parenthesis to the power 1 / n , a greater than 0 , n not equal to 0")[$a^(m / n) = ( a^(1 / n) )^(m) = ( a^(m) )^(1 / n) , " " " " " " " " a > 0 "," " " " " n ≠ 0$] ] #math.equation(block: false, alt: "x to the power the fraction 3 over 4")[$x^(frac(3, 4))$] means \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "x to the power the fraction 3 over 4")[$x^(frac(3, 4))$] means the fourth root of #math.equation(block: false, alt: "x")[$x$] cubed ] #math.equation(block: false, alt: "x to the power 3 / 4")[$x^(3 / 4)$] means + the fourth root of #math.equation(block: false, alt: "x")[$x$] cubed + the cube root of #math.equation(block: false, alt: "x")[$x$] to the fourth + the 3/4 root of #math.equation(block: false, alt: "x")[$x$] + the cube root of the fourth root of #math.equation(block: false, alt: "x")[$x$] To compute #math.equation(block: false, alt: "a to the power m / n")[$a^(m / n)$], we can compute the #math.equation(block: false, alt: "n")[$n$]th root first, or the #math.equation(block: false, alt: "m")[$m$]th power, whichever is easier. For example, #math.equation(block: true, alt: "8 to the power 2 / 3 equals open parenthesis 8 squared close parenthesis to the power 1 / 3 equals 64 to the power 1 / 3 equals 4")[$8^(2 / 3) = attach(( 8^(2) ), t: 1 / 3) = 64^(1 / 3) = 4$] or #math.equation(block: true, alt: "8 to the power 2 / 3 equals open parenthesis 8 to the power 1 / 3 close parenthesis squared equals 2 squared equals 4")[$8^(2 / 3) = attach(( 8^(1 / 3) ), t: 2) = 2^(2) = 4$] #examplebox("Example 1")[][ + #math.equation(block: true, alt: "81 to the power 3 / 4, equals open parenthesis 81 to the power 1 / 4 close parenthesis cubed; equals 3 cubed equals 27")[$81^(3 / 4) & = attach(( 81^(1 / 4) ), t: 3) \ & = 3^(3) = 27$] + #math.equation(block: true, alt: "minus 27 to the power 5 / 3, equals minus open parenthesis 27 to the power 1 / 3 close parenthesis to the power 5; equals minus 3 to the power 5 equals minus 243")[$− 27^(5 / 3) & = − attach(( 27^(1 / 3) ), t: 5) \ & = − 3^(5) = − 243$] + #math.equation(block: true, alt: "27 to the power minus 2 / 3, equals the fraction 1 over open parenthesis 27 to the power 1 / 3 close parenthesis squared; equals the fraction 1 over 3 squared equals the fraction 1 over 9")[$27^(− 2 / 3) & = frac(1, attach(( 27^(1 / 3) ), t: 2)) \ & = frac(1, 3^(2)) = frac(1, 9)$] + #math.equation(block: true, alt: "5 to the power 3 / 2, equals open parenthesis 5 to the power 1 / 2 close parenthesis cubed; approximately equals open parenthesis 2.236 close parenthesis cubed approximately equals 11.180")[$5^(3 / 2) & = attach(( 5^(1 / 2) ), t: 3) \ & ≈ ( 2.236 )^(3) ≈ 11.180$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ You can verify all the calculations in Example on your calculator. For example, to evaluate #math.equation(block: false, alt: "81 to the power 3 / 4")[$81^(3 / 4)$], key in 81 ^ ( 3 ÷ 4 ) ENTER or simply 81 ^ 0.75 ENTER ] Evaluate each power. + #math.equation(block: false, alt: "32 to the power minus 3 / 5 equals")[$32^(− 3 / 5) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "minus 81 to the power 1.25 equals")[$− 81^(1.25) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] + #math.equation(block: false, alt: "minus 243")[$− 243$] ] Evaluate each power. + #math.equation(block: false, alt: "32 to the power minus 3 / 5")[$32^(− 3 / 5)$] + #math.equation(block: false, alt: "minus 81 to the power 1.25")[$− 81^(1.25)$] #solutionbox[ + #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] + #math.equation(block: false, alt: "minus 243")[$− 243$] ] Explain how to evaluate #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power the fraction minus 3 over 4")[$f ( x ) = x^(frac(− 3, 4))$] for #math.equation(block: false, alt: "x equals 625")[$x = 625$] by hand. \_\_\_\_\_ Explain how to evaluate #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power minus 3 / 4")[$f ( x ) = x^(− 3 / 4)$] for #math.equation(block: false, alt: "x equals 625")[$x = 625$] by hand. === Power Functions The graphs of power functions #math.equation(block: false, alt: "y equals x to the power m / n")[$y = x^(m / n)$], where #math.equation(block: false, alt: "m / n")[$m / n$] is positive, are all increasing for #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$]. If #math.equation(block: false, alt: "m / n greater than 1")[$m / n > 1$], the graph is concave up. If #math.equation(block: false, alt: "0 less than m / n less than 1")[$0 < m / n < 1$], the graph is concave down. Some examples are shown below. #figure(figph[graphs of power functions], alt: "graphs of power functions", caption: none) Perhaps the single most useful piece of information a scientist can have about an animal is its metabolic rate. The metabolic rate is the amount of energy the animal uses per unit of time for its usual activities, including locomotion, growth, and reproduction. The basal metabolic rate, or BMR, sometimes called the resting metabolic rate, is the minimum amount of energy the animal can expend in order to survive. #examplebox("Example 2")[][ A revised form of Kleiber's rule states that the basal metabolic rate for many groups of animals is given by #math.equation(block: true, alt: "B open parenthesis m close parenthesis equals 70 m to the power 0.75")[$B ( m ) = 70 m^(0.75)$] where #math.equation(block: false, alt: "m")[$m$] is the mass of the animal in kilograms and the BMR is measured in kilocalories per day. + Calculate the BMR for various animals whose masses are given in the table.#figure(table( columns: 8, align: left, inset: 6pt, table.header([Animal], [Bat], [Squirrel], [Raccoon], [Lynx], [Human], [Moose], [Rhinoceros]), [Weight (kg)], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "360")[$360$]], [#math.equation(block: false, alt: "3500")[$3500$]], [BMR (kcal/day)], [], [], [], [], [], [], [], )) + Sketch a graph of Kleiber’s rule for #math.equation(block: false, alt: "0 less than m less than or equal to 400")[$0 < m ≤ 400$]. + Do larger species eat more or less, relative to their body mass, than smaller ones? #solutionbox[ + We evaluate the function for the values of #math.equation(block: false, alt: "m")[$m$] given. For example, to calculate the BMR of a bat, we compute #math.equation(block: true, alt: "B open parenthesis 0.1 close parenthesis equals 70 open parenthesis 0.1 close parenthesis to the power 0.75 equals 12.4")[$B ( 0.1 ) = 70 ( 0.1 )^(0.75) = 12.4$] A bat expends, and hence must consume, at least #math.equation(block: false, alt: "12")[$12$] kilocalories per day. We evaluate the function to complete the rest of the table.#figure(table( columns: 8, align: left, inset: 6pt, table.header([Animal], [Bat], [Squirrel], [Raccoon], [Lynx], [Human], [Moose], [Rhinoceros]), [Weight (kg)], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "360")[$360$]], [#math.equation(block: false, alt: "3500")[$3500$]], [BMR (kcal/day)], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "333")[$333$]], [#math.equation(block: false, alt: "897")[$897$]], [#math.equation(block: false, alt: "1694")[$1694$]], [#math.equation(block: false, alt: "5785")[$5785$]], [#math.equation(block: false, alt: "31 , 853")[$31 , 853$]], )) + We plot the data from the table to obtain the graph below.#figure(figph[graph of BMR vs mass], alt: "graph of BMR vs mass", caption: none) + If energy consumption were proportional to body weight, the graph would be a straight line. But because the exponent in Kleiber’s rule, #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$], is less than #math.equation(block: false, alt: "1")[$1$], the graph is concave down, or bends downward. Therefore, larger species eat less than smaller ones, relative to their body weight. ] ] + Complete the table of values for the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power minus 3 / 4")[$f ( x ) = x^(− 3 / 4)$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) + Sketch the graph of the function. #solutionbox[ + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "5.623")[$5.623$]], [#math.equation(block: false, alt: "3.344")[$3.344$]], [#math.equation(block: false, alt: "1.682")[$1.682$]], [#math.equation(block: false, alt: "1")[$1$]], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0.595")[$0.595$]], [#math.equation(block: false, alt: "0.299")[$0.299$]], [#math.equation(block: false, alt: "0.210")[$0.210$]], [#math.equation(block: false, alt: "0.178")[$0.178$]], )) + A graph is below. ] #figure(figph[power function], alt: "power function", caption: none) + Complete the table of values for the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power minus 3 / 4")[$f ( x ) = x^(− 3 / 4)$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graph of the function. #solutionbox[ + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "5.623")[$5.623$]], [#math.equation(block: false, alt: "3.344")[$3.344$]], [#math.equation(block: false, alt: "1.682")[$1.682$]], [#math.equation(block: false, alt: "1")[$1$]], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0.595")[$0.595$]], [#math.equation(block: false, alt: "0.299")[$0.299$]], [#math.equation(block: false, alt: "0.210")[$0.210$]], [#math.equation(block: false, alt: "0.178")[$0.178$]], )) + #figure(figph[power function], alt: "power function", caption: none) ] Describe the concavity of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power the fraction a over b")[$f ( x ) = x^(frac(a, b))$], where #math.equation(block: false, alt: "the fraction a over b greater than 0")[$display(frac(a, b)) > 0$]. \_\_\_\_\_ #solutionbox[ The graph is concave up if #math.equation(block: false, alt: "the fraction a over b greater than 1")[$display(frac(a, b)) > 1$], and concave down if #math.equation(block: false, alt: "the fraction a over b less than 1")[$display(frac(a, b)) < 1$]. ] Describe the concavity of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power a / b")[$f ( x ) = x^(a / b)$], where #math.equation(block: false, alt: "the fraction a over b greater than 0")[$display(frac(a, b)) > 0$]. + It is concave up. + It is concave down. + It is concave up if #math.equation(block: false, alt: "the fraction a over b greater than 1")[$display(frac(a, b)) > 1$], and concave down if #math.equation(block: false, alt: "the fraction a over b less than 1")[$display(frac(a, b)) < 1$]. + It is concave up if #math.equation(block: false, alt: "the fraction a over b less than 1")[$display(frac(a, b)) < 1$], and concave down if #math.equation(block: false, alt: "the fraction a over b greater than 1")[$display(frac(a, b)) > 1$]. === More about Scaling In Example we saw that large animals eat less than smaller ones, relative to their body weight. This is because the scaling exponent in Kleiber's rule is less than #math.equation(block: false, alt: "1")[$1$]. For example, let #math.equation(block: false, alt: "s")[$s$] represent the mass of a squirrel. The mass of a moose is then #math.equation(block: false, alt: "600 s")[$600 s$], and its metabolic rate is #math.equation(block: true, alt: "B open parenthesis 600 s close parenthesis, equals 70 open parenthesis 600 s close parenthesis to the power 0.75; equals 600 to the power 0.75 times 70 s to the power 0.75 equals 121 B open parenthesis s close parenthesis")[$B ( 600 s ) & = 70 ( 600 s )^(0.75) \ & = 600^(0.75) ⋅ 70 s^(0.75) = 121 B ( s )$] or #math.equation(block: false, alt: "121")[$121$] times the metabolic rate of the squirrel. Metabolic rate scales as #math.equation(block: false, alt: "k to the power 0.75")[$k^(0.75)$], compared to the mass of the animal. In a famous experiment in the 1960s, an elephant was given LSD. The dose was determined from a previous experiment in which a #math.equation(block: false, alt: "2.6")[$2.6$]-kg cat was given #math.equation(block: false, alt: "0.26")[$0.26$] gram of LSD. Because the elephant weighed #math.equation(block: false, alt: "2970")[$2970$] kg, the experimenters used a direct proportion to calculate the dose for the elephant: #math.equation(block: true, alt: "the fraction 0.26 g over 2.6 kg equals the fraction x g over 2970 kg")[$frac(0.26 " g", 2.6 " kg") = frac(x " g", 2970 " kg")$] and arrived at the figure #math.equation(block: false, alt: "297")[$297$] g of LSD. Unfortunately, the elephant did not survive the experiment. #examplebox("Example 3")[][ Use Kleiber's rule and the dosage for a cat to estimate the corresponding dose for an elephant. #solutionbox[ If the experimenters had taken into account the scaling exponent of #math.equation(block: false, alt: "0.75")[$0.75$] in metabolic rate, they would have used a smaller dose. Because the elephant weighs #math.equation(block: false, alt: "the fraction 2970 over 2.6")[$display(frac(2970, 2.6))$] , or about #math.equation(block: false, alt: "1142")[$1142$] times as much as the cat, the dose would be #math.equation(block: false, alt: "1142 to the power 0.75 equals 196")[$1142^(0.75) = 196$] times the dosage for a cat, or about #math.equation(block: false, alt: "51")[$51$] grams. ] ] A human being weighs about #math.equation(block: false, alt: "70")[$70$] kg, and #math.equation(block: false, alt: "0.2")[$0.2$] mg of LSD is enough to induce severe psychotic symptoms. Use these data and Kleiber's rule to predict what dosage would produce a similar effect in an elephant. About \_\_\_\_\_ mg #solutionbox[ About #math.equation(block: false, alt: "3.3")[$3.3$] mg ] A human being weighs about #math.equation(block: false, alt: "70")[$70$] kg, and #math.equation(block: false, alt: "0.2")[$0.2$] mg of LSD is enough to induce severe psychotic symptoms. Use these data and Kleiber's rule to predict what dosage would produce a similar effect in an elephant. #solutionbox[ #math.equation(block: false, alt: "open parenthesis the fraction 2970 over 70 close parenthesis to the power 0.75 open parenthesis 0.2 close parenthesis approximately equals 3.3")[$attach(( display(frac(2970, 70)) ), t: 0.75) ( 0.2 ) ≈ 3.3$] mg ] === Radical Notation Because #math.equation(block: false, alt: "a to the power 1 / n equals the nth root of a")[$a^(1 / n) = root(n, a)$], we can write any power with a fractional exponent in radical form as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Rational Exponents and Radicals] #math.equation(block: true, alt: "a to the power m / n equals the nth root of a to the power m equals open parenthesis the nth root of a close parenthesis to the power m")[$a^(m / n) = root(n, a^(m)) = attach(( root(n, a) ), t: m)$] ] #examplebox("Example 4")[][ + #math.equation(block: false, alt: "125 to the power 4 / 3 equals the cube root of 125 to the power 4 or open parenthesis the cube root of 125 close parenthesis to the power 4")[$125^(4 / 3) = root(3, 125^(4)) " or " attach(( root(3, 125) ), t: 4)$] + #math.equation(block: false, alt: "x to the power 0.4 equals x to the power 2 / 5 equals the 5th root of x squared")[$x^(0.4) = x^(2 / 5) = root(5, x^(2))$] + #math.equation(block: false, alt: "6 w to the power minus 3 / 4 equals the fraction 6 over the 4th root of w cubed")[$6 w^(− 3 / 4) = display(frac(6, root(4, w^(3))))$] ] Write each expression in radical notation. + #math.equation(block: false, alt: "5 t to the power 1.25")[$5 t^(1.25)$] #linebreak() \_\_\_\_\_ + #math.equation(block: false, alt: "3 m to the power minus 5 / 3")[$3 m^(− 5 / 3)$] #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "5 t to the power 1.25 equals 5 the 4th root of t to the power 5")[$5 t^(1.25) = 5 root(4, t^(5))$] + #math.equation(block: false, alt: "3 m to the power minus 5 / 3 equals the fraction 3 over the cube root of m to the power 5")[$3 m^(− 5 / 3) = display(frac(3, root(3, m^(5))))$] ] Write each expression in radical notation. + #math.equation(block: false, alt: "5 t to the power 1.25")[$5 t^(1.25)$] + #math.equation(block: false, alt: "3 m to the power minus 5 / 3")[$3 m^(− 5 / 3)$] #solutionbox[ + #math.equation(block: false, alt: "5 t to the power 1.25 equals 5 the 4th root of t to the power 5")[$5 t^(1.25) = 5 root(4, t^(5))$] + #math.equation(block: false, alt: "3 m to the power minus 5 / 3 equals the fraction 3 over the cube root of m to the power 5")[$3 m^(− 5 / 3) = display(frac(3, root(3, m^(5))))$] ] The notation #math.equation(block: false, alt: "a to the power 0.6")[$a^(0.6)$] means \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "a to the power 0.6 equals a to the power the fraction 3 over 5")[$a^(0.6) = a^(frac(3, 5))$] ] The notation #math.equation(block: false, alt: "a to the power 0.6")[$a^(0.6)$] means + #math.equation(block: false, alt: "the 6th root of a")[$root(6, a)$] + #math.equation(block: false, alt: "the fraction 1 over a to the power 6")[$display(frac(1, a^(6)))$] + #math.equation(block: false, alt: "a to the power 1 / 6")[$a^(1 / 6)$] + #math.equation(block: false, alt: "a to the power 3 / 5")[$a^(3 / 5)$] Usually, we will want to convert from radical notation to fractional exponents, since exponential notation is easier to use. #examplebox("Example 5")[][ + #math.equation(block: false, alt: "the square root of x to the power 5 equals x to the power 5 / 2")[$sqrt(x^(5)) = x^(5 / 2)$] + #math.equation(block: false, alt: "5 the 4th root of p cubed equals 5 p to the power 3 / 4")[$5 root(4, p^(3)) = 5 p^(3 / 4)$] + #math.equation(block: false, alt: "the fraction 3 over the 5th root of t squared equals 3 t to the power minus 2 / 5")[$display(frac(3, root(5, t^(2)))) = 3 t^(− 2 / 5)$] + #math.equation(block: false, alt: "the cube root of 2 y squared equals open parenthesis 2 y squared close parenthesis to the power 1 / 3 equals 2 to the power 1 / 3 y to the power 2 / 3")[$root(3, 2 y^(2)) = attach(( 2 y^(2) ), t: 1 / 3) = 2^(1 / 3) y^(2 / 3)$] ] Convert to exponential notation. + #math.equation(block: false, alt: "the cube root of 6 w squared equals")[$root(3, 6 w^(2)) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "the 4th root of the fraction v cubed over s to the power 5 equals")[$root(4, display(frac(v^(3), s^(5)))) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "6 to the power 1 / 3 w to the power 2 / 3")[$6^(1 / 3) w^(2 / 3)$] + #math.equation(block: false, alt: "v to the power 3 / 4 s to the power minus 5 / 4")[$v^(3 / 4) s^(− 5 / 4)$] ] Convert to exponential notation. + #math.equation(block: false, alt: "the cube root of 6 w squared")[$root(3, 6 w^(2))$] + #math.equation(block: false, alt: "the 4th root of the fraction v cubed over s to the power 5")[$root(4, display(frac(v^(3), s^(5))))$] #solutionbox[ + #math.equation(block: false, alt: "6 to the power 1 / 3 w to the power 2 / 3")[$6^(1 / 3) w^(2 / 3)$] + #math.equation(block: false, alt: "v to the power 3 / 4 s to the power minus 5 / 4")[$v^(3 / 4) s^(− 5 / 4)$] ] === Operations with Rational Exponents Powers with rational exponents—positive, negative, or zero—obey the laws of exponents, which we discussed in Variation. You may want to review those laws before studying the following examples. #examplebox("Example 6")[][ + #math.equation(block: true, alt: "the fraction 7 to the power 0.75 over 7 to the power 0.5, equals 7 to the power 0.75 minus 0.5 equals 7 to the power 0.25, Apply the second law of exponents.")[$frac(7^(0.75), 7^(0.5)) & = 7^(0.75 − 0.5) = 7^(0.25) & & "Apply the second law of exponents."$] + #math.equation(block: true, alt: "v times v to the power minus 2 / 3, equals v to the power 1 plus open parenthesis minus 2 / 3 close parenthesis, Apply the first law of exponents.; equals v to the power 1 / 3")[$v ⋅ v^(− 2 / 3) & = v^(1 + ( − 2 / 3 )) & & "Apply the first law of exponents." \ & = v^(1 / 3)$] + #math.equation(block: true, alt: "open parenthesis x to the power 8 close parenthesis to the power 0.5, equals x to the power 8 open parenthesis 0.5 close parenthesis equals x to the power 4, Apply the third law of exponents.")[$attach(( x^(8) ), t: 0.5) & = x^(8 ( 0.5 )) = x^(4) & & "Apply the third law of exponents."$] + #math.equation(block: true, alt: "the fraction open parenthesis 5 to the power 1 / 2 y squared close parenthesis squared over open parenthesis 5 to the power 2 / 3 y close parenthesis cubed, equals the fraction 5 y to the power 4 over 5 squared y cubed, Apply the fourth law of exponents.; equals the fraction y to the power 4 minus 3 over 5 to the power 2 minus 1 equals the fraction y over 5, Apply the second law of exponents.")[$frac(attach(( 5^(1 / 2) y^(2) ), t: 2), attach(( 5^(2 / 3) y ), t: 3)) & = frac(5 y^(4), 5^(2) y^(3)) & & "Apply the fourth law of exponents." \ & = frac(y^(4 − 3), 5^(2 − 1)) = frac(y, 5) & & "Apply the second law of exponents."$] ] Simplify by applying the laws of exponents. + #math.equation(block: false, alt: "x to the power 1 / 3 open parenthesis x plus x to the power 2 / 3 close parenthesis equals")[$x^(1 / 3) ( x + x^(2 / 3) ) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "the fraction n to the power 9 / 4 over 4 n to the power 3 / 4 equals")[$display(frac(n^(9 / 4), 4 n^(3 / 4))) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "x to the power 4 / 3 plus x")[$x^(4 / 3) + x$] + #math.equation(block: false, alt: "the fraction n to the power 3 / 2 over 4")[$display(frac(n^(3 / 2), 4))$] ] Simplify by applying the laws of exponents. + #math.equation(block: false, alt: "x to the power 1 / 3 open parenthesis x plus x to the power 2 / 3 close parenthesis")[$x^(1 / 3) ( x + x^(2 / 3) )$] + #math.equation(block: false, alt: "the fraction n to the power 9 / 4 over 4 n to the power 3 / 4")[$display(frac(n^(9 / 4), 4 n^(3 / 4)))$] #solutionbox[ + #math.equation(block: false, alt: "x to the power 4 / 3 plus x")[$x^(4 / 3) + x$] + #math.equation(block: false, alt: "the fraction n to the power 3 / 2 over 4")[$display(frac(n^(3 / 2), 4))$] ] Which of the following is the correct way to evaluate #math.equation(block: false, alt: "15 to the power the fraction 3 over 5")[$15^(frac(3, 5))$] on a calculator? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "15 ∧ open parenthesis 3 divided by 5 close parenthesis")[$15 ∧ ( 3 ÷ 5 )$] ] Which of the following is the solution to #math.equation(block: false, alt: "x to the power minus 3 / 5 equals 8")[$x^(− 3 / 5) = 8$] ? + #math.equation(block: false, alt: "8 to the power minus 2 / 5")[$8^(− 2 / 5)$] + #math.equation(block: false, alt: "8 to the power 3 / 5")[$8^(3 / 5)$] + #math.equation(block: false, alt: "8 to the power minus 5 / 3")[$8^(− 5 / 3)$] + #math.equation(block: false, alt: "8 to the power 5 / 3")[$8^(5 / 3)$] === Solving Equations According to the third law of exponents, when we raise a power to another power, we multiply the exponents together. In particular, if the two exponents are reciprocals, then their product is #math.equation(block: false, alt: "1")[$1$]. For example, #math.equation(block: true, alt: "open parenthesis x to the power 2 / 3 close parenthesis to the power 3 / 2 equals x to the power open parenthesis 2 / 3 close parenthesis open parenthesis 3 / 2 close parenthesis equals x to the power 1 equals x")[$attach(( x^(2 / 3) ), t: 3 / 2) = x^(( 2 / 3 ) ( 3 / 2 )) = x^(1) = x$] This observation can help us to solve equations involving fractional exponents. For instance, to solve the equation #math.equation(block: true, alt: "x to the power 2 / 3 equals 4")[$x^(2 / 3) = 4$] we raise both sides of the equation to the reciprocal power, #math.equation(block: false, alt: "3 / 2")[$3 / 2$]. This gives us #math.equation(block: true, alt: "open parenthesis x to the power 2 / 3 close parenthesis to the power 3 / 2, equals 4 to the power 3 / 2; x, equals 8")[$attach(( x^(2 / 3) ), t: 3 / 2) & = 4^(3 / 2) \ x & = 8$] The solution is #math.equation(block: false, alt: "8")[$8$]. #examplebox("Example 7")[][ Solve #math.equation(block: false, alt: "open parenthesis 2 x plus 1 close parenthesis to the power 3 / 4 equals 27")[$" " " " ( 2 x + 1 )^(3 / 4) = 27$] #solutionbox[ We raise both sides of the equation to the reciprocal power, #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$]. #math.equation(block: true, alt: "open bracket open parenthesis 2 x plus 1 close parenthesis to the power 3 / 4 close bracket to the power 4 / 3, equals 27 to the power 4 / 3, Apply the third law of exponents.; 2 x plus 1, equals 81, Solve as usual.; x, equals 40")[$attach([ ( 2 x + 1 )^(3 / 4) ], t: 4 / 3) & = 27^(4 / 3) & & "Apply the third law of exponents." \ 2 x + 1 & = 81 & & "Solve as usual." \ x & = 40$] ] ] To solve the equation #math.equation(block: false, alt: "x to the power the fraction 2 over 3 equals k")[$x^(frac(2, 3)) = k$] we should \_\_\_\_\_ #solutionbox[ Raise both sides to the reciprocal of the exponent ] To solve the equation #math.equation(block: false, alt: "x to the power 2 / 3 equals k")[$x^(2 / 3) = k$] we should + Raise both sides to the negative of the exponent. + Divide both sides by the exponent. + Raise both sides to the reciprocal of the exponent. + Raise the right side to the given exponent. Solve the equation #math.equation(block: false, alt: "3.2 z to the power 0.6 minus 9.7 equals 8.7")[$3.2 z^(0.6) − 9.7 = 8.7$]. Round your answer to two decimal places. Answer: \_\_\_\_\_ #math.equation(block: true, alt: "Isolate the power.")[$"Isolate the power."$] #math.equation(block: true, alt: "Raise both sides to the reciprocal power .")[$"Raise both sides to the reciprocal power" .$] #solutionbox[ #math.equation(block: true, alt: "18.45")[$18.45$] ] Solve the equation #math.equation(block: false, alt: "3.2 z to the power 0.6 minus 9.7 equals 8.7")[$" " 3.2 z^(0.6) − 9.7 = 8.7 " "$]. Round your answer to two decimal places. #solutionbox[ Isolate the power, then raise both sides to the reciprocal power to get #math.equation(block: false, alt: "18.45")[$18.45$] ] Explain why #math.equation(block: false, alt: "x the square root of x equals x to the power 1.5")[$x sqrt(x) = x^(1.5)$]. \_\_\_\_\_ Explain why #math.equation(block: false, alt: "x the square root of x equals x to the power 1.5")[$x sqrt(x) = x^(1.5)$]. #notebox("Investigation", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vampire Bats] Small animals such as bats cannot survive for long without eating. The graph below shows how the weight, #math.equation(block: false, alt: "W")[$W$], of a typical vampire bat decreases over time until its next meal, until the bat reaches the point of starvation. The curve is the graph of the function #math.equation(block: true, alt: "W open parenthesis h close parenthesis equals 130.25 h to the power minus 0.126")[$W ( h ) = 130.25 h^(− 0.126)$] where #math.equation(block: false, alt: "h")[$h$] is the number of hours since the bat’s most recent meal. (Source: Wilkinson, 1984) #figure(figph[graph of bat weight], alt: "graph of bat weight", caption: none) + Use the graph to estimate answers to the following questions: How long can the bat survive after eating until its next meal? What is the bat’s weight at the point of starvation? + Use the formula for #math.equation(block: false, alt: "W open parenthesis h close parenthesis")[$W ( h )$] to verify your answers. + Write and solve an equation to answer the question: When the bat's weight has dropped to #math.equation(block: false, alt: "90")[$90$] grams, how long can it survive before eating again? + Complete the table showing the number of hours since the bat last ate when its weight has dropped to the given values.#figure(table( columns: 5, align: left, inset: 6pt, table.header([Weight (grams)], [#math.equation(block: false, alt: "97.5")[$97.5$]], [#math.equation(block: false, alt: "92.5")[$92.5$]], [#math.equation(block: false, alt: "85")[$85$]], [#math.equation(block: false, alt: "80")[$80$]]), [Hours since eating], [], [], [], [], [Point on graph], [#math.equation(block: false, alt: "A")[$A$]], [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "D")[$D$]], )) + Compute the slope of the line segments from point #math.equation(block: false, alt: "A")[$A$] to point #math.equation(block: false, alt: "B")[$B$], and from point #math.equation(block: false, alt: "C")[$C$] to point #math.equation(block: false, alt: "D")[$D$]. Include units in your answers.#figure(figph[graph of bat weight], alt: "graph of bat weight", caption: none) + What happens to the slope of the curve as #math.equation(block: false, alt: "h")[$h$] increases? What does this tell you about the concavity of the curve? + Suppose a bat that weighs #math.equation(block: false, alt: "80")[$80$] grams consumes #math.equation(block: false, alt: "5")[$5$] grams of blood. How many hours of life does it gain? Suppose a bat that weighs #math.equation(block: false, alt: "97.5")[$97.5$] grams gives up a meal of #math.equation(block: false, alt: "5")[$5$] grams of blood. How many hours of life does it forfeit? + Vampire bats sometimes donate blood (through regurgitation) to other bats that are close to starvation. Suppose a bat at point #math.equation(block: false, alt: "A")[$A$] on the curve donates #math.equation(block: false, alt: "5")[$5$] grams of blood to a bat at point #math.equation(block: false, alt: "D")[$D$]. Explain why this strategy is effective for the survival of the bat community. ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Rational exponent ==== CONCEPTS + Rational exponents: #math.equation(block: false, alt: "a to the power m / n equals open parenthesis a to the power 1 / n close parenthesis to the power m equals open parenthesis a to the power m close parenthesis to the power 1 / n , a greater than 0 , n not equal to 0.")[$a^(m / n) = attach(( a^(1 / n) ), t: m) = attach(( a^(m) ), t: 1 / n) , " " " " " " a > 0 , " " " " n ≠ 0.$] + To compute #math.equation(block: false, alt: "a to the power m / n")[$a^(m / n)$], we can compute the #math.equation(block: false, alt: "n")[$n$]th root first, or the #math.equation(block: false, alt: "m")[$m$]th power, whichever is easier. + The graphs of power functions #math.equation(block: false, alt: "y equals x to the power m / n")[$y = x^(m / n)$], where #math.equation(block: false, alt: "m / n")[$m / n$] is positive, are all increasing for #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$]. If #math.equation(block: false, alt: "m / n greater than 1")[$m / n > 1$], the graph is concave up. If #math.equation(block: false, alt: "0 less than m / n less than 1")[$0 < m / n < 1$], the graph is concave down. + Radical notation: #math.equation(block: false, alt: "a to the power m / n equals the nth root of a to the power m equals open parenthesis the nth root of a close parenthesis to the power m")[$a^(m / n) = root(n, a^(m)) = attach(( root(n, a) ), t: m)$]. + Powers with rational exponents—positive, negative, or zero—obey the laws of exponents. + To solve the equation #math.equation(block: false, alt: "x to the power m / n equals k")[$x^(m / n) = k$], we raise both sides to the power #math.equation(block: false, alt: "n / m")[$n / m$]. ==== STUDY QUESTIONS + What does the notation #math.equation(block: false, alt: "a to the power 0.98")[$a^(0.98)$] mean? + Explain how to evaluate the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power minus 3 / 4")[$f ( x ) = x^(− 3 / 4)$] for #math.equation(block: false, alt: "x equals 625")[$x = 625$], without using a calculator. + Explain why #math.equation(block: false, alt: "x the square root of x equals x to the power 1.5")[$x sqrt(x) = x^(1.5)$]. + What is the first step in solving the equation #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis to the power minus 5 / 2 equals 1.8")[$( x − 2 )^(− 5 / 2) = 1.8$]? + If the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power a / b")[$f ( x ) = x^(a / b)$] is concave down, and #math.equation(block: false, alt: "a / b greater than 0")[$a / b > 0$], what else can you say about #math.equation(block: false, alt: "a / b")[$a / b$]? ==== SKILLS Practice each skill in the Homework problems listed. + Simplify and evaluate powers with rational exponents: \#1–4, 13–18 + Graph power functions with rational exponents: \#19–22 + Solve radical equations: \#23–38, 59 and 60 + Analyze power functions with rational exponents: \#23–36 + Simplify expressions using the laws of exponents: \#37–44, 57–70 + Solve equations involving rational exponents: \#45–56 === Homework 3.4 For the problems in Homework 3.4, assume that all variables represent positive numbers. For Problems 1-4, evaluate each power. + #math.equation(block: false, alt: "81 to the power 3 / 4")[$81^(3 / 4)$] + #math.equation(block: false, alt: "125 to the power 2 / 3")[$125^(2 / 3)$] + #math.equation(block: false, alt: "625 to the power 0.75")[$625^(0.75)$] + #math.equation(block: false, alt: "27")[$27$] + #math.equation(block: false, alt: "25")[$25$] + #math.equation(block: false, alt: "125")[$125$] + #math.equation(block: false, alt: "minus 8 to the power 2 / 3")[$− 8^(2 / 3)$] + #math.equation(block: false, alt: "minus 64 to the power 2 / 3")[$− 64^(2 / 3)$] + #math.equation(block: false, alt: "243 to the power 0.4")[$243^(0.4)$] + #math.equation(block: false, alt: "16 to the power minus 3 / 2")[$16^(− 3 / 2)$] + #math.equation(block: false, alt: "8 to the power minus 4 / 3")[$8^(− 4 / 3)$] + #math.equation(block: false, alt: "32 to the power minus 1.6")[$32^(− 1.6)$] + #math.equation(block: false, alt: "the fraction 1 over 64")[$display(frac(1, 64))$] + #math.equation(block: false, alt: "the fraction 1 over 16")[$display(frac(1, 16))$] + #math.equation(block: false, alt: "the fraction 1 over 256")[$display(frac(1, 256))$] + #math.equation(block: false, alt: "minus 125 to the power minus 4 / 3")[$− 125^(− 4 / 3)$] + #math.equation(block: false, alt: "minus 32 to the power minus 3 / 5")[$− 32^(− 3 / 5)$] + #math.equation(block: false, alt: "100 to the power minus 2.5")[$100^(− 2.5)$] For Problems 5–8, write each power in radical form. + #math.equation(block: false, alt: "x to the power 4 / 5")[$x^(4 / 5)$] + #math.equation(block: false, alt: "b to the power minus 5 / 6")[$b^(− 5 / 6)$] + #math.equation(block: false, alt: "open parenthesis p q close parenthesis to the power minus 2 / 3")[$( p q )^(− 2 / 3)$] + #math.equation(block: false, alt: "the 5th root of x to the power 4")[$root(5, x^(4))$] + #math.equation(block: false, alt: "the fraction 1 over the 6th root of b to the power 5")[$display(frac(1, root(6, b^(5))))$] + #math.equation(block: false, alt: "the fraction 1 over the cube root of open parenthesis p q close parenthesis squared")[$display(frac(1, root(3, ( p q )^(2))))$] + #math.equation(block: false, alt: "y to the power 3 / 4")[$y^(3 / 4)$] + #math.equation(block: false, alt: "a to the power minus 2 / 7")[$a^(− 2 / 7)$] + #math.equation(block: false, alt: "open parenthesis s t close parenthesis to the power minus 3 / 5")[$( s t )^(− 3 / 5)$] + #math.equation(block: false, alt: "3 x to the power 0.4")[$3 x^(0.4)$] + #math.equation(block: false, alt: "4 z to the power minus 4 / 3")[$4 z^(− 4 / 3)$] + #math.equation(block: false, alt: "minus 2 x to the power 0.25 y to the power 0.75")[$− 2 x^(0.25) y^(0.75)$] + #math.equation(block: false, alt: "3 the 5th root of x squared")[$3 root(5, x^(2))$] + #math.equation(block: false, alt: "the fraction 4 over the cube root of z to the power 4")[$display(frac(4, root(3, z^(4))))$] + #math.equation(block: false, alt: "minus 2 the 4th root of x y cubed")[$− 2 root(4, x y^(3))$] + #math.equation(block: false, alt: "5 y to the power 2 / 3")[$5 y^(2 / 3)$] + #math.equation(block: false, alt: "6 w to the power minus 1.5")[$6 w^(− 1.5)$] + #math.equation(block: false, alt: "minus 3 x to the power 0.4 y to the power 0.6")[$− 3 x^(0.4) y^(0.6)$] For Problems 9–12, write each expression with fractional exponents. + #math.equation(block: false, alt: "the cube root of x squared")[$root(3, x^(2))$] + #math.equation(block: false, alt: "2 the 5th root of a b cubed")[$2 root(5, a b^(3))$] + #math.equation(block: false, alt: "the fraction minus 4 m over the 6th root of p to the power 7")[$display(frac(− 4 m, root(6, p^(7))))$] + #math.equation(block: false, alt: "x to the power 2 / 3")[$x^(2 / 3)$] + #math.equation(block: false, alt: "2 a to the power 1 / 5 b to the power 3 / 5")[$2 a^(1 / 5) b^(3 / 5)$] + #math.equation(block: false, alt: "minus 4 m p to the power minus 7 / 6")[$− 4 m p^(− 7 / 6)$] + #math.equation(block: false, alt: "the square root of y cubed")[$sqrt(y^(3))$] + #math.equation(block: false, alt: "6 the 5th root of open parenthesis a b close parenthesis cubed")[$6 root(5, ( a b )^(3))$] + #math.equation(block: false, alt: "the fraction minus 2 n over the 8th root of q to the power 11")[$display(frac(− 2 n, root(8, q^(11))))$] + #math.equation(block: false, alt: "the cube root of open parenthesis a b close parenthesis squared")[$root(3, ( a b )^(2))$] + #math.equation(block: false, alt: "the fraction 8 over the 4th root of x cubed")[$display(frac(8, root(4, x^(3))))$] + #math.equation(block: false, alt: "the fraction R over 3 the square root of T K to the power 5")[$display(frac(R, 3 sqrt(T K^(5))))$] + #math.equation(block: false, alt: "open parenthesis a b close parenthesis to the power 2 / 3")[$( a b )^(2 / 3)$] + #math.equation(block: false, alt: "8 x to the power minus 3 / 4")[$8 x^(− 3 / 4)$] + #math.equation(block: false, alt: "the fraction 1 over 3 R T to the power minus 1 / 2 K to the power minus 5 / 2")[$display(frac(1, 3)) R T^(− 1 / 2) K^(− 5 / 2)$] + #math.equation(block: false, alt: "the cube root of a b squared")[$root(3, a b^(2))$] + #math.equation(block: false, alt: "the fraction 5 over the cube root of y squared")[$display(frac(5, root(3, y^(2))))$] + #math.equation(block: false, alt: "the fraction S over 4 the square root of V H cubed")[$display(frac(S, 4 sqrt(V H^(3))))$] For Problems 13–16, evaluate each root without using a calculator. + #math.equation(block: false, alt: "the 5th root of 32 cubed")[$root(5, 32^(3))$] + #math.equation(block: false, alt: "minus the cube root of 27 to the power 4")[$− root(3, 27^(4))$] + #math.equation(block: false, alt: "the 4th root of 16 y to the power 12")[$root(4, 16 y^(12))$] + #math.equation(block: false, alt: "8")[$8$] + #math.equation(block: false, alt: "minus 81")[$− 81$] + #math.equation(block: false, alt: "2 y cubed")[$2 y^(3)$] + #math.equation(block: false, alt: "the 4th root of 16 to the power 5")[$root(4, 16^(5))$] + #math.equation(block: false, alt: "minus the cube root of 125 squared")[$− root(3, 125^(2))$] + #math.equation(block: false, alt: "the 5th root of 243 x to the power 10")[$root(5, 243 x^(10))$] + #math.equation(block: false, alt: "minus the square root of a to the power 8 b to the power 16")[$− sqrt(a^(8) b^(16))$] + #math.equation(block: false, alt: "the cube root of 8 x to the power 9 y to the power 27")[$root(3, 8 x^(9) y^(27))$] + #math.equation(block: false, alt: "minus the 4th root of 81 a to the power 8 b to the power 12")[$− root(4, 81 a^(8) b^(12))$] + #math.equation(block: false, alt: "minus a to the power 4 b to the power 8")[$− a^(4) b^(8)$] + #math.equation(block: false, alt: "2 x cubed y to the power 9")[$2 x^(3) y^(9)$] + #math.equation(block: false, alt: "minus 3 a squared b cubed")[$− 3 a^(2) b^(3)$] + #math.equation(block: false, alt: "minus the square root of a to the power 10 b to the power 36")[$− sqrt(a^(10) b^(36))$] + #math.equation(block: false, alt: "the cube root of 64 x to the power 6 y to the power 18")[$root(3, 64 x^(6) y^(18))$] + #math.equation(block: false, alt: "minus the 5th root of 32 x to the power 25 y to the power 5")[$− root(5, 32 x^(25) y^(5))$] For Problems 17–18, use a calculator to approximate each power or root to the nearest thousandth. + #math.equation(block: false, alt: "12 to the power 5 / 6")[$12^(5 / 6)$] + #math.equation(block: false, alt: "the cube root of 6 to the power 4")[$root(3, 6^(4))$] + #math.equation(block: false, alt: "37 to the power minus 2 / 3")[$37^(− 2 / 3)$] + #math.equation(block: false, alt: "4.7 to the power 2.3")[$4.7^(2.3)$] + #math.equation(block: false, alt: "7.931")[$7.931$] + #math.equation(block: false, alt: "10.903")[$10.903$] + #math.equation(block: false, alt: "0.090")[$0.090$] + #math.equation(block: false, alt: "35.142")[$35.142$] + #math.equation(block: false, alt: "20 to the power 5 / 4")[$20^(5 / 4)$] + #math.equation(block: false, alt: "the 5th root of 8 cubed")[$root(5, 8^(3))$] + #math.equation(block: false, alt: "128 to the power minus 3 / 4")[$128^(− 3 / 4)$] + #math.equation(block: false, alt: "16.1 to the power 0.29")[$16.1^(0.29)$] During a flu epidemic in a small town, health officials estimate that the number of people infected #math.equation(block: false, alt: "t")[$t$] days after the first case was discovered is given by #math.equation(block: true, alt: "I open parenthesis t close parenthesis equals 50 t to the power 3 / 5")[$I ( t ) = 50 t^(3 / 5)$] + Make a table of values for #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$] on the domain #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 20")[$0 ≤ t ≤ 20$]. What is the range of the function on that domain?#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) + How long will it be before #math.equation(block: false, alt: "300")[$300$] people are ill? + Graph the function #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$] and verify your answer to part (b) on your graph. + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$]], [#math.equation(block: false, alt: "131")[$131$]], [#math.equation(block: false, alt: "199")[$199$]], [#math.equation(block: false, alt: "254")[$254$]], [#math.equation(block: false, alt: "302")[$302$]], )) #linebreak() Range: #math.equation(block: false, alt: "open bracket 0 , 302 close bracket")[$[ 0 , 302 ]$] + #math.equation(block: false, alt: "approximately equals 19.812")[$≈ 19.812$] or about #math.equation(block: false, alt: "20")[$20$] days + #figure(figph[power function], alt: "power function", caption: none) The research division of an advertising firm estimates that the number of people who have seen their ads #math.equation(block: false, alt: "t")[$t$] days after the campaign begins is given by the function #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 2000 t to the power 5 / 4")[$N ( t ) = 2000 t^(5 / 4)$] + Make a table of values for #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$] on the domain #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 20")[$0 ≤ t ≤ 20$]. What is the range of the function on that domain?#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "14")[$14$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) + How long will it be before #math.equation(block: false, alt: "75 , 000")[$75 , 000$] people have seen the ads? + Graph the function #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$] and verify your answer to part (b) on your graph. In Problems 21–22, graph each set of power functions in the suggested window and compare the graphs. #math.equation(block: true, alt: "y sub 1 equals x , y sub 2 equals x to the power 5 / 4 , y sub 3 equals x to the power 3 / 2 , y sub 4 equals x squared , y sub 5 equals x to the power 5 / 2")[$y_(1) = x , " " " " y_(2) = x^(5 / 4) , " " " " y_(3) = x^(3 / 2) , " " " " y_(4) = x^(2) , " " " " y_(5) = x^(5 / 2)$] #math.equation(block: true, alt: "Xmin equals 0 , Xmax equals 6 , Ymin equals 0 , Ymax equals 10")[$"Xmin" = 0 , " " " " " " "Xmax" = 6 , " " " " " " "Ymin" = 0 , " " " " " " "Ymax" = 10$] All the graphs are increasing and concave up. For #math.equation(block: false, alt: "x greater than 1")[$x > 1$], each graph increases more quickly than the previous one. #math.equation(block: true, alt: "y sub 1 equals x to the power 2 / 5 , y sub 2 equals x to the power 1 / 2 , y sub 3 equals x to the power 2 / 3 , y sub 4 equals x to the power 3 / 4 , y sub 5 equals x")[$y_(1) = x^(2 / 5) , " " " " y_(2) = x^(1 / 2) , " " " " y_(3) = x^(2 / 3) , " " " " y_(4) = x^(3 / 4) , " " " " y_(5) = x$] #math.equation(block: true, alt: "Xmin equals 0 , Xmax equals 6 , Ymin equals 0 , Ymax equals 4")[$"Xmin" = 0 , " " " " " " "Xmax" = 6 , " " " " " " "Ymin" = 0 , " " " " " " "Ymax" = 4$] The #emph[surface to volume ratio] is important in studying how organisms grow and why animals of different sizes have different characteristics. + Write formulas for the volume, #math.equation(block: false, alt: "V")[$V$], and the surface area, #math.equation(block: false, alt: "A")[$A$], of a cube in terms of its length, #math.equation(block: false, alt: "L")[$L$]. + Express the length of the cube as a function of its volume. Express the length of the cube as a function of its surface area. + Express the surface area of the cube as a function of its volume. + Express the surface to volume ratio of a cube in terms of its length. What happens to the surface to volume ratio as #math.equation(block: false, alt: "L")[$L$] increases? + #math.equation(block: false, alt: "V equals L cubed")[$V = L^(3)$], #math.equation(block: false, alt: "A equals 6 L squared")[$A = 6 L^(2)$] + #math.equation(block: false, alt: "L equals V to the power 1 / 3")[$L = V^(1 / 3)$], #math.equation(block: false, alt: "L equals open parenthesis the fraction A over 6 close parenthesis to the power 1 / 2")[$L = attach(( display(frac(A, 6)) ), t: 1 / 2)$] + #math.equation(block: false, alt: "A equals 6 V to the power 2 / 3")[$A = 6 V^(2 / 3)$] + #math.equation(block: false, alt: "the fraction A over V equals the fraction 6 over L")[$display(frac(A, V)) = frac(6, L)$]. As #math.equation(block: false, alt: "L")[$L$] increases, the surface-to-volume ratio decreases. Repeat Problem 23 for the volume and surface area of a sphere in terms of its radius, #math.equation(block: false, alt: "R")[$R$]. + Write formulas for the volume, #math.equation(block: false, alt: "V")[$V$], and the surface area, #math.equation(block: false, alt: "A")[$A$], of a sphere in terms of its radius, #math.equation(block: false, alt: "R")[$R$]. + Express the radius of the sphere as a function of its volume. Express the radius of the sphere as a function of its surface area. + Express the surface area of the sphere as a function of its volume. + Express the surface to volume ratio of a sphere in terms of its radius. What happens to the surface to volume ratio as #math.equation(block: false, alt: "R")[$R$] increases? A brewery wants to replace its old vats with larger ones. To estimate the cost of the new equipment, the accountant uses the #math.equation(block: false, alt: "0.6")[$0.6$] rule for industrial costs, which states that the cost of a new container is approximately #math.equation(block: false, alt: "N equals C r to the power 0.6")[$N = C r^(0.6)$], where #math.equation(block: false, alt: "C")[$C$] is the cost of the old container and #math.equation(block: false, alt: "r")[$r$] is the ratio of the capacity of the new container to the old one. + If an old vat cost \$#math.equation(block: false, alt: "5000")[$5000$], graph #math.equation(block: false, alt: "N")[$N$] as a function of #math.equation(block: false, alt: "r")[$r$]. + How much should the accountant budget for a new vat that holds #math.equation(block: false, alt: "1.8")[$1.8$] times as much as the old one? + #figure(figph[power function], alt: "power function", caption: none) + \$#math.equation(block: false, alt: "7114.32")[$7114.32$] If a quantity of air expands without changing temperature, its pressure, in pounds per square inch, is given by #math.equation(block: false, alt: "P equals k V to the power minus 1.4")[$P = k V^(− 1.4)$], where #math.equation(block: false, alt: "V")[$V$] is the volume of the air in cubic inches and #math.equation(block: false, alt: "k equals 2.79 times 10 to the power 4")[$k = 2.79 × 10^(4)$]. + Graph #math.equation(block: false, alt: "P")[$P$] as a function of #math.equation(block: false, alt: "V")[$V$]. + Find the air pressure of an air sample when its volume is #math.equation(block: false, alt: "50")[$50$] cubic inches. In the 1970s, Jared Diamond studied the number of bird species on small islands near New Guinea. He found that larger islands support a larger number of different species, according to the formula #math.equation(block: true, alt: "S equals 15.1 A to the power 0.22")[$S = 15.1 A^(0.22)$] where #math.equation(block: false, alt: "S")[$S$] is the number of species on an island of area #math.equation(block: false, alt: "A")[$A$] square kilometers. (Source: Chapman and Reiss, 1992) + Fill in the table.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "A")[$A$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "5000")[$5000$]], [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]]), [#math.equation(block: false, alt: "S")[$S$]], [$#hide($000000$)$], [$#hide($000000$)$], [$#hide($000000$)$], [$#hide($000000$)$], [$#hide($000000$)$], )) + Graph the function on the domain #math.equation(block: false, alt: "0 less than A less than or equal to 10 , 000")[$0 < A ≤ 10 , 000$]. + How many species of birds would you expect to find on Manus Island, with an area of #math.equation(block: false, alt: "2100")[$2100$] square kilometers? On Lavongai, whose area is #math.equation(block: false, alt: "1140")[$1140$] square kilometers? + How large must an island be in order to support #math.equation(block: false, alt: "200")[$200$] different species of bird? + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "A")[$A$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "5000")[$5000$]], [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]]), [#math.equation(block: false, alt: "S")[$S$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "42")[$42$]], [#math.equation(block: false, alt: "69")[$69$]], [#math.equation(block: false, alt: "98")[$98$]], [#math.equation(block: false, alt: "115")[$115$]], )) + #figure(figph[power function], alt: "power function", caption: none) + #math.equation(block: false, alt: "81")[$81$], #math.equation(block: false, alt: "71")[$71$] + #math.equation(block: false, alt: "126 , 000")[$126 , 000$] sq km The drainage basin of a river channel is the area of land that contributes water to the river. The table gives the lengths in miles of some of the world’s largest rivers and the areas of their drainage basins in square miles. (Source: Leopold, Wolman, and Miller 1992) + Plot the data, using units of #math.equation(block: false, alt: "100 , 000")[$100 , 000$] on the horizontal axis and units of #math.equation(block: false, alt: "500")[$500$] on the vertical axis. + The length, #math.equation(block: false, alt: "L")[$L$], of the channel is related to the area, #math.equation(block: false, alt: "A")[$A$], of its drainage basin according to the formula #math.equation(block: true, alt: "L equals 1.05 A to the power 0.58")[$L = 1.05 A^(0.58)$] Graph this function on top of the data points. + The drainage basin for the Congo covers about #math.equation(block: false, alt: "1 , 600 , 000")[$1 , 600 , 000$] square miles. Estimate the length of the Congo River. + The Rio Grande is #math.equation(block: false, alt: "1700")[$1700$] miles long. What is the area of its drainage basin? #figure(table( columns: 3, align: left, inset: 6pt, table.header([River], [Area of #linebreak() drainage basin #linebreak()], [Length]), [Amazon], [#math.equation(block: false, alt: "2 , 700 , 000")[$2 , 700 , 000$]], [#math.equation(block: false, alt: "4300")[$4300$]], [Nile], [#math.equation(block: false, alt: "1 , 400 , 000")[$1 , 400 , 000$]], [#math.equation(block: false, alt: "4200")[$4200$]], [Mississippi], [#math.equation(block: false, alt: "1 , 300 , 000")[$1 , 300 , 000$]], [#math.equation(block: false, alt: "4100")[$4100$]], [Yangtze], [#math.equation(block: false, alt: "580 , 000")[$580 , 000$]], [#math.equation(block: false, alt: "2900")[$2900$]], [Volga], [#math.equation(block: false, alt: "480 , 000")[$480 , 000$]], [#math.equation(block: false, alt: "2300")[$2300$]], [St. Lawrence], [#math.equation(block: false, alt: "460 , 000")[$460 , 000$]], [#math.equation(block: false, alt: "1900")[$1900$]], [Ganges], [#math.equation(block: false, alt: "440 , 000")[$440 , 000$]], [#math.equation(block: false, alt: "1400")[$1400$]], [Orinoco], [#math.equation(block: false, alt: "380 , 000")[$380 , 000$]], [#math.equation(block: false, alt: "1400")[$1400$]], [Indus], [#math.equation(block: false, alt: "360 , 000")[$360 , 000$]], [#math.equation(block: false, alt: "2000")[$2000$]], [Danube], [#math.equation(block: false, alt: "350 , 000")[$350 , 000$]], [#math.equation(block: false, alt: "1800")[$1800$]], [Colorado], [#math.equation(block: false, alt: "250 , 000")[$250 , 000$]], [#math.equation(block: false, alt: "1700")[$1700$]], [Platte], [#math.equation(block: false, alt: "72 , 000")[$72 , 000$]], [#math.equation(block: false, alt: "800")[$800$]], [Rhine], [#math.equation(block: false, alt: "63 , 000")[$63 , 000$]], [#math.equation(block: false, alt: "900")[$900$]], [Seine], [#math.equation(block: false, alt: "48 , 000")[$48 , 000$]], [#math.equation(block: false, alt: "500")[$500$]], [Delaware], [#math.equation(block: false, alt: "12 , 000")[$12 , 000$]], [#math.equation(block: false, alt: "200")[$200$]], )) The table at right shows the exponent, #math.equation(block: false, alt: "p")[$p$], in the allometric equation #math.equation(block: true, alt: "variable equals k open parenthesis body mass close parenthesis to the power p")[$"variable" = k ( "body mass" )^(p)$] for some variables related to mammals. (Source: Chapman and Reiss, 1992) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Variable], [Exponent, #math.equation(block: false, alt: "p")[$p$]]), [Home range size], [#math.equation(block: false, alt: "1.26")[$1.26$]], [Lung volume], [#math.equation(block: false, alt: "1.02")[$1.02$]], [Brain mass], [#math.equation(block: false, alt: "0.70")[$0.70$]], [Respiration rate], [#math.equation(block: false, alt: "minus 0.26")[$− 0.26$]], )) + Match each equation to one of the graphs shown in the figure.#figure(figph[four power functions], alt: "four power functions", caption: none) + Explain how the value of #math.equation(block: false, alt: "p")[$p$] in the allometric equation determines the shape of the graph. Consider the cases #math.equation(block: false, alt: "p greater than 1")[$p > 1$], #math.equation(block: false, alt: "0 less than p less than 1")[$0 < p < 1$], and #math.equation(block: false, alt: "p less than 0")[$p < 0$]. + Home range size: II, lung volume: III, brain mass: I, respiration rate: IV + If #math.equation(block: false, alt: "p greater than 1")[$p > 1$], the graph is increasing and concave up. If #math.equation(block: false, alt: "0 less than p less than 1")[$0 < p < 1$], the graph is increasing and concave down. If #math.equation(block: false, alt: "p less than 0")[$p < 0$], the graph is decreasing and concave up. The average body mass of a dolphin is about #math.equation(block: false, alt: "140")[$140$] kilograms, twice the body mass of an average human male. + Using the allometric equations in Problem 29, calculate the ratio of the brain mass of a dolphin to that of a human. + A good-sized brown bear weighs about #math.equation(block: false, alt: "280")[$280$] kilograms, twice the weight of a dolphin. Calculate the ratio of the brain mass of a brown bear to that of a dolphin. + Use a ratio to compare the heartbeat frequencies of a dolphin and a human, and those of a brown bear and a dolphin. (See Example of Roots and Radicals.) The gourd species #emph[Tricosanthes] grows according to the formula #math.equation(block: false, alt: "L equals a d to the power 2.2")[$L = a d^(2.2)$], where #math.equation(block: false, alt: "L")[$L$] is its length and #math.equation(block: false, alt: "d")[$d$] is its width. The species #emph[Lagenaria] has the growth law #math.equation(block: false, alt: "L equals a d to the power 0.81")[$L = a d^(0.81)$]. (Source: Burton, 1998) + By comparing the exponents, predict which gourd grows into a long, thin shape, and which is relatively fatter. Which species is called the snake gourd, and which is the bottle gourd? + The snake gourd reaches a length of #math.equation(block: false, alt: "2")[$2$] meters (#math.equation(block: false, alt: "200")[$200$] cm), with a diameter of only #math.equation(block: false, alt: "4")[$4$] cm. Find the value of #math.equation(block: false, alt: "a")[$a$] in its growth law. + The bottle gourd is #math.equation(block: false, alt: "10")[$10$] cm long and #math.equation(block: false, alt: "7")[$7$] cm in diameter at maturity. Find the value of #math.equation(block: false, alt: "a")[$a$] in its growth law. + The giant bottle gourd grows to a length of #math.equation(block: false, alt: "23")[$23$] cm with a diameter of #math.equation(block: false, alt: "20")[$20$] cm. Does it grow according to the same law as standard bottle gourds? + Tricosanthes is the snake gourd and Lagenaria is the bottle gourd. Tricosanthes is thinner and Lagenaria is fatter. + #math.equation(block: false, alt: "a approximately equals 9.5")[$a ≈ 9.5$] + #math.equation(block: false, alt: "a approximately equals 2")[$a ≈ 2$] + Yes As a fiddler crab grows, one claw (called the chela) grows much faster than the rest of the body. The table shows the mass of the chela, #math.equation(block: false, alt: "C")[$C$], versus the mass of the rest of the body, #math.equation(block: false, alt: "b")[$b$], for a number of fiddler crabs. (Source: Burton, 1998) #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "b")[$b$]], [#math.equation(block: false, alt: "65")[$65$]], [#math.equation(block: false, alt: "110")[$110$]], [#math.equation(block: false, alt: "170")[$170$]], [#math.equation(block: false, alt: "205")[$205$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "360")[$360$]], [#math.equation(block: false, alt: "615")[$615$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "68")[$68$]], [#math.equation(block: false, alt: "110")[$110$]], [#math.equation(block: false, alt: "240")[$240$]], )) + Plot the data. + On the same axes, graph the function #math.equation(block: false, alt: "C equals 0.007 b to the power 1.63")[$C = 0.007 b^(1.63)$]. How well does the function fit the data? + Using the function in part (b), predict the chela mass of a fiddler crab if the rest of its body weighs #math.equation(block: false, alt: "400")[$400$] mg. + The chela from a fiddler crab weighs #math.equation(block: false, alt: "250")[$250$] mg. How much does the rest of its body weigh? + As the body mass of a fiddler crab doubles from #math.equation(block: false, alt: "100")[$100$] mg to #math.equation(block: false, alt: "200")[$200$] mg, by what factor does the mass of its chela increase? As the body mass doubles from #math.equation(block: false, alt: "200")[$200$] mg to #math.equation(block: false, alt: "400")[$400$] mg? The climate of a region has a great influence on the types of animals that can survive there. Extreme temperatures create difficult living conditions, so the diversity of wildlife decreases as the annual temperature range increases. Along the west coast of North America, the number of species of mammals, #math.equation(block: false, alt: "M")[$M$], is approximately related to the temperature range, #math.equation(block: false, alt: "R")[$R$], (in degrees Celsius) by the function #math.equation(block: false, alt: "M equals f open parenthesis R close parenthesis equals 433.8 R to the power minus 0.742")[$M = f ( R ) = 433.8 R^(− 0.742)$]. (Source: Chapman and Reiss, 1992) + Graph the function for temperature ranges up to #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]C. + How many species would you expect to find in a region where the temperature range is #math.equation(block: false, alt: "10 degrees")[$10^(∘)$]C? Label the corresponding point on your graph. + If #math.equation(block: false, alt: "50")[$50$] different species are found in a certain region, what temperature range would you expect the region to experience? Label the corresponding point on your graph. + Evaluate the function to find #math.equation(block: false, alt: "f open parenthesis 9 close parenthesis")[$f ( 9 )$], #math.equation(block: false, alt: "f open parenthesis 10 close parenthesis")[$f ( 10 )$], #math.equation(block: false, alt: "f open parenthesis 19 close parenthesis")[$f ( 19 )$], and #math.equation(block: false, alt: "f open parenthesis 20 close parenthesis")[$f ( 20 )$]. What do these values represent? Calculate the change in the number of species as the temperature range increases from #math.equation(block: false, alt: "9 degrees")[$9^(∘)$]C to #math.equation(block: false, alt: "10 degrees")[$10^(∘)$]C and from #math.equation(block: false, alt: "19 degrees")[$19^(∘)$]C to #math.equation(block: false, alt: "20 degrees")[$20^(∘)$]C. Which #math.equation(block: false, alt: "1 degrees")[$1^(∘)$] increase results in a greater decrease in diversity? Explain your answer in terms of slopes on your graph. + #figure(figph[powe function], alt: "powe function", caption: none) + #math.equation(block: false, alt: "79")[$79$] species + #math.equation(block: false, alt: "18.4 degrees")[$18.4^(∘)$]C + #math.equation(block: false, alt: "f open parenthesis 9 close parenthesis approximately equals 85")[$f ( 9 ) ≈ 85$], #math.equation(block: false, alt: "f open parenthesis 10 close parenthesis approximately equals 79")[$f ( 10 ) ≈ 79$], #math.equation(block: false, alt: "f open parenthesis 19 close parenthesis approximately equals 49")[$f ( 19 ) ≈ 49$], #math.equation(block: false, alt: "f open parenthesis 20 close parenthesis approximately equals 47")[$f ( 20 ) ≈ 47$]; from #math.equation(block: false, alt: "9 degrees")[$9^(∘)$]C to #math.equation(block: false, alt: "10 degrees")[$10^(∘)$]C has the greater decrease, corresponding to the steeper slope. A bicycle ergometer is used to measure the amount of power generated by a cyclist. The scatterplot shows how long an athlete was able to sustain various levels of power output. The curve is the graph of #math.equation(block: false, alt: "y equals 500 x to the power minus 0.29")[$y = 500 x^(− 0.29)$], which approximately models the data. (Source: Alexander, 1992) #figure(figph[scatterplot and power function], alt: "scatterplot and power function", caption: none) + In this graph, which variable is independent and which is dependent? + The athlete maintained #math.equation(block: false, alt: "650")[$650$] watts of power for #math.equation(block: false, alt: "40")[$40$] seconds. What power output does the equation predict for #math.equation(block: false, alt: "40")[$40$] seconds? + The athlete maintained #math.equation(block: false, alt: "300")[$300$] watts of power for #math.equation(block: false, alt: "10")[$10$] minutes. How long does the equation predict that power output can be maintained? + In 1979, a remarkable pedal-powered aircraft called the Gossamer Albatross was successfully flown across the English Channel. The flight took #math.equation(block: false, alt: "3")[$3$] hours. According to the equation, what level of power can be maintained for #math.equation(block: false, alt: "3")[$3$] hours? + The Gossamer Albatross needed #math.equation(block: false, alt: "250")[$250$] watts of power to keep it airborne. For how long can #math.equation(block: false, alt: "250")[$250$] watts be maintained according to the given equation? Inflating a Balloon at the start of this chapter gives data for the pressure inside April and Tolu's balloon as a function of its diameter. As the diameter of the balloon increases from #math.equation(block: false, alt: "5")[$5$] cm to #math.equation(block: false, alt: "20")[$20$] cm, the pressure inside decreases. Can we find a function that describes this portion of the graph? + Pressure is the force per unit area exerted by the balloon on the air inside, or #math.equation(block: false, alt: "P equals the fraction F over A")[$P = display(frac(F, A))$]. Because the balloon is spherical, its surface area, #math.equation(block: false, alt: "A")[$A$], is given by #math.equation(block: false, alt: "A equals π d squared")[$A = π d^(2)$]. Because the force increases as the balloon expands, we will try a power function #math.equation(block: false, alt: "F equals k d to the power p")[$F = k d^(p)$], where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "p")[$p$] are constants, and see if this fits the data. Combine the three equations, #math.equation(block: false, alt: "P equals the fraction F over A")[$P = display(frac(F, A))$], #math.equation(block: false, alt: "A equals π d squared")[$A = π d^(2)$], and #math.equation(block: false, alt: "F equals k d to the power p")[$F = k d^(p)$], to express #math.equation(block: false, alt: "P")[$P$] as a power function of #math.equation(block: false, alt: "d")[$d$]. + Use your calculator's power regression feature to find a power function that fits the data. Graph the function #math.equation(block: false, alt: "P equals 211 d to the power minus 0.7")[$P = 211 d^(− 0.7)$] on top of the data. Do the data support the hypothesis that #math.equation(block: false, alt: "P")[$P$] is a power function of #math.equation(block: false, alt: "d")[$d$]? + What is the value of the exponent #math.equation(block: false, alt: "p")[$p$] in #math.equation(block: false, alt: "F equals k d to the power p")[$F = k d^(p)$]? + #math.equation(block: false, alt: "P equals the fraction k over π d to the power p minus 2")[$P = display(frac(k, π)) d^(p − 2)$] + #figure(figph[scatterplot and power function], alt: "scatterplot and power function", caption: none) #linebreak() The power function is a good fit on this interval. + #math.equation(block: false, alt: "1.3")[$1.3$] The table shows the total number of frequent flyer miles redeemed by customers through the given year. (Source: www.hotelnewsresource.com) + Plot the data, with #math.equation(block: false, alt: "t equals 0")[$t = 0$] in 1980. What type of function might model the data? + Graph the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 3.13 t to the power 2.33")[$f ( t ) = 3.13 t^(2.33)$] on top of the data. + Evaluate #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis")[$f ( 5 )$] and #math.equation(block: false, alt: "f open parenthesis 25 close parenthesis")[$f ( 25 )$]. What do those values mean in this context? + Use the regression equation to predict when the total number of miles redeemed will reach 10 trillion. Hint: How many billions make a trillion? #figure(table( columns: 2, align: left, inset: 6pt, table.header([Year], [Cumulative #linebreak() miles redeemed #linebreak() (billions) #linebreak()]), [#math.equation(block: false, alt: "1982")[$1982$]], [#math.equation(block: false, alt: "14.8")[$14.8$]], [#math.equation(block: false, alt: "1984")[$1984$]], [#math.equation(block: false, alt: "85.3")[$85.3$]], [#math.equation(block: false, alt: "1986")[$1986$]], [#math.equation(block: false, alt: "215.4")[$215.4$]], [#math.equation(block: false, alt: "1988")[$1988$]], [#math.equation(block: false, alt: "387.5")[$387.5$]], [#math.equation(block: false, alt: "1990")[$1990$]], [#math.equation(block: false, alt: "641.3")[$641.3$]], [#math.equation(block: false, alt: "1992")[$1992$]], [#math.equation(block: false, alt: "975.2")[$975.2$]], [#math.equation(block: false, alt: "1994")[$1994$]], [#math.equation(block: false, alt: "1455.9")[$1455.9$]], [#math.equation(block: false, alt: "1996")[$1996$]], [#math.equation(block: false, alt: "1996")[$1996$]], [#math.equation(block: false, alt: "1998")[$1998$]], [#math.equation(block: false, alt: "2670.8")[$2670.8$]], [#math.equation(block: false, alt: "2000")[$2000$]], [#math.equation(block: false, alt: "3379.1")[$3379.1$]], [#math.equation(block: false, alt: "2002")[$2002$]], [#math.equation(block: false, alt: "4123.6")[$4123.6$]], )) For Problems 37–42, simplify by applying the laws of exponents. Write your answers with positive exponents only. + #math.equation(block: false, alt: "4 a to the power 6 / 5 a to the power 4 / 5")[$4 a^(6 / 5) a^(4 / 5)$] + #math.equation(block: false, alt: "9 b to the power 4 / 3 b to the power 1 / 3")[$9 b^(4 / 3) b^(1 / 3)$] + #math.equation(block: false, alt: "4 a squared")[$4 a^(2)$] + #math.equation(block: false, alt: "9 b to the power 5 / 3")[$9 b^(5 / 3)$] + #math.equation(block: false, alt: "open parenthesis minus 2 m to the power 2 / 3 close parenthesis to the power 4")[$( − 2 m^(2 / 3) )^(4)$] + #math.equation(block: false, alt: "open parenthesis minus 5 n to the power 3 / 4 close parenthesis cubed")[$( − 5 n^(3 / 4) )^(3)$] + #math.equation(block: false, alt: "the fraction 8 w to the power 9 / 4 over 2 w to the power 3 / 4")[$display(frac(8 w^(9 / 4), 2 w^(3 / 4)))$] + #math.equation(block: false, alt: "the fraction 12 z to the power 11 / 3 over 4 z to the power 5 / 3")[$display(frac(12 z^(11 / 3), 4 z^(5 / 3)))$] + #math.equation(block: false, alt: "4 w to the power 3 / 2")[$4 w^(3 / 2)$] + #math.equation(block: false, alt: "3 z squared")[$3 z^(2)$] + #math.equation(block: false, alt: "open parenthesis minus 3 u to the power 5 / 3 close parenthesis open parenthesis 5 u to the power minus 2 / 3 close parenthesis")[$( − 3 u^(5 / 3) ) ( 5 u^(− 2 / 3) )$] + #math.equation(block: false, alt: "open parenthesis minus 2 v to the power 7 / 8 close parenthesis open parenthesis minus 3 v to the power minus 3 / 8 close parenthesis")[$( − 2 v^(7 / 8) ) ( − 3 v^(− 3 / 8) )$] + #math.equation(block: false, alt: "the fraction k to the power 3 / 4 over 2 k")[$display(frac(k^(3 / 4), 2 k))$] + #math.equation(block: false, alt: "the fraction 4 h to the power 2 / 3 over 3 h")[$display(frac(4 h^(2 / 3), 3 h))$] + #math.equation(block: false, alt: "the fraction 1 over 2 k to the power 1 / 4")[$display(frac(1, 2 k^(1 / 4)))$] + #math.equation(block: false, alt: "the fraction 4 over 3 h to the power 1 / 3")[$display(frac(4, 3 h^(1 / 3)))$] + #math.equation(block: false, alt: "c to the power minus 2 / 3 open parenthesis the fraction 2 over 3 c squared close parenthesis")[$c^(− 2 / 3) ( display(frac(2, 3)) c^(2) )$] + #math.equation(block: false, alt: "the fraction r cubed over 4 open parenthesis r to the power minus 5 / 2 close parenthesis")[$display(frac(r^(3), 4)) ( r^(− 5 / 2) )$] The incubation time for a bird's egg is a function of the mass, #math.equation(block: false, alt: "m")[$m$], of the egg, and has been experimentally determined as #math.equation(block: true, alt: "I open parenthesis m close parenthesis equals 12.0 m to the power 0.217")[$I ( m ) = 12.0 m^(0.217)$] where #math.equation(block: false, alt: "m")[$m$] is measured in grams and #math.equation(block: false, alt: "I")[$I$] is in days. (Source: Burton, 1998) + Calculate the incubation time (to the nearest day) for the wren, whose eggs weigh about #math.equation(block: false, alt: "2.5")[$2.5$] grams, and the greylag goose, whose eggs weigh #math.equation(block: false, alt: "46")[$46$] grams. + During incubation, birds' eggs lose water vapor through their porous shells. The rate of water loss from the egg is also a function of its mass, and it appears to follow the rule #math.equation(block: true, alt: "W open parenthesis m close parenthesis equals 0.015 m to the power 0.742")[$W ( m ) = 0.015 m^(0.742)$] in grams per day. Combine the functions #math.equation(block: false, alt: "I open parenthesis m close parenthesis")[$I ( m )$] and #math.equation(block: false, alt: "W open parenthesis m close parenthesis")[$W ( m )$] to calculate the fraction of the initial egg mass that is lost during the entire incubation period. + Explain why your result shows that most eggs lose about #math.equation(block: false, alt: "18 %")[$18 upright(%)$] of their mass during incubation. + Wren: #math.equation(block: false, alt: "15")[$15$] days, greylag goose: #math.equation(block: false, alt: "28")[$28$] days + #math.equation(block: false, alt: "the fraction I open parenthesis m close parenthesis times W open parenthesis m close parenthesis over m equals 0.18 m to the power minus 0.041")[$display(frac(I ( m ) ⋅ W ( m ), m)) = 0.18 m^(− 0.041)$] + Because #math.equation(block: false, alt: "m to the power minus 0.041")[$m^(− 0.041)$] is close to #math.equation(block: false, alt: "m to the power 0")[$m^(0)$], the fraction lost is close to #math.equation(block: false, alt: "0.18")[$0.18$]. The incubation time for birds' eggs is given by #math.equation(block: true, alt: "I open parenthesis m close parenthesis equals 12.0 m to the power 0.217")[$I ( m ) = 12.0 m^(0.217)$] where #math.equation(block: false, alt: "m")[$m$] is the weight of the egg in grams, and #math.equation(block: false, alt: "I")[$I$] is in days. (See Problem 43.) Before hatching, the eggs take in oxygen at the rate of #math.equation(block: true, alt: "O open parenthesis m close parenthesis equals 22.2 m to the power 0.77")[$O ( m ) = 22.2 m^(0.77)$] in milliliters per day. (Source: Burton, 1998) + Combine the functions #math.equation(block: false, alt: "I open parenthesis m close parenthesis")[$I ( m )$] and #math.equation(block: false, alt: "O open parenthesis m close parenthesis")[$O ( m )$] to calculate the total amount of oxygen taken in by the egg during its incubation. + Use your result from part (a) to explain why daily oxygen consumption per unit mass is approximately inversely proportional to incubation time. + Predict the daily oxygen consumption per gram of a herring gull's eggs, given that their incubation time is #math.equation(block: false, alt: "26")[$26$] days. (The actual value is #math.equation(block: false, alt: "11")[$11$] milliliters per day.) For Problems 45–50, solve. Round your answers to the nearest thousandth if necessary. #math.equation(block: true, alt: "x to the power 2 / 3 minus 1 equals 15")[$x^(2 / 3) − 1 = 15$] #math.equation(block: true, alt: "x equals 64")[$x = 64$] #math.equation(block: true, alt: "x to the power 3 / 4 plus 3 equals 11")[$x^(3 / 4) + 3 = 11$] #math.equation(block: true, alt: "x to the power minus 2 / 5 equals 9")[$x^(− 2 / 5) = 9$] #math.equation(block: true, alt: "x equals the fraction 1 over 243")[$x = display(frac(1, 243))$] #math.equation(block: true, alt: "x to the power minus 3 / 2 equals 8")[$x^(− 3 / 2) = 8$] #math.equation(block: true, alt: "2 open parenthesis 5.2 minus x to the power 5 / 3 close parenthesis equals 1.4")[$2 ( 5.2 − x^(5 / 3) ) = 1.4$] #math.equation(block: true, alt: "x approximately equals 2.466")[$x ≈ 2.466$] #math.equation(block: true, alt: "3 open parenthesis 8.6 minus x to the power 5 / 2 close parenthesis equals 6.5")[$3 ( 8.6 − x^(5 / 2) ) = 6.5$] Kepler's law gives a relation between the period, #math.equation(block: false, alt: "p")[$p$], of a planet's revolution, in years, and its average distance, #math.equation(block: false, alt: "a")[$a$], from the sun: #math.equation(block: true, alt: "p squared equals K a cubed")[$p^(2) = K a^(3)$] where #math.equation(block: false, alt: "K equals 1.243 times 10 to the power minus 24")[$K = 1.243 × 10^(− 24)$], #math.equation(block: false, alt: "a")[$a$] is measured in miles, and #math.equation(block: false, alt: "p")[$p$] is in years. + Solve Kepler's law for #math.equation(block: false, alt: "p")[$p$] as a function of #math.equation(block: false, alt: "a")[$a$]. + Find the period of Mars if its average distance from the sun is #math.equation(block: false, alt: "1.417 times 10 to the power 8")[$1.417 × 10^(8)$] miles. + #math.equation(block: false, alt: "p equals 1.115 times 10 to the power minus 12 a to the power 3 / 2")[$p = 1.115 × 10^(− 12) a^(3 / 2)$] + #math.equation(block: false, alt: "1.88")[$1.88$] years Refer to Kepler's law, #math.equation(block: false, alt: "p squared equals K a cubed")[$p^(2) = K a^(3)$], in Problem 51. + Solve Kepler's law for #math.equation(block: false, alt: "a")[$a$] as a function of #math.equation(block: false, alt: "p")[$p$]. + Find the average distance from Venus to the sun if its period is #math.equation(block: false, alt: "0.615")[$0.615$] years. If #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis 3 x minus 4 close parenthesis to the power 3 / 2")[$f ( x ) = ( 3 x − 4 )^(3 / 2)$], find #math.equation(block: false, alt: "x")[$x$] so that #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 27")[$f ( x ) = 27$]. #math.equation(block: true, alt: "the fraction 13 over 3")[$display(frac(13, 3))$] If #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis 6 x minus 2 close parenthesis to the power 5 / 3")[$g ( x ) = ( 6 x − 2 )^(5 / 3)$], find #math.equation(block: false, alt: "x")[$x$] so that #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 32")[$g ( x ) = 32$]. If #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals 12 x to the power minus 5 / 4")[$S ( x ) = 12 x^(− 5 / 4)$], find #math.equation(block: false, alt: "x")[$x$] so that #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals 20")[$S ( x ) = 20$]. #math.equation(block: true, alt: "0.665")[$0.665$] If #math.equation(block: false, alt: "T open parenthesis x close parenthesis equals 9 x to the power minus 6 / 5")[$T ( x ) = 9 x^(− 6 / 5)$], find #math.equation(block: false, alt: "x")[$x$] so that #math.equation(block: false, alt: "T open parenthesis x close parenthesis equals 15")[$T ( x ) = 15$]. For Problems 57–64, use the distributive law to find the product. #math.equation(block: true, alt: "2 x to the power 1 / 2 open parenthesis x minus x to the power 1 / 2 close parenthesis")[$2 x^(1 / 2) ( x − x^(1 / 2) )$] #math.equation(block: true, alt: "2 x to the power 3 / 2 minus 2 x")[$2 x^(3 / 2) − 2 x$] #math.equation(block: true, alt: "x to the power 1 / 3 open parenthesis 2 x to the power 2 / 3 minus x to the power 1 / 3 close parenthesis")[$x^(1 / 3) ( 2 x^(2 / 3) − x^(1 / 3) )$] #math.equation(block: true, alt: "the fraction 1 over 2 y to the power minus 1 / 3 open parenthesis y to the power 2 / 3 plus 3 y to the power minus 5 / 6 close parenthesis")[$display(frac(1, 2)) y^(− 1 / 3) ( y^(2 / 3) + 3 y^(− 5 / 6) )$] #math.equation(block: true, alt: "the fraction 1 over 2 y to the power 1 / 3 plus the fraction 3 over 2 y to the power minus 7 / 6")[$display(frac(1, 2)) y^(1 / 3) + display(frac(3, 2)) y^(− 7 / 6)$] #math.equation(block: true, alt: "3 y to the power minus 3 / 8 open parenthesis the fraction 1 over 4 y to the power minus 1 / 4 plus y to the power 3 / 4 close parenthesis")[$3 y^(− 3 / 8) ( display(frac(1, 4)) y^(− 1 / 4) + y^(3 / 4) )$] #math.equation(block: true, alt: "open parenthesis 2 x to the power 1 / 4 plus 1 close parenthesis open parenthesis x to the power 1 / 4 minus 1 close parenthesis")[$( 2 x^(1 / 4) + 1 ) ( x^(1 / 4) − 1 )$] #math.equation(block: true, alt: "2 x to the power 1 / 2 minus x to the power 1 / 4 minus 1")[$2 x^(1 / 2) − x^(1 / 4) − 1$] #math.equation(block: true, alt: "open parenthesis 2 x to the power 1 / 3 minus 1 close parenthesis open parenthesis x to the power 1 / 3 plus 1 close parenthesis")[$( 2 x^(1 / 3) − 1 ) ( x^(1 / 3) + 1 )$] #math.equation(block: true, alt: "open parenthesis a to the power 3 / 4 minus 2 close parenthesis squared")[$( a^(3 / 4) − 2 )^(2)$] #math.equation(block: true, alt: "a to the power 3 / 2 minus 4 a to the power 3 / 4 plus 4")[$a^(3 / 2) − 4 a^(3 / 4) + 4$] #math.equation(block: true, alt: "open parenthesis a to the power 2 / 3 plus 3 close parenthesis squared")[$( a^(2 / 3) + 3 )^(2)$] For Problems 65–70, factor out the smallest power from each expression. Write your answers with positive exponents only. #math.equation(block: true, alt: "x to the power 3 / 2 plus x equals x open parenthesis ? close parenthesis")[$x^(3 / 2) + x = x ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "x open parenthesis x to the power 1 / 2 plus 1 close parenthesis")[$x ( x^(1 / 2) + 1 )$] #math.equation(block: true, alt: "y minus y to the power 2 / 3 equals y to the power 2 / 3 open parenthesis ? close parenthesis")[$y − y^(2 / 3) = y^(2 / 3) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "y to the power 3 / 4 minus y to the power minus 1 / 4 equals y to the power minus 1 / 4 open parenthesis ? close parenthesis")[$y^(3 / 4) − y^(− 1 / 4) = y^(− 1 / 4) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "the fraction y minus 1 over y to the power 1 / 4")[$display(frac(y − 1, y^(1 / 4)))$] #math.equation(block: true, alt: "x to the power minus 3 / 2 plus x to the power minus 1 / 2 equals x to the power minus 3 / 2 open parenthesis ? close parenthesis")[$x^(− 3 / 2) + x^(− 1 / 2) = x^(− 3 / 2) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "a to the power 1 / 3 plus 3 minus a to the power minus 1 / 3 equals a to the power minus 1 / 3 open parenthesis ? close parenthesis")[$a^(1 / 3) + 3 − a^(− 1 / 3) = a^(− 1 / 3) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "the fraction a to the power 2 / 3 plus a to the power 1 / 3 minus 1 over a to the power 1 / 3")[$display(frac(a^(2 / 3) + a^(1 / 3) − 1, a^(1 / 3)))$] #math.equation(block: true, alt: "3 b minus b to the power 3 / 4 plus 4 b to the power minus 3 / 4 equals b to the power minus 3 / 4 open parenthesis ? close parenthesis")[$3 b − b^(3 / 4) + 4 b^(− 3 / 4) = b^(− 3 / 4) ( " " " " "?" " " " " )$]