#set document(title: "3.2 Integer 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")) == 3.2#h(0.6em)Integer Exponents Recall that a positive integer exponent tells us how many times its base occurs as a factor in an expression. For example, #math.equation(block: true, alt: "4 a cubed b squared means 4 a a a b b")[$4 a^(3) b^(2) " " " " " means " " " " " 4 a a a b b$] What meaning can we assign to a negative exponent? === Negative Exponents Study the list of powers of #math.equation(block: false, alt: "2")[$2$] shown in Table (a) and observe the pattern as we move up the list from bottom to top. Each time the exponent increases by 1 we multiply by another factor of #math.equation(block: false, alt: "2")[$2$]. We can continue up the list as far as we like. #figure(figph[table of positive exponents], alt: "table of positive exponents", caption: none) $#hide($000000000000$)$a. If we move back down the list, we divide by #math.equation(block: false, alt: "2")[$2$] at each step, until we get to the bottom of the list, #math.equation(block: false, alt: "2 to the power 1 equals 2")[$2^(1) = 2$]. What if we continue the list in the same way, dividing by #math.equation(block: false, alt: "2")[$2$] each time we decrease the exponent? The results are shown in Table (b). As we continue to divide by #math.equation(block: false, alt: "2")[$2$], we generate fractions whose denominators are powers of #math.equation(block: false, alt: "2")[$2$]. In particular, #figure(figph[table of integer exponents], alt: "table of integer exponents", caption: none) $#hide($000000000000$)$b. #math.equation(block: true, alt: "2 to the power minus 1 equals the fraction 1 over 2 equals the fraction 1 over 2 to the power 1 and 2 to the power minus 2 equals the fraction 1 over 4 equals the fraction 1 over 2 squared")[$2^(− 1) = frac(1, 2) = frac(1, 2^(1)) " " " " " " " " " and " " " " " " " " " 2^(− 2) = frac(1, 4) = frac(1, 2^(2))$] Based on these observations, we make the following definitions. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Definition of Negative and Zero Exponents] #math.equation(block: true, alt: "a to the power minus n, equals the fraction 1 over a to the power n, open parenthesis a not equal to 0 close parenthesis; a to the power 0, equals 1, open parenthesis a not equal to 0 close parenthesis")[$a^(− n) & = frac(1, a^(n)) & & ( a ≠ 0 ) \ a^(0) & = 1 & & ( a ≠ 0 )$] ] These definitions tell us that if the base #math.equation(block: false, alt: "a")[$a$] is not zero, then any number raised to the zero power is #math.equation(block: false, alt: "1")[$1$], and that a negative exponent denotes a reciprocal. #examplebox("Example 1")[][ + #math.equation(block: false, alt: "2 to the power minus 3 equals the fraction 1 over 2 cubed equals the fraction 1 over 8")[$2^(− 3) = display(frac(1, 2^(3))) = display(frac(1, 8))$] + #math.equation(block: false, alt: "9 x to the power minus 2 equals 9 times the fraction 1 over x squared equals the fraction 9 over x squared")[$9 x^(− 2) = 9 ⋅ display(frac(1, x^(2))) = display(frac(9, x^(2)))$] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + A negative exponent does #emph[not] mean that the power is negative! For example, #math.equation(block: true, alt: "2 to the power minus 3 not equal to minus 2 cubed")[$2^(− 3) ≠ − 2^(3)$] + In Exampleb, note that #math.equation(block: true, alt: "9 x to the power minus 2 not equal to the fraction 1 over 9 x squared")[$9 x^(− 2) ≠ frac(1, 9 x^(2))$] The exponent, #math.equation(block: false, alt: "minus 2")[$− 2$], applies only to the base #math.equation(block: false, alt: "x")[$x$], not to #math.equation(block: false, alt: "9")[$9$]. ] Which of these is equivalent to #math.equation(block: false, alt: "2 x to the power minus 4")[$2 x^(− 4)$]? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "the fraction 2 over x to the power 4")[$display(frac(2, x^(4)))$]is equivalent to #math.equation(block: false, alt: "2 x to the power minus 4")[$2 x^(− 4)$]. ] Which of these expressions is equivalent to #math.equation(block: false, alt: "2 x to the power minus 4")[$2 x^(− 4)$]? + #math.equation(block: false, alt: "minus 2 x to the power 4")[$− 2 x^(4)$] + #math.equation(block: false, alt: "the fraction 1 over 2 x to the power 4")[$display(frac(1, 2 x^(4)))$] + #math.equation(block: false, alt: "2 minus x to the power 4")[$2 − x^(4)$] + #math.equation(block: false, alt: "the fraction 2 over x to the power 4")[$display(frac(2, x^(4)))$] Write each expression without using negative exponents. + #math.equation(block: false, alt: "5 to the power minus 4 equals")[$5^(− 4) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "5 x to the power minus 4 equals")[$5 x^(− 4) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "the fraction 1 over 5 to the power 4")[$display(frac(1, 5^(4)))$] + #math.equation(block: false, alt: "the fraction 5 over x to the power 4")[$display(frac(5, x^(4)))$] ] Write each expression without using negative exponents. + #math.equation(block: false, alt: "5 to the power minus 4")[$5^(− 4)$] + #math.equation(block: false, alt: "5 x to the power minus 4")[$5 x^(− 4)$] #solutionbox[ + #math.equation(block: false, alt: "the fraction 1 over 5 to the power 4")[$display(frac(1, 5^(4)))$] + #math.equation(block: false, alt: "the fraction 5 over x to the power 4")[$display(frac(5, x^(4)))$] ] In the next example, we see how to evaluate expressions that contain negative exponents and how to solve equations involving negative exponents. #examplebox("Example 2")[][ The body mass index, or BMI, is one measure of a person’s physical fitness. Your body mass index is defined by #math.equation(block: true, alt: "B M I equals w h to the power minus 2")[$B M I = w h^(− 2)$] where #math.equation(block: false, alt: "w")[$w$] is your weight in kilograms and #math.equation(block: false, alt: "h")[$h$] is your height in meters. The World Health Organization classifies a person as obese if his or her BMI is #math.equation(block: false, alt: "25")[$25$] or higher. + Calculate the BMI for a woman who is #math.equation(block: false, alt: "1.625")[$1.625$] meters (#math.equation(block: false, alt: "64")[$64$] inches) tall and weighs #math.equation(block: false, alt: "54")[$54$] kilograms (#math.equation(block: false, alt: "120")[$120$] pounds). + For a fixed weight, how does BMI vary with height? + The world’s heaviest athlete is the amateur sumo wrestler Emanuel Yarbrough, who weighs #math.equation(block: false, alt: "319")[$319$] kg (#math.equation(block: false, alt: "704")[$704$] pounds). What height would Yarbrough have to be to have a BMI under #math.equation(block: false, alt: "25")[$25$]? #solutionbox[ + #math.equation(block: false, alt: "B M I equals 54 open parenthesis 1.625 to the power minus 2 close parenthesis equals 54 open parenthesis the fraction 1 over 1.625 squared close parenthesis equals 20.45")[$B M I = 54 ( 1.625^(− 2) ) = 54 ( display(frac(1, 1.625^(2))) ) = 20.45$] + #math.equation(block: false, alt: "B M I equals w h to the power minus 2 equals the fraction w over h squared")[$B M I = w h^(− 2) = display(frac(w, h^(2)))$], so BMI varies inversely with the square of height. That is, for a fixed weight, BMI decreases as height increases. + To find the height that gives a BMI of #math.equation(block: false, alt: "25")[$25$], we solve the equation #math.equation(block: false, alt: "25 equals 319 h to the power minus 2")[$" " 25 = 319 h^(− 2)$]. Note that the variable #math.equation(block: false, alt: "h")[$h$] appears in the denominator of a fraction, so we begin by clearing the denominator—in this case we multiply both sides of the equation by #math.equation(block: false, alt: "h squared")[$h^(2)$]. #math.equation(block: true, alt: "25, equals the fraction 319 over h squared, Multiply both sides by h squared .; 25 h squared, equals 319, Divide both sides by 25.; h squared, equals 12.76, Extract square roots.; h, approximately equals 3.57")[$25 & = frac(319, h^(2)) & & "Multiply both sides by " bold(italic(h))^(2) . \ 25 h^(2) & = 319 & & "Divide both sides by " 25. \ h^(2) & = 12.76 & & "Extract square roots." \ h & ≈ 3.57$] To have a BMI under #math.equation(block: false, alt: "25")[$25$], Yarbrough would have to be over #math.equation(block: false, alt: "3.57")[$3.57$] meters, or #math.equation(block: false, alt: "11")[$11$] feet #math.equation(block: false, alt: "8")[$8$] inches tall. (In fact, he is #math.equation(block: false, alt: "6")[$6$] feet #math.equation(block: false, alt: "8")[$8$] inches tall.) ] ] Solve the equation #math.equation(block: false, alt: "0.2 x to the power minus 3 equals 1.5")[$0.2 x^(− 3) = 1.5$] #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ Enter "root(n,x)" for #math.equation(block: false, alt: "the nth root of x")[$root(n, x)$]. #math.equation(block: true, alt: "Rewrite without a negative exponent.")[$"Rewrite without a negative exponent."$] #math.equation(block: true, alt: "Clear the fraction.")[$"Clear the fraction."$] #math.equation(block: true, alt: "Isolate the variable.")[$"Isolate the variable."$] #solutionbox[ #math.equation(block: true, alt: "x equals the cube root of the fraction 2 over 15 approximately equals 0.51")[$x = root(3, display(frac(2, 15))) ≈ 0.51$] ] Solve the equation #math.equation(block: false, alt: "0.2 x to the power minus 3 equals 1.5")[$" " 0.2 x^(− 3) = 1.5$] #solutionbox[ Rewrite without a negative exponent, clear the fraction, then isolate the variable, to find #math.equation(block: true, alt: "x equals the cube root of the fraction 2 over 15 approximately equals 0.51")[$x = root(3, display(frac(2, 15))) ≈ 0.51$] ] Explain why #math.equation(block: false, alt: "the fraction 1 over a to the power minus n equals a to the power n")[$display(frac(1, a^(− n))) = a^(n)$], if #math.equation(block: false, alt: "a not equal to 0.")[$a ≠ 0.$] \_\_\_\_\_ Explain why #math.equation(block: false, alt: "the fraction 1 over a to the power minus n equals a to the power n")[$" " display(frac(1, a^(− n))) = a^(n) " "$], if #math.equation(block: false, alt: "a not equal to 0.")[$a ≠ 0.$] === Power Functions The functions that describe direct and inverse variation are part of a larger family of functions called #strong[power functions]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Power Function] A function of the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals k x to the power p")[$f ( x ) = k x^(p)$] where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "p")[$p$] are nonzero constants, is called a #strong[power function]. ] Examples of power functions are #math.equation(block: true, alt: "V open parenthesis r close parenthesis equals the fraction 4 over 3 π r cubed and L open parenthesis T close parenthesis equals 0.8125 T squared")[$V ( r ) = frac(4, 3) π r^(3) " " " " " and " " " " " L ( T ) = 0.8125 T^(2)$] In addition, the basic functions #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x and g open parenthesis x close parenthesis equals the fraction 1 over x squared")[$f ( x ) = frac(1, x) " " " " " and " " " " " g ( x ) = frac(1, x^(2))$] which we studied in Modeling with Functions can be written as #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x to the power minus 1 and g open parenthesis x close parenthesis equals x to the power minus 2")[$f ( x ) = x^(− 1) " " " " " and " " " " " g ( x ) = x^(− 2)$] Their graphs are shown below. Note that the domains of power functions with negative exponents do not include zero. #figure(figph[graphs of the two basic reciprocal functions], alt: "graphs of the two basic reciprocal functions", caption: none) #examplebox("Example 3")[][ Which of the following are power functions? + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over 3 x to the power 4 plus 2")[$f ( x ) = display(frac(1, 3)) x^(4) + 2$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 3 x to the power 4")[$g ( x ) = display(frac(1, 3 x^(4)))$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction x plus 6 over x cubed")[$h ( x ) = display(frac(x + 6, x^(3)))$] #solutionbox[ + This is not a power function, because of the addition of the constant term. + We can write #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 3 x to the power minus 4")[$g ( x ) = frac(1, 3) x^(− 4)$], so #math.equation(block: false, alt: "g")[$g$] is a power function. + This is not a power function, but it can be treated as the sum of two power functions, because #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals x to the power minus 2 plus 6 x to the power minus 3")[$h ( x ) = x^(− 2) + 6 x^(− 3)$]. ] ] Write each function as a power function in the form #math.equation(block: false, alt: "y equals k x to the power p")[$y = k x^(p)$]. For this exercise, enter rational numbers in decimal form. For example, enter "0.5" rather than "1/2". + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 12 over x squared equals")[$f ( x ) = display(frac(12, x^(2))) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 4 x equals")[$g ( x ) = display(frac(1, 4 x)) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction 2 over 5 x to the power 6 equals")[$h ( x ) = display(frac(2, 5 x^(6))) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 12 x to the power minus 2")[$f ( x ) = 12 x^(− 2)$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 4 x to the power minus 1 equals 0.25 x to the power minus 1")[$g ( x ) = display(frac(1, 4)) x^(− 1) = 0.25 x^(− 1)$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction 2 over 5 x to the power minus 6 equals 0.4 x to the power minus 6")[$h ( x ) = display(frac(2, 5)) x^(− 6) = 0.4 x^(− 6)$] ] Write each function as a power function in the form #math.equation(block: false, alt: "y equals k x to the power p")[$y = k x^(p)$]. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 12 over x squared")[$f ( x ) = display(frac(12, x^(2)))$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 4 x")[$g ( x ) = display(frac(1, 4 x))$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction 2 over 5 x to the power 6")[$h ( x ) = display(frac(2, 5 x^(6)))$] #solutionbox[ + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 12 x to the power minus 2")[$f ( x ) = 12 x^(− 2)$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 4 x to the power minus 1")[$g ( x ) = display(frac(1, 4)) x^(− 1)$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction 2 over 5 x to the power minus 6")[$h ( x ) = display(frac(2, 5)) x^(− 6)$] ] Which statement is true about power functions? \_\_\_\_\_ #solutionbox[ They can describe direct or inverse variation. ] Which statement is true about power functions? + They can describe direct or inverse variation. + They involve a power of the output variable. + The #math.equation(block: false, alt: "y")[$y$]-intercept must be a positive number. + They include all linear and quadratic functions. Most applications are concerned with positive variables only, so many models use only the portion of the graph in the first quadrant. #examplebox("Example 4")[][ In the Middle Ages in Europe, castles were built as defensive strongholds. An attacking force would build a huge catapult called a trebuchet to hurl rocks and scrap metal inside the castle walls. The engineers could adjust its range by varying the mass of the projectiles. The mass, #math.equation(block: false, alt: "m")[$m$], of the projectile should be inversely proportional to the square of the distance, #math.equation(block: false, alt: "d")[$d$], to the target. + Use a negative exponent to write #math.equation(block: false, alt: "m")[$m$] as a function of #math.equation(block: false, alt: "d")[$d$], #math.equation(block: false, alt: "m equals f open parenthesis d close parenthesis")[$m = f ( d )$]. + The engineers test the trebuchet with a #math.equation(block: false, alt: "20")[$20$]-kilogram projectile, which lands #math.equation(block: false, alt: "250")[$250$] meters away. Find the constant of proportionality; then rewrite your formula for #math.equation(block: false, alt: "m")[$m$]. + Graph #math.equation(block: false, alt: "m equals f open parenthesis d close parenthesis")[$m = f ( d )$]. + The trebuchet is #math.equation(block: false, alt: "180")[$180$] meters from the courtyard within the castle. What size projectile will hit the target? + The attacking force would like to hurl a #math.equation(block: false, alt: "100")[$100$]-kilogram projectile at the castle. How close must the attackers bring their trebuchet? #solutionbox[ + If we use #math.equation(block: false, alt: "k")[$k$] for the constant of proportionality, then #math.equation(block: false, alt: "m equals the fraction k over d squared")[$m = display(frac(k, d^(2)))$]. Rewriting this equation with a negative exponent gives #math.equation(block: false, alt: "m equals k d to the power minus 2")[$m = k d^(− 2)$]. + We substitute #math.equation(block: false, alt: "m equals 20")[$m = 20$] and #math.equation(block: false, alt: "d equals 250")[$d = 250$] to obtain #math.equation(block: true, alt: "20, equals k open parenthesis 250 close parenthesis to the power minus 2, Multiply both sides by 250 squared .; 1 , 250 , 000, equals k")[$20 & = k ( 250 )^(− 2) & & "Multiply both sides by " 250^(2) . \ 1 , 250 , 000 & = k$] Thus, #math.equation(block: false, alt: "m equals 1 , 250 , 000 d to the power minus 2")[$m = 1 , 250 , 000 d^(− 2)$]. + We evaluate the function for several values of #math.equation(block: false, alt: "m")[$m$], or use a calculator to obtain the graph below.#figure(figph[graph of mass vs distance], alt: "graph of mass vs distance", caption: none) + We substitute #math.equation(block: false, alt: "d equals 180")[$d = 180$] into the formula: #math.equation(block: true, alt: "m, equals 1 , 250 , 000 open parenthesis 180 close parenthesis to the power minus 2; equals the fraction 1 , 250 , 000 over 32 , 400; approximately equals 38.58")[$m & = 1 , 250 , 000 ( 180 )^(− 2) \ & = frac(1 "," 250 "," 000, 32 "," 400) \ & ≈ 38.58$] The attackers should use a mass of approximately #math.equation(block: false, alt: "38.6")[$38.6$] kilograms. + We substitute #math.equation(block: false, alt: "m equals 100")[$m = 100$] into the formula and solve for #math.equation(block: false, alt: "d")[$d$]. #math.equation(block: true, alt: "100, equals 1 , 250 , 000 d to the power minus 2, Multiply by d squared .; 100 d squared, equals 1 , 250 , 000, Divide by 100.; d squared, equals 12 , 500, Take sqwuare roots.; d, equals plus or minus the square root of 12 , 500")[$100 & = 1 , 250 , 000 d^(− 2) & & "Multiply by " bold(italic(d))^(2) . \ 100 d^(2) & = 1 , 250 , 000 & & "Divide by " 100. \ d^(2) & = 12 , 500 & & "Take sqwuare roots." \ d & = ± sqrt(12 "," 500)$] They must locate the trebuchet #math.equation(block: false, alt: "the square root of 12 , 500 approximately equals 111.8")[$sqrt(12 "," 500) ≈ 111.8$] meters from the castle. ] ] The function #math.equation(block: false, alt: "m equals the fraction k over d squared")[$" " m = display(frac(k, d^(2))) " "$] is an example of an #strong[inverse square law], because #math.equation(block: false, alt: "m")[$m$] varies inversely with the square of #math.equation(block: false, alt: "d")[$d$]. Such laws are fairly common in physics and its applications, because gravitational and other forces behave in this way. Here is a more modern example of an inverse square law. Cell phone towers typically transmit signals at 10 watts of power. The signal strength varies inversely with the square of distance from the tower, and 1 kilometer away the signal strength is 0.8 picowatt. (A picowatt is #math.equation(block: false, alt: "10 to the power minus 12")[$10^(− 12)$] watt.) Cell phones can receive a signal as small as 0.01 picowatt. How far can you be from the nearest tower and still hope to have cell phone reception? About \_\_\_\_\_ km #solutionbox[ About 9 km ] Cell phone towers typically transmit signals at 10 watts of power. The signal strength varies inversely with the square of distance from the tower, and 1 kilometer away the signal strength is 0.8 picowatt. (A picowatt is #math.equation(block: false, alt: "10 to the power minus 12")[$10^(− 12)$] watt.) Cell phones can receive a signal as small as 0.01 picowatt. How far can you be from the nearest tower and still hope to have cell phone reception? #solutionbox[ #math.equation(block: false, alt: "S equals the fraction 0.8 over d squared ,")[$S = display(frac(0.8, d^(2))) ,$] where #math.equation(block: false, alt: "S")[$S$] is in picowatts and #math.equation(block: false, alt: "d")[$d$] is in kilometers. If #math.equation(block: false, alt: "S equals 0.01")[$S = 0.01$], we solve for #math.equation(block: false, alt: "d")[$d$] to find #math.equation(block: false, alt: "d")[$d$] is about 9 km. ] What is an inverse square law? Give an example. \_\_\_\_\_ What is an inverse square law? Give an example. === Working with Negative Exponents A negative exponent denotes the #emph[reciprocal] of a power. Thus, to simplify a fraction with a negative exponent, we compute the positive power of its reciprocal. #examplebox("Example 5")[][ + #math.equation(block: false, alt: "open parenthesis the fraction 3 over 5 close parenthesis to the power minus 2 equals the fraction 1 over open parenthesis the fraction 3 over 5 close parenthesis squared equals open parenthesis the fraction 5 over 3 close parenthesis squared equals the fraction 25 over 9")[$attach(( display(frac(3, 5)) ), t: − 2) = display(frac(1, attach(( display(frac(3, 5)) ), t: 2))) = attach(( display(frac(5, 3)) ), t: 2) = display(frac(25, 9))$] + #math.equation(block: false, alt: "open parenthesis the fraction x cubed over 4 close parenthesis to the power minus 3 equals open parenthesis the fraction 4 over x cubed close parenthesis cubed equals the fraction open parenthesis 4 close parenthesis cubed over open parenthesis x cubed close parenthesis cubed equals the fraction 64 over x to the power 9")[$attach(( display(frac(x^(3), 4)) ), t: − 3) = attach(( display(frac(4, x^(3))) ), t: 3) = display(frac(( 4 )^(3), attach(( x^(3) ), t: 3))) = display(frac(64, x^(9)))$] ] Simplify #math.equation(block: false, alt: "open parenthesis the fraction 2 over x squared close parenthesis to the power minus 4 equals")[$attach(( display(frac(2, x^(2))) ), t: − 4) =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "the fraction x to the power 8 over 16")[$display(frac(x^(8), 16))$] ] Simplify #math.equation(block: false, alt: "open parenthesis the fraction 2 over x squared close parenthesis to the power minus 4")[$attach(( display(frac(2, x^(2))) ), t: − 4)$] #solutionbox[ #math.equation(block: true, alt: "the fraction x to the power 8 over 16")[$display(frac(x^(8), 16))$] ] Dividing by a power with a negative exponent is equivalent to multiplying by a power with a positive exponent. #examplebox("Example 6")[][ + #math.equation(block: false, alt: "the fraction 1 over 5 to the power minus 3, equals 1 divided by 5 to the power minus 3; equals 1 divided by the fraction 1 over 5 cubed; equals 1 times 5 cubed equals 125")[$frac(1, 5^(− 3)) & = 1 ÷ 5^(− 3) \ & = 1 ÷ frac(1, 5^(3)) \ & = 1 × 5^(3) = 125$] + #math.equation(block: false, alt: "the fraction k squared over m to the power minus 4, equals k squared divided by m to the power minus 4; equals k squared divided by the fraction 1 over m to the power 4; equals k squared m to the power 4")[$frac(k^(2), m^(− 4)) & = k^(2) ÷ m^(− 4) \ & = k^(2) ÷ frac(1, m^(4)) \ & = k^(2) m^(4)$] ] Write each expression without using negative exponents. + #math.equation(block: false, alt: "open parenthesis the fraction 3 over b to the power 4 close parenthesis to the power minus 2 equals")[$attach(( display(frac(3, b^(4))) ), t: − 2) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "the fraction 12 over x to the power minus 6 equals")[$display(frac(12, x^(− 6))) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "the fraction b to the power 8 over 9")[$display(frac(b^(8), 9))$] + #math.equation(block: false, alt: "12 x to the power 6")[$12 x^(6)$] ] Write each expression without using negative exponents. + #math.equation(block: false, alt: "open parenthesis the fraction 3 over b to the power 4 close parenthesis to the power minus 2")[$attach(( display(frac(3, b^(4))) ), t: − 2)$] + #math.equation(block: false, alt: "the fraction 12 over x to the power minus 6")[$display(frac(12, x^(− 6)))$] #solutionbox[ + #math.equation(block: false, alt: "the fraction b to the power 8 over 9")[$display(frac(b^(8), 9))$] + #math.equation(block: false, alt: "12 x to the power 6")[$12 x^(6)$] ] === Laws of Exponents The laws of exponents apply to all integer exponents, positive, negative, and zero. When we allow negative exponents, we can simplify the rule for computing quotients of powers. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quotient of Powers] II. #math.equation(block: false, alt: "the fraction a to the power m over a to the power n equals a to the power m minus n open parenthesis a not equal to 0 close parenthesis")[$display(frac(a^(m), a^(n)) = a^(m − n) #hide($b l a n k$) ( a ≠ 0 ))$] ] For example, by applying this new version of the law for quotients, we find #math.equation(block: true, alt: "the fraction x squared over x to the power 5 equals x to the power 2 minus 5 equals x to the power minus 3")[$frac(x^(2), x^(5)) = x^(2 − 5) = x^(− 3)$] which is consistent with our previous version of the rule, #math.equation(block: true, alt: "the fraction x squared over x to the power 5 equals the fraction 1 over x to the power 5 minus 2 equals the fraction 1 over x cubed")[$frac(x^(2), x^(5)) = frac(1, x^(5 − 2)) = frac(1, x^(3))$] Which of these is equivalent to #math.equation(block: false, alt: "the fraction m squared over m to the power 6")[$display(frac(m^(2), m^(6)))$]? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "m to the power minus 4")[$m^(− 4)$] ] Which of these is equivalent to #math.equation(block: false, alt: "the fraction m squared over m to the power 6")[$display(frac(m^(2), m^(6)))$]? + #math.equation(block: false, alt: "the fraction 1 over m cubed")[$display(frac(1, m^(3)))$] + #math.equation(block: false, alt: "minus m to the power 4")[$− m^(4)$] + #math.equation(block: false, alt: "the fraction 1 over m to the power minus 4")[$display(frac(1, m^(− 4)))$] + #math.equation(block: false, alt: "m to the power minus 4")[$m^(− 4)$] For reference, we restate the laws of exponents below. The laws are valid for all integer exponents #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "n")[$n$], and for #math.equation(block: false, alt: "a , b not equal to 0")[$a , b ≠ 0$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Laws of Exponents] + #math.equation(block: false, alt: "a to the power m times a to the power n equals a to the power m plus n")[$" " display(a^(m) ⋅ a^(n) = a^(m + n))$] + #math.equation(block: false, alt: "the fraction a to the power m over a to the power n equals a to the power m minus n")[$" " display(frac(a^(m), a^(n)) = a^(m − n))$] + #math.equation(block: false, alt: "open parenthesis a to the power m close parenthesis to the power n equals a to the power m n")[$" " display(attach(( a^(m) ), t: n) = a^(m n))$] + #math.equation(block: false, alt: "open parenthesis a b close parenthesis to the power n equals a to the power n b to the power n")[$" " display(attach(( a b ), t: n) = a^(n) b^(n))$] + #math.equation(block: false, alt: "open parenthesis the fraction a over b close parenthesis to the power n equals the fraction a to the power n over b to the power n")[$" " display(attach(( frac(a, b) ), t: n) = frac(a^(n), b^(n)))$] ] #examplebox("Example 7")[][ #math.equation(block: true, alt: "a, . x cubed times x to the power minus 5 equals x to the power 3 minus 5 equals x to the power minus 2, Apply the first law: Add exponents.; b, . the fraction 8 x to the power minus 2 over 4 x to the power minus 6 equals the fraction 8 over 4 x to the power minus 2 minus open parenthesis minus 6 close parenthesis equals 2 x to the power 4, Apply the second law: Subtract exponents.; c, . open parenthesis 5 x to the power minus 3 close parenthesis to the power minus 2 equals 5 to the power minus 2 open parenthesis x to the power minus 3 close parenthesis to the power minus 2 equals the fraction x to the power 6 over 25, Apply laws IV and III.")[$"a" & . " " x^(3) ⋅ x^(− 5) = x^(3 − 5) = x^(− 2) & & "Apply the first law: Add exponents." \ "b" & . " " frac(8 x^(− 2), 4 x^(− 6)) = frac(8, 4) x^(− 2 − ( − 6 )) = 2 x^(4) & & "Apply the second law: Subtract exponents." \ "c" & . " " attach(( 5 x^(− 3) ), t: − 2) = 5^(− 2) ( x^(− 3) )^(− 2) = frac(x^(6), 25) & & "Apply laws IV and III."$] ] You can check that each of the calculations in Example is shorter when we use negative exponents instead of converting the expressions into algebraic fractions. Which of the following is not one of the laws of exponents? \_\_\_\_\_ #solutionbox[ The equation #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power n equals x to the power n plus y to the power n")[$( x + y )^(n) = x^(n) + y^(n)$] is not one of the laws of exponents. ] Which of the following is not one of the laws of exponents? + #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power n equals x to the power n plus y to the power n")[$( x + y )^(n) = x^(n) + y^(n)$] + #math.equation(block: false, alt: "open parenthesis the fraction x over y close parenthesis to the power n equals the fraction x to the power n over y to the power n")[$attach(( display(frac(x, y)) ), t: n) = display(frac(x^(n), y^(n)))$] + #math.equation(block: false, alt: "open parenthesis x y close parenthesis to the power n equals x to the power n y to the power n")[$( x y )^(n) = x^(n) y^(n)$] + #math.equation(block: false, alt: "x to the power n x to the power m equals x to the power n plus m")[$x^(n) x^(m) = x^(n + m)$] Simplify by applying the laws of exponents. Write without negative exponents. + #math.equation(block: false, alt: "open parenthesis 2 a to the power minus 4 close parenthesis open parenthesis minus 4 a squared close parenthesis equals")[$( 2 a^(− 4) ) ( − 4 a^(2) ) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "the fraction open parenthesis r squared close parenthesis to the power minus 3 over 3 r to the power minus 4 equals")[$display(frac(( r^(2) )^(− 3), 3 r^(− 4))) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "the fraction minus 8 over a squared")[$display(frac(− 8, a^(2)))$] + #math.equation(block: false, alt: "the fraction 1 over 3 r squared")[$display(frac(1, 3 r^(2)))$] ] Simplify by applying the laws of exponents. Write without negative exponents. + #math.equation(block: false, alt: "open parenthesis 2 a to the power minus 4 close parenthesis open parenthesis minus 4 a squared close parenthesis")[$( 2 a^(− 4) ) ( − 4 a^(2) )$] + #math.equation(block: false, alt: "the fraction open parenthesis r squared close parenthesis to the power minus 3 over 3 r to the power minus 4")[$display(frac(( r^(2) )^(− 3), 3 r^(− 4)))$] #solutionbox[ + #math.equation(block: false, alt: "the fraction minus 8 over a squared")[$display(frac(− 8, a^(2)))$] + #math.equation(block: false, alt: "the fraction 1 over 3 r squared")[$display(frac(1, 3 r^(2)))$] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The laws of exponents do not apply to sums or differences of powers. We can add or subtract like terms, that is, powers with the same exponent. For example, #math.equation(block: true, alt: "6 x to the power minus 2 plus 3 x to the power minus 2 equals 9 x to the power minus 2")[$6 x^(− 2) + 3 x^(− 2) = 9 x^(− 2)$] but #emph[we cannot add or subtract terms with different exponents]. Thus, for example, #math.equation(block: true, alt: "4 x squared, minus 3 x to the power minus 2, cannot be simplified; x to the power minus 1, plus x to the power minus 3, cannot be simplified")[$4 x^(2) & − 3 x^(− 2) & & "cannot be simplified" \ x^(− 1) & + x^(− 3) & & "cannot be simplified"$] ] At the start of this section, we saw that #math.equation(block: false, alt: "2 to the power 0 equals 1")[$2^(0) = 1$], and in fact #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$] as long as #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. Now we can see that this definition is consistent with the laws of exponents. The quotient of any (nonzero) number divided by itself is #math.equation(block: false, alt: "1")[$1$]. But by applying the second law of exponents, we also have #math.equation(block: true, alt: "1 equals the fraction a to the power m over a to the power m equals a to the power m minus m equals a to the power 0")[$1 = frac(a^(m), a^(m)) = a^(m − m) = a^(0)$] Thus, #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Zero as Exponent] #math.equation(block: true, alt: "a to the power 0 equals 1 , if a not equal to 0")[$a^(0) = 1 , " " " " " if " a ≠ 0$] ] For example, #math.equation(block: true, alt: "3 to the power 0 equals 1 , open parenthesis minus 528 close parenthesis to the power 0 equals 1 , and open parenthesis 0.024 close parenthesis to the power 0 equals 1")[$3^(0) = 1 , " " " " ( − 528 )^(0) = 1 , " " " " " and " " " " " ( 0.024 )^(0) = 1$] Give a numerical example to show why we cannot add or subtract terms with the same variable but different exponents. \_\_\_\_\_ Give a numerical example to show why we cannot add or subtract terms with the same variable but different exponents. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Power function - Inverse square law ==== CONCEPTS + A negative exponent denotes a reciprocal: #math.equation(block: false, alt: "a to the power minus n equals the fraction 1 over a to the power n")[$a^(− n) = display(frac(1, a^(n)))$], if #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. + Any number (except zero) raised to the zero power is #math.equation(block: false, alt: "1")[$1$]: #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$], if #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. + A function of the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals k x to the power p")[$f ( x ) = k x^(p)$], where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "p")[$p$] are constants, is called a #strong[power function]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Laws of Exponents] + #math.equation(block: false, alt: "a to the power m times a to the power n equals a to the power m plus n")[$display(a^(m) ⋅ a^(n) = a^(m + n))$] + #math.equation(block: false, alt: "the fraction a to the power m over a to the power n equals a to the power m minus n")[$display(frac(a^(m), a^(n)) = a^(m − n))$] + #math.equation(block: false, alt: "open parenthesis a to the power m close parenthesis to the power n equals a to the power m n")[$display(attach(( a^(m) ), t: n) = a^(m n))$] + #math.equation(block: false, alt: "open parenthesis a b close parenthesis to the power n equals a to the power n b to the power n")[$display(attach(( a b ), t: n) = a^(n) b^(n))$] + #math.equation(block: false, alt: "open parenthesis the fraction a over b close parenthesis to the power n equals the fraction a to the power n over b to the power n")[$display(attach(( frac(a, b) ), t: n) = frac(a^(n), b^(n)))$] ] ==== STUDY QUESTIONS + Explain the difference between each pair of expressions. + #math.equation(block: false, alt: "minus 2 cubed")[$− 2^(3)$] and #math.equation(block: false, alt: "2 to the power minus 3")[$2^(− 3)$] + #math.equation(block: false, alt: "minus x to the power 4")[$− x^(4)$] and #math.equation(block: false, alt: "x to the power minus 4")[$x^(− 4)$] + #math.equation(block: false, alt: "minus 2 to the power n")[$− 2^(n)$] and #math.equation(block: false, alt: "2 to the power minus n")[$2^(− n)$] + Write a power function for "#math.equation(block: false, alt: "y")[$y$] varies inversely with the cube of #math.equation(block: false, alt: "x")[$x$]."? + Explain why it makes sense to define #math.equation(block: false, alt: "10 to the power 0 equals 1")[$10^(0) = 1$]. + Why is zero excluded from the domain of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x to the power minus 2")[$f ( x ) = 3 x^(− 2)$]? + Choose a value for to show that the following statement is false: #math.equation(block: true, alt: "2 x to the power minus 2 plus 4 to the power x minus 1 equals 6 to the power x minus 3 False!")[$2 x^(− 2) + 4^(x − 1) = 6^(x − 3) #hide($00000$) "False!"$] ==== SKILLS Practice each skill in the Homework problems listed. + Simplify expressions with negative exponents: \#1–12 + Solve equations involving negative exponents: \#19–24 + Write formulas for power functions: \#17 and 18, 25–34 + Evaluate and analyze power functions: \#13–16, 25–34 + Apply the laws of exponents to simplify expressions: \#35–62 === Homework 3.2 Make a table showing powers of #math.equation(block: false, alt: "3")[$3$] from #math.equation(block: false, alt: "3 to the power minus 5")[$3^(− 5)$] to #math.equation(block: false, alt: "3 to the power 5")[$3^(5)$]. Illustrate why defining #math.equation(block: false, alt: "3 to the power 0 equals 1")[$3^(0) = 1$] makes sense. #figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]), [#math.equation(block: false, alt: "3 to the power n")[$3^(n)$]], [#math.equation(block: false, alt: "the fraction 1 over 243")[$frac(1, 243)$]], [#math.equation(block: false, alt: "the fraction 1 over 81")[$frac(1, 81)$]], [#math.equation(block: false, alt: "the fraction 1 over 27")[$frac(1, 27)$]], [#math.equation(block: false, alt: "the fraction 1 over 9")[$frac(1, 9)$]], [#math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "81")[$81$]], [#math.equation(block: false, alt: "243")[$243$]], )) Each time #math.equation(block: false, alt: "n")[$n$] increases by #math.equation(block: false, alt: "1")[$1$], we multiply the power in the bottom row by #math.equation(block: false, alt: "3")[$3$]. Make a table showing powers of #math.equation(block: false, alt: "5")[$5$] from #math.equation(block: false, alt: "5 to the power minus 4")[$5^(− 4)$] to #math.equation(block: false, alt: "5 to the power 4")[$5^(4)$]. Illustrate why defining #math.equation(block: false, alt: "5 to the power 0 equals 1")[$5^(0) = 1$] makes sense. For Problems 3–6, compute each power. + #math.equation(block: false, alt: "2 cubed")[$2^(3)$] + #math.equation(block: false, alt: "open parenthesis minus 2 close parenthesis cubed")[$( − 2 )^(3)$] + #math.equation(block: false, alt: "2 to the power minus 3")[$2^(− 3)$] + #math.equation(block: false, alt: "open parenthesis minus 2 close parenthesis to the power minus 3")[$( − 2 )^(− 3)$] + #math.equation(block: false, alt: "8")[$8$] + #math.equation(block: false, alt: "minus 8")[$− 8$] + #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] + #math.equation(block: false, alt: "the fraction minus 1 over 8")[$display(frac(− 1, 8))$] + #math.equation(block: false, alt: "4 squared")[$4^(2)$] + #math.equation(block: false, alt: "open parenthesis minus 4 close parenthesis squared")[$( − 4 )^(2)$] + #math.equation(block: false, alt: "4 to the power minus 2")[$4^(− 2)$] + #math.equation(block: false, alt: "open parenthesis minus 4 close parenthesis to the power minus 2")[$( − 4 )^(− 2)$] + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis cubed")[$attach(( display(frac(1, 2)) ), t: 3)$] + #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 2 close parenthesis cubed")[$attach(( − display(frac(1, 2)) ), t: 3)$] + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power minus 3")[$attach(( display(frac(1, 2)) ), t: − 3)$] + #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 2 close parenthesis to the power minus 3")[$attach(( − display(frac(1, 2)) ), t: − 3)$] + #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] + #math.equation(block: false, alt: "the fraction minus 1 over 8")[$display(frac(− 1, 8))$] + #math.equation(block: false, alt: "8")[$8$] + #math.equation(block: false, alt: "minus 8")[$− 8$] + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 4 close parenthesis squared")[$attach(( display(frac(1, 4)) ), t: 2)$] + #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 4 close parenthesis squared")[$attach(( − display(frac(1, 4)) ), t: 2)$] + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 4 close parenthesis to the power minus 2")[$attach(( display(frac(1, 4)) ), t: − 2)$] + #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 4 close parenthesis to the power minus 2")[$attach(( − display(frac(1, 4)) ), t: − 2)$] For Problems 7–12, write without negative exponents and simplify. + #math.equation(block: false, alt: "2 to the power minus 1")[$2^(− 1)$] + #math.equation(block: false, alt: "open parenthesis minus 5 close parenthesis to the power minus 2")[$( − 5 )^(− 2)$] + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power minus 3")[$attach(( display(frac(1, 3)) ), t: − 3)$] + #math.equation(block: false, alt: "the fraction 1 over open parenthesis minus 2 close parenthesis to the power minus 4")[$display(frac(1, ( − 2 )^(− 4)))$] + #math.equation(block: false, alt: "the fraction 1 over 2 to the power 1 equals the fraction 1 over 2")[$display(frac(1, 2^(1))) = display(frac(1, 2))$] + #math.equation(block: false, alt: "the fraction 1 over open parenthesis minus 5 close parenthesis squared equals the fraction 1 over 25")[$display(frac(1, ( − 5 )^(2))) = display(frac(1, 25))$] + #math.equation(block: false, alt: "3 cubed equals 27")[$3^(3) = 27$] + #math.equation(block: false, alt: "open parenthesis minus 2 close parenthesis to the power 4 equals 16")[$( − 2 )^(4) = 16$] + #math.equation(block: false, alt: "3 to the power minus 2")[$3^(− 2)$] + #math.equation(block: false, alt: "open parenthesis minus 2 close parenthesis to the power minus 3")[$( − 2 )^(− 3)$] + #math.equation(block: false, alt: "open parenthesis the fraction 3 over 5 close parenthesis to the power minus 2")[$attach(( display(frac(3, 5)) ), t: − 2)$] + #math.equation(block: false, alt: "the fraction 1 over open parenthesis minus 3 close parenthesis to the power minus 3")[$display(frac(1, ( − 3 )^(− 3)))$] + #math.equation(block: false, alt: "the fraction 5 over 4 to the power minus 3")[$display(frac(5, 4^(− 3)))$] + #math.equation(block: false, alt: "open parenthesis 2 q close parenthesis to the power minus 5")[$( 2 q )^(− 5)$] + #math.equation(block: false, alt: "minus 4 x to the power minus 2")[$− 4 x^(− 2)$] + #math.equation(block: false, alt: "the fraction 8 over b to the power minus 3")[$display(frac(8, b^(− 3)))$] + #math.equation(block: false, alt: "5 times 4 cubed equals 320")[$5 ⋅ 4^(3) = 320$] + #math.equation(block: false, alt: "the fraction 1 over open parenthesis 2 q close parenthesis to the power 5 equals the fraction 1 over 32 q to the power 5")[$display(frac(1, ( 2 q )^(5))) = display(frac(1, 32 q^(5)))$] + #math.equation(block: false, alt: "the fraction minus 4 over x squared")[$display(frac(− 4, x^(2)))$] + #math.equation(block: false, alt: "8 b cubed")[$8 b^(3)$] + #math.equation(block: false, alt: "the fraction 3 over 2 to the power minus 6")[$display(frac(3, 2^(− 6)))$] + #math.equation(block: false, alt: "open parenthesis 4 k close parenthesis to the power minus 3")[$( 4 k )^(− 3)$] + #math.equation(block: false, alt: "minus 7 x to the power minus 4")[$− 7 x^(− 4)$] + #math.equation(block: false, alt: "the fraction 5 over a to the power minus 5")[$display(frac(5, a^(− 5)))$] + #math.equation(block: false, alt: "open parenthesis m minus n close parenthesis to the power minus 2")[$( m − n )^(− 2)$] + #math.equation(block: false, alt: "y to the power minus 2 plus y to the power minus 3")[$y^(− 2) + y^(− 3)$] + #math.equation(block: false, alt: "2 p q to the power minus 4")[$2 p q^(− 4)$] + #math.equation(block: false, alt: "the fraction minus 5 y to the power minus 2 over x to the power minus 5")[$display(frac(− 5 y^(− 2), x^(− 5)))$] + #math.equation(block: false, alt: "the fraction 1 over open parenthesis m minus n close parenthesis squared")[$display(frac(1, ( m − n )^(2)))$] + #math.equation(block: false, alt: "the fraction 1 over y squared plus the fraction 1 over y cubed")[$display(frac(1, y^(2))) + display(frac(1, y^(3)))$] + #math.equation(block: false, alt: "the fraction 2 p over q to the power 4")[$display(frac(2 p, q^(4)))$] + #math.equation(block: false, alt: "the fraction minus 5 x to the power 5 over y squared")[$display(frac(− 5 x^(5), y^(2)))$] + #math.equation(block: false, alt: "open parenthesis p plus q close parenthesis to the power minus 3")[$( p + q )^(− 3)$] + #math.equation(block: false, alt: "z to the power minus 1 minus z to the power minus 2")[$z^(− 1) − z^(− 2)$] + #math.equation(block: false, alt: "8 m to the power minus 2 n squared")[$8 m^(− 2) n^(2)$] + #math.equation(block: false, alt: "the fraction minus 6 y to the power minus 3 over x to the power minus 3")[$display(frac(− 6 y^(− 3), x^(− 3)))$] Use your calculator to fill in the tables in Problems 13 and 14. Round your answers to two decimal places. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x to the power minus 2")[$f ( x ) = x^(− 2)$] + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "16")[$16$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What happens to the values of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as the values of #math.equation(block: false, alt: "x")[$x$] increase? Explain why. + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.125")[$0.125$]], [#math.equation(block: false, alt: "0.0625")[$0.0625$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What happens to the values of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as the values of #math.equation(block: false, alt: "x")[$x$] decrease toward #math.equation(block: false, alt: "0")[$0$]? Explain why. + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "16")[$16$]]), [#math.equation(block: false, alt: "x to the power minus 2")[$x^(− 2)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.06")[$0.06$]], [#math.equation(block: false, alt: "0.02")[$0.02$]], [#math.equation(block: false, alt: "0.00")[$0.00$]], )) + The values of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] decrease, because #math.equation(block: false, alt: "x to the power minus 2")[$x^(− 2)$] is the reciprocal of #math.equation(block: false, alt: "x squared")[$x^(2)$]. + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.125")[$0.125$]], [#math.equation(block: false, alt: "0.0625")[$0.0625$]]), [#math.equation(block: false, alt: "x to the power minus 2")[$x^(− 2)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "64")[$64$]], [#math.equation(block: false, alt: "256")[$256$]], )) + The values of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] increase toward infinity, because #math.equation(block: false, alt: "x to the power minus 2")[$x^(− 2)$] is the reciprocal of #math.equation(block: false, alt: "x squared")[$x^(2)$]. #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals x to the power minus 3")[$g ( x ) = x^(− 3)$] + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4.5")[$4.5$]], [#math.equation(block: false, alt: "6.2")[$6.2$]], [#math.equation(block: false, alt: "9.3")[$9.3$]]), [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What happens to the values of #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] as the values of #math.equation(block: false, alt: "x")[$x$] increase? Explain why. + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.03")[$0.03$]], [#math.equation(block: false, alt: "0.002")[$0.002$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What happens to the values of #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] as the values of #math.equation(block: false, alt: "x")[$x$] decrease toward #math.equation(block: false, alt: "0")[$0$]? Explain why. + Use your calculator to graph each of the following functions on the window #math.equation(block: true, alt: "Xmin, equals minus 5, Xmax equals 5; Ymin, equals minus 2, Ymax equals 10")[$"Xmin" & = − 5 & & "Xmax" = 5 \ "Ymin" & = − 2 & & "Ymax" = 10$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared")[$f ( x ) = x^(2)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power minus 2")[$f ( x ) = x^(− 2)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x squared")[$f ( x ) = display(frac(1, x^(2)))$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over x close parenthesis squared")[$f ( x ) = attach(( display(frac(1, x)) ), t: 2)$] + Which functions have the same graph? Explain your results. b. (ii), (iii), and (iv) have the same graph, because they represent the same function. + Use your calculator to graph each of the following functions on the window #math.equation(block: true, alt: "Xmin, equals minus 3, Xmax equals 5; Ymin, equals minus 5, Ymax equals 5")[$"Xmin" & = − 3 & & "Xmax" = 5 \ "Ymin" & = − 5 & & "Ymax" = 5$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power minus 3")[$f ( x ) = x^(− 3)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x cubed")[$f ( x ) = display(frac(1, x^(3)))$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over x close parenthesis cubed")[$f ( x ) = attach(( display(frac(1, x)) ), t: 3)$] + Which functions have the same graph? Explain your results. For Problems 17–18, write each expression as a power function using negative exponents. + #math.equation(block: false, alt: "F open parenthesis r close parenthesis equals the fraction 3 over r to the power 4")[$F ( r ) = display(frac(3, r^(4)))$] + #math.equation(block: false, alt: "G open parenthesis w close parenthesis equals the fraction 2 over 5 w cubed")[$G ( w ) = display(frac(2, 5 w^(3)))$] + #math.equation(block: false, alt: "H open parenthesis z close parenthesis equals the fraction 1 over open parenthesis 3 z close parenthesis squared")[$H ( z ) = display(frac(1, ( 3 z )^(2)))$] + #math.equation(block: false, alt: "F open parenthesis r close parenthesis equals 3 r to the power minus 4")[$F ( r ) = 3 r^(− 4)$] + #math.equation(block: false, alt: "G open parenthesis w close parenthesis equals the fraction 2 over 5 w to the power minus 3")[$G ( w ) = display(frac(2, 5)) w^(− 3)$] + #math.equation(block: false, alt: "H open parenthesis z close parenthesis equals the fraction 1 over 9 z to the power minus 2")[$H ( z ) = display(frac(1, 9)) z^(− 2)$] + #math.equation(block: false, alt: "h open parenthesis s close parenthesis equals the fraction 9 over s cubed")[$h ( s ) = display(frac(9, s^(3)))$] + #math.equation(block: false, alt: "f open parenthesis v close parenthesis equals the fraction 3 over 8 v to the power 6")[$f ( v ) = display(frac(3, 8 v^(6)))$] + #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals the fraction 1 over open parenthesis 5 t close parenthesis to the power 4")[$g ( t ) = display(frac(1, ( 5 t )^(4)))$] For Problems 19–24, solve. #math.equation(block: true, alt: "6 x to the power minus 2 equals 3.84")[$6 x^(− 2) = 3.84$] #math.equation(block: false, alt: "x equals minus 1.25")[$x = − 1.25$] or #math.equation(block: false, alt: "x equals 1.25")[$x = 1.25$] #math.equation(block: true, alt: "0.8 w to the power minus 2 equals 1.25")[$0.8 w^(− 2) = 1.25$] #math.equation(block: true, alt: "12 plus 0.04 t to the power minus 3 equals 175.84")[$12 + 0.04 t^(− 3) = 175.84$] #math.equation(block: true, alt: "t equals the fraction 1 over 16")[$t = display(frac(1, 16))$] #math.equation(block: true, alt: "854 minus 48 z to the power minus 3 equals 104")[$854 − 48 z^(− 3) = 104$] #math.equation(block: true, alt: "100 minus 0.15 v to the power minus 4 equals 6.25")[$100 − 0.15 v^(− 4) = 6.25$] #math.equation(block: false, alt: "v equals the fraction 1 over 5")[$v = display(frac(1, 5))$] or #math.equation(block: false, alt: "v equals the fraction minus 1 over 5")[$v = display(frac(− 1, 5))$] #math.equation(block: true, alt: "8100 p to the power minus 4 minus 250 equals 3656.25")[$8100 p^(− 4) − 250 = 3656.25$] When an automobile accelerates, the power, #math.equation(block: false, alt: "P")[$P$], needed to overcome air resistance varies directly with a power of the speed, #math.equation(block: false, alt: "v")[$v$]. + Use the data and the graph to find the scaling exponent and the constant of variation. Then write a formula for #math.equation(block: false, alt: "P")[$P$] as a power function of #math.equation(block: false, alt: "v")[$v$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$] (mph)], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]]), [#math.equation(block: false, alt: "P")[$P$] (watts)], [#math.equation(block: false, alt: "355")[$355$]], [#math.equation(block: false, alt: "2840")[$2840$]], [#math.equation(block: false, alt: "9585")[$9585$]], [#math.equation(block: false, alt: "22 , 720")[$22 , 720$]], )) #figure(figph[power function], alt: "power function", caption: none) + Find the speed that requires #math.equation(block: false, alt: "50 , 000")[$50 , 000$] watts of power. + If you increase your speed by #math.equation(block: false, alt: "50 %")[$50 upright(%)$], by what factor does the power requirement increase? + #math.equation(block: false, alt: "P equals 0.355 v cubed")[$P = 0.355 v^(3)$] + #math.equation(block: false, alt: "v approximately equals 52.03")[$v ≈ 52.03$] mph + #math.equation(block: false, alt: "3.375")[$3.375$] The power, #math.equation(block: false, alt: "P")[$P$], generated by a windmill varies directly with a power of wind velocity, #math.equation(block: false, alt: "v")[$v$]. + Use the data and the graph to find the scaling exponent and the constant of variation. Then write a formula for #math.equation(block: false, alt: "P")[$P$] as a power function of #math.equation(block: false, alt: "v")[$v$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$] (mph)], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]]), [#math.equation(block: false, alt: "P")[$P$] (watts)], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "405")[$405$]], [#math.equation(block: false, alt: "960")[$960$]], )) #figure(figph[power function], alt: "power function", caption: none) + Find the wind velocity needed to generate #math.equation(block: false, alt: "500")[$500$] watts of power. + If the wind speed drops by half, what happens to the power generated? The “Rule of 70” is used to estimate how long it takes an investment to double in value when interest is compounded annually. The doubling time, #math.equation(block: false, alt: "D")[$D$], is inversely proportional to the interest rate, #math.equation(block: false, alt: "i")[$i$]. (Note that #math.equation(block: false, alt: "i")[$i$] is expressed as a percent, not as a decimal fraction. For example, if the interest rate is #math.equation(block: false, alt: "8 %")[$8 upright(%)$], then #math.equation(block: false, alt: "i equals 8")[$i = 8$].) + Use the data and the graph to find the constant of proportionality and write #math.equation(block: false, alt: "D")[$D$] as a power function of #math.equation(block: false, alt: "i")[$i$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "i")[$i$] (mph)], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "D")[$D$] (watts)], [#math.equation(block: false, alt: "17.5")[$17.5$]], [#math.equation(block: false, alt: "11.67")[$11.67$]], [#math.equation(block: false, alt: "8.75")[$8.75$]], [#math.equation(block: false, alt: "7")[$7$]], )) #figure(figph[power function], alt: "power function", caption: none) + If the interest rate increases from #math.equation(block: false, alt: "5 %")[$5 upright(%)$] to #math.equation(block: false, alt: "6 %")[$6 upright(%)$], how will the doubling time change? + #math.equation(block: false, alt: "D equals the fraction 70 over i")[$D = display(frac(70, i))$] + It decreases by about #math.equation(block: false, alt: "2.3")[$2.3$] years. The f-stop setting on a camera regulates the size of the aperture and thus the amount of light entering the camera. The f-stop #math.equation(block: false, alt: "f")[$f$] is inversely proportional to the diameter, #math.equation(block: false, alt: "d")[$d$], of the aperture. + Use the data and the graph to find the constant of proportionality and write #math.equation(block: false, alt: "d")[$d$] as a power function of #math.equation(block: false, alt: "f")[$f$]. Values of #math.equation(block: false, alt: "d")[$d$] have been rounded to one decimal place.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "f")[$f$]], [#math.equation(block: false, alt: "2.8")[$2.8$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5.6")[$5.6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "11")[$11$]]), [#math.equation(block: false, alt: "d")[$d$]], [#math.equation(block: false, alt: "17.9")[$17.9$]], [#math.equation(block: false, alt: "12.5")[$12.5$]], [#math.equation(block: false, alt: "8.9")[$8.9$]], [#math.equation(block: false, alt: "6.3")[$6.3$]], [#math.equation(block: false, alt: "4.5")[$4.5$]], )) #figure(figph[power function], alt: "power function", caption: none) + Why are the f-stop settings labeled with the values given in the table? Hint: As you stop down the aperture from one f-value to the next, by what factor does #math.equation(block: false, alt: "d")[$d$] increase? The Stefan-Boltzmann law relates the total amount of radiation emitted by a star to its temperature, #math.equation(block: false, alt: "T")[$T$], in kelvins, by the following formula: #math.equation(block: true, alt: "s T to the power 4 equals the fraction L over 4 π R squared")[$s T^(4) = frac(L, 4 π R^(2))$] where #math.equation(block: false, alt: "R")[$R$] is the radius of the star, #math.equation(block: false, alt: "L")[$L$] is its luminosity, and #math.equation(block: false, alt: "s equals 5.7 times 10 to the power minus 8 watt/m squared")[$s = 5.7 × 10^(− 8) attach(" watt/m", t: 2)$] is a constant governing radiation. (See Algebra Skills Refresher Scientific Notation to review scientific notation.) + Write a formula for luminosity as a power function of temperature for a fixed radius. + The radius of the Sun is #math.equation(block: false, alt: "R equals 9.96 times 10 to the power 8")[$R = 9.96 × 10^(8)$] meters, and its luminosity is #math.equation(block: false, alt: "L equals 3.9 times 10 to the power 26")[$L = 3.9 × 10^(26)$] watts. Calculate the temperature of the Sun. + #math.equation(block: false, alt: "L equals open parenthesis 4 π s R squared close parenthesis T to the power 4 approximately equals 7.2 times 10 to the power minus 7 R squared T to the power 4")[$L = ( 4 π s R^(2) ) T^(4) ≈ 7.2 × 10^(− 7) R^(2) T^(4)$] + #math.equation(block: false, alt: "4840")[$4840$] K Poiseuille's law for the flow of liquid through a tube can be used to describe blood flow through an artery. The rate of flow, #math.equation(block: false, alt: "F")[$F$], in liters per minute is proportional to the fourth power of the radius, #math.equation(block: false, alt: "r")[$r$], divided by the length, #math.equation(block: false, alt: "L")[$L$], of the artery. + Write a formula for the rate of flow as a power function of radius. + If the radius and length of the artery are measured in centimeters, then the constant of variation, #math.equation(block: false, alt: "k equals 7.8 times 10 to the power 5")[$k = 7.8 × 10^(5)$], is determined by blood pressure and viscosity. If a certain artery is #math.equation(block: false, alt: "20")[$20$] centimeters long, what should its radius be in order to allow a blood flow of #math.equation(block: false, alt: "5")[$5$] liters per minute? Airplanes use radar to detect the distances to other objects. A radar unit transmits a pulse of energy, which bounces off a distant object, and the echo of the pulse returns to the sender. The power, #math.equation(block: false, alt: "P")[$P$], of the returning echo is inversely proportional to the fourth power of the distance, #math.equation(block: false, alt: "d")[$d$], to the object. A radar operator receives an echo of #math.equation(block: false, alt: "5 times 10 to the power minus 10")[$5 × 10^(− 10)$] watts from an aircraft #math.equation(block: false, alt: "2")[$2$] nautical miles away. + Express the power of the echo received in picowatts. (#math.equation(block: false, alt: "1")[$1$] picowatt #math.equation(block: false, alt: "equals 10 to the power minus 12")[$= 10^(− 12)$] watts.) + Write a function that expresses #math.equation(block: false, alt: "P")[$P$] in terms of d using negative exponents. Use picowatts for the units of power. + Complete the table of values for the power of the echo received from objects at various distances.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "d")[$d$] (nautical miles)], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "P")[$P$] (picowatts)], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Radar units can typically detect signals as low as #math.equation(block: false, alt: "10 to the power minus 13")[$10^(− 13)$] watts. How far away is an aircraft whose echo is #math.equation(block: false, alt: "10 to the power minus 13")[$10^(− 13)$] watts? Hint: Convert #math.equation(block: false, alt: "10 to the power minus 13")[$10^(− 13)$] watts to picowatts. + Sketch a graph of #math.equation(block: false, alt: "P")[$P$] as a function of #math.equation(block: false, alt: "d")[$d$]. Use units of picowatts on the vertical axis. + #math.equation(block: false, alt: "500")[$500$] picowatts + #math.equation(block: false, alt: "P equals 8000 d to the power minus 4")[$P = 8000 d^(− 4)$] + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "d")[$d$] (nautical miles)], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "P")[$P$] (picowatts)], [#math.equation(block: false, alt: "31.3")[$31.3$]], [#math.equation(block: false, alt: "12.8")[$12.8$]], [#math.equation(block: false, alt: "3.3")[$3.3$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], )) + #math.equation(block: false, alt: "16.8")[$16.8$] nautical miles + #figure(figph[inverse-square], alt: "inverse-square", caption: none) The lifetime of a star is roughly inversely proportional to the cube of its mass. Our Sun, which has a mass of one solar mass, will last for approximately #math.equation(block: false, alt: "10")[$10$] billion years. + Write a power function for the lifetime, #math.equation(block: false, alt: "L")[$L$], of a star in terms of its mass, #math.equation(block: false, alt: "m")[$m$]. + Sketch a graph of the function using units of solar mass on the horizontal axis. + How long will a star that is #math.equation(block: false, alt: "10")[$10$] times as massive as the Sun last? + One solar mass is about #math.equation(block: false, alt: "2 times 10 to the power 30")[$2 × 10^(30)$] kilograms. Rewrite your formula for #math.equation(block: false, alt: "L")[$L$] with the units of mass in kilograms. + How long will a star that is half as massive as the Sun last? The amount of force or thrust generated by the propeller of a ship is a function of two variables: the diameter of the propeller and its speed, in rotations per minute. The thrust, #math.equation(block: false, alt: "T")[$T$], in pounds, is proportional to the square of the speed, #math.equation(block: false, alt: "r")[$r$], and the fourth power of the diameter, #math.equation(block: false, alt: "d")[$d$], in feet. + Write a formula for the thrust in terms of the speed if the diameter of the propeller is #math.equation(block: false, alt: "2")[$2$] feet. + A propeller of diameter #math.equation(block: false, alt: "2")[$2$] feet generates a thrust of #math.equation(block: false, alt: "1000")[$1000$] pounds at #math.equation(block: false, alt: "100")[$100$] rotations per minute. Find the constant of variation in the formula for thrust. + Sketch a graph of the thrust as a function of the propeller speed for a propellor of diameter #math.equation(block: false, alt: "4")[$4$] feet. If the speed of the propeller is doubled, by what factor does the thrust increase? + #math.equation(block: false, alt: "T equals 16 k r squared")[$T = 16 k r^(2)$] + #math.equation(block: false, alt: "T equals 0.1 r squared")[$T = 0.1 r^(2)$] + #figure(figph[parabola], alt: "parabola", caption: none) Refer to Problem 33. + Write a formula for the thrust, #math.equation(block: false, alt: "T")[$T$], in terms of the diameter of the propeller if its speed is #math.equation(block: false, alt: "100")[$100$] rotations per minute. + A propeller of diameter #math.equation(block: false, alt: "4")[$4$] feet generates a thrust of #math.equation(block: false, alt: "32 , 000")[$32 , 000$] pounds at #math.equation(block: false, alt: "100")[$100$] rotations per minute. Find the constant of variation in the formula for thrust. + Sketch a graph of the thrust as a function of the diameter of the propeller at a speed of #math.equation(block: false, alt: "100")[$100$] rotations per minute. If the diameter of the propeller is doubled, by what factor does the thrust increase? For Problems 35–40, use the laws of exponents to simplify and write without negative exponents. + #math.equation(block: false, alt: "a to the power minus 3 times a to the power 8")[$a^(− 3) ⋅ a^(8)$] + #math.equation(block: false, alt: "5 to the power minus 4 times 5 to the power minus 3")[$5^(− 4) ⋅ 5^(− 3)$] + #math.equation(block: false, alt: "the fraction p to the power minus 7 over p to the power minus 4")[$display(frac(p^(− 7), p^(− 4)))$] + #math.equation(block: false, alt: "open parenthesis 7 to the power minus 2 close parenthesis to the power 5")[$attach(( 7^(− 2) ), t: 5)$] + #math.equation(block: false, alt: "a to the power 5")[$a^(5)$] + #math.equation(block: false, alt: "the fraction 1 over 5 to the power 7")[$display(frac(1, 5^(7)))$] + #math.equation(block: false, alt: "the fraction 1 over p cubed")[$display(frac(1, p^(3)))$] + #math.equation(block: false, alt: "the fraction 1 over 7 to the power 10")[$display(frac(1, 7^(10)))$] + #math.equation(block: false, alt: "b squared times b to the power minus 6")[$b^(2) ⋅ b^(− 6)$] + #math.equation(block: false, alt: "4 to the power minus 2 times 4 to the power minus 6")[$4^(− 2) ⋅ 4^(− 6)$] + #math.equation(block: false, alt: "the fraction w to the power minus 9 over w squared")[$display(frac(w^(− 9), w^(2)))$] + #math.equation(block: false, alt: "open parenthesis 9 to the power minus 4 close parenthesis cubed")[$attach(( 9^(− 4) ), t: 3)$] + #math.equation(block: false, alt: "open parenthesis 4 x to the power minus 5 close parenthesis open parenthesis 5 x squared close parenthesis")[$( 4 x^(− 5) ) ( 5 x^(2) )$] + #math.equation(block: false, alt: "the fraction 3 u to the power minus 3 over 9 u to the power 9")[$display(frac(3 u^(− 3), 9 u^(9)))$] + #math.equation(block: false, alt: "the fraction 5 to the power 6 t to the power 0 over 5 to the power minus 2 t to the power minus 1")[$display(frac(5^(6) t^(0), 5^(− 2) t^(− 1)))$] + #math.equation(block: false, alt: "the fraction 20 over x cubed")[$display(frac(20, x^(3)))$] + #math.equation(block: false, alt: "the fraction 1 over 3 u to the power 12")[$display(frac(1, 3 u^(12)))$] + #math.equation(block: false, alt: "5 to the power 8 t")[$5^(8) t$] + #math.equation(block: false, alt: "open parenthesis 3 y to the power minus 8 close parenthesis open parenthesis 2 y to the power 4 close parenthesis")[$( 3 y^(− 8) ) ( 2 y^(4) )$] + #math.equation(block: false, alt: "the fraction 4 c to the power minus 4 over 8 c to the power minus 8")[$display(frac(4 c^(− 4), 8 c^(− 8)))$] + #math.equation(block: false, alt: "the fraction 3 to the power 10 s to the power minus 1 over 3 to the power minus 5 s to the power 0")[$display(frac(3^(10) s^(− 1), 3^(− 5) s^(0)))$] + #math.equation(block: false, alt: "open parenthesis 3 x to the power minus 2 y cubed close parenthesis to the power minus 2")[$attach(( 3 x^(− 2) y^(3) ), t: − 2)$] + #math.equation(block: false, alt: "open parenthesis the fraction 6 a to the power minus 3 over b squared close parenthesis to the power minus 2")[$attach(( display(frac(6 a^(− 3), b^(2))) ), t: − 2)$] + #math.equation(block: false, alt: "the fraction 5 h to the power minus 3 open parenthesis h to the power 4 close parenthesis to the power minus 2 over 6 h to the power minus 5")[$display(frac(5 h^(− 3) ( h^(4) )^(− 2), 6 h^(− 5)))$] + #math.equation(block: false, alt: "the fraction x to the power 4 over 9 y to the power 6")[$display(frac(x^(4), 9 y^(6)))$] + #math.equation(block: false, alt: "the fraction a to the power 6 b to the power 4 over 36")[$display(frac(a^(6) b^(4), 36))$] + #math.equation(block: false, alt: "the fraction 5 over 6 h to the power 6")[$display(frac(5, 6 h^(6)))$] + #math.equation(block: false, alt: "open parenthesis 2 x cubed y to the power minus 4 close parenthesis to the power minus 3")[$attach(( 2 x^(3) y^(− 4) ), t: − 3)$] + #math.equation(block: false, alt: "open parenthesis the fraction a to the power 4 over 4 b to the power minus 5 close parenthesis to the power minus 3")[$attach(( display(frac(a^(4), 4 b^(− 5))) ), t: − 3)$] + #math.equation(block: false, alt: "the fraction 4 v to the power minus 5 open parenthesis v to the power minus 2 close parenthesis to the power minus 4 over 3 v to the power minus 8")[$display(frac(4 v^(− 5) ( v^(− 2) )^(− 4), 3 v^(− 8)))$] For Problems 41–44, write each expression as a sum of terms of the form #math.equation(block: false, alt: "k x to the power p")[$k x^(p)$]. + #math.equation(block: false, alt: "the fraction x over 3 plus the fraction 3 over x")[$display(frac(x, 3)) + display(frac(3, x))$] + #math.equation(block: false, alt: "the fraction x minus 6 x squared over 4 x cubed")[$display(frac(x − 6 x^(2), 4 x^(3)))$] + #math.equation(block: false, alt: "the fraction 1 over 3 x plus 3 x to the power minus 1")[$display(frac(1, 3)) x + 3 x^(− 1)$] + #math.equation(block: false, alt: "the fraction 1 over 4 x to the power minus 2 minus the fraction 3 over 2 x to the power minus 1")[$display(frac(1, 4)) x^(− 2) − display(frac(3, 2)) x^(− 1)$] + #math.equation(block: false, alt: "the fraction 2 over x squared minus the fraction x squared over 2")[$display(frac(2, x^(2))) − display(frac(x^(2), 2))$] + #math.equation(block: false, alt: "the fraction 5 x plus 1 over open parenthesis 3 x close parenthesis squared")[$display(frac(5 x + 1, ( 3 x )^(2)))$] + #math.equation(block: false, alt: "the fraction 2 over x to the power 4 open parenthesis the fraction x squared over 4 plus the fraction x over 2 minus the fraction 1 over 4 close parenthesis")[$display(frac(2, x^(4))) ( display(frac(x^(2), 4)) + display(frac(x, 2)) − display(frac(1, 4)) )$] + #math.equation(block: false, alt: "the fraction x squared over 3 open parenthesis the fraction 2 over x to the power 4 minus the fraction 1 over 3 x squared plus the fraction 1 over 2 close parenthesis")[$display(frac(x^(2), 3)) ( display(frac(2, x^(4))) − display(frac(1, 3 x^(2))) + display(frac(1, 2)) )$] + #math.equation(block: false, alt: "the fraction 1 over 2 x to the power minus 2 plus x to the power minus 3 minus the fraction 1 over 2 x to the power minus 4")[$display(frac(1, 2)) x^(− 2) + x^(− 3) − display(frac(1, 2)) x^(− 4)$] + #math.equation(block: false, alt: "the fraction 2 over 3 x to the power minus 2 minus the fraction 1 over 9 plus the fraction 1 over 6 x squared")[$display(frac(2, 3)) x^(− 2) − display(frac(1, 9)) + display(frac(1, 6)) x^(2)$] + #math.equation(block: false, alt: "the fraction 9 over x cubed open parenthesis the fraction x cubed over 3 minus 1 minus the fraction 1 over x cubed close parenthesis")[$display(frac(9, x^(3))) ( display(frac(x^(3), 3)) − 1 − display(frac(1, x^(3))) )$] + #math.equation(block: false, alt: "the fraction x squared over 2 open parenthesis the fraction 3 over x minus the fraction 5 over x cubed plus the fraction 7 over x to the power 5 close parenthesis")[$display(frac(x^(2), 2)) ( display(frac(3, x)) − display(frac(5, x^(3))) + display(frac(7, x^(5))) )$] For Problems 45–50, use the distributive law to write each product as a sum of power functions. #math.equation(block: true, alt: "x to the power minus 1 open parenthesis x squared minus 3 x plus 2 close parenthesis")[$x^(− 1) ( x^(2) − 3 x + 2 )$] #math.equation(block: true, alt: "x minus 3 plus 2 x to the power minus 1")[$x − 3 + 2 x^(− 1)$] #math.equation(block: true, alt: "3 x to the power minus 2 open parenthesis 2 x to the power 4 plus x squared minus 4 close parenthesis")[$3 x^(− 2) ( 2 x^(4) + x^(2) − 4 )$] #math.equation(block: true, alt: "minus 3 t to the power minus 2 open parenthesis t squared minus 2 minus 4 t to the power minus 2 close parenthesis")[$− 3 t^(− 2) ( t^(2) − 2 − 4 t^(− 2) )$] #math.equation(block: true, alt: "minus 3 plus 6 t to the power minus 2 plus 12 t to the power minus 4")[$− 3 + 6 t^(− 2) + 12 t^(− 4)$] #math.equation(block: true, alt: "minus t to the power minus 3 open parenthesis 3 t squared minus 1 minus t to the power minus 2 close parenthesis")[$− t^(− 3) ( 3 t^(2) − 1 − t^(− 2) )$] #math.equation(block: true, alt: "2 u to the power minus 3 open parenthesis minus 2 u cubed minus u squared plus 3 u close parenthesis")[$2 u^(− 3) ( − 2 u^(3) − u^(2) + 3 u )$] #math.equation(block: true, alt: "minus 4 minus 2 u to the power minus 1 plus 6 u to the power minus 2")[$− 4 − 2 u^(− 1) + 6 u^(− 2)$] #math.equation(block: true, alt: "2 u to the power minus 1 open parenthesis minus 1 minus u minus 2 u squared close parenthesis")[$2 u^(− 1) ( − 1 − u − 2 u^(2) )$] For Problems 51–54, factor as indicated, writing the second factor with positive exponents only. #math.equation(block: true, alt: "4 x squared plus 16 x to the power minus 2 equals 4 x to the power minus 2 open parenthesis ? close parenthesis")[$4 x^(2) + 16 x^(− 2) = 4 x^(− 2) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "4 x to the power minus 2 open parenthesis x to the power 4 plus 4 close parenthesis")[$4 x^(− 2) ( x^(4) + 4 )$] #math.equation(block: true, alt: "20 y minus 15 y to the power minus 1 equals 5 y to the power minus 1 open parenthesis ? close parenthesis")[$20 y − 15 y^(− 1) = 5 y^(− 1) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "3 a to the power minus 3 minus 3 a plus a cubed equals a to the power minus 3 open parenthesis ? close parenthesis")[$3 a^(− 3) − 3 a + a^(3) = a^(− 3) ( " " " " "?" " " " " )$] #math.equation(block: true, alt: "a to the power minus 3 open parenthesis 3 minus 3 a to the power 4 plus a to the power 6 close parenthesis")[$a^(− 3) ( 3 − 3 a^(4) + a^(6) )$] #math.equation(block: true, alt: "2 minus 4 q to the power minus 2 minus 8 q to the power minus 4 equals 2 q to the power minus 4 open parenthesis ? close parenthesis")[$2 − 4 q^(− 2) − 8 q^(− 4) = 2 q^(− 4) ( " " " " "?" " " " " )$] + Is it true that #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power minus 2 equals x to the power minus 2 plus y to the power minus 2")[$( x + y )^(− 2) = x^(− 2) + y^(− 2)$]? Explain why or why not. + Give a numerical example to support your answer. + No, because #math.equation(block: false, alt: "the fraction 1 over open parenthesis x plus y close parenthesis squared")[$frac(1, ( x + y )^(2))$] is not #math.equation(block: false, alt: "the fraction 1 over x squared plus the fraction 1 over y squared")[$frac(1, x^(2)) + frac(1, y^(2))$]. + Let #math.equation(block: false, alt: "x equals 1")[$x = 1$], #math.equation(block: false, alt: "y equals 2")[$y = 2$], then #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power minus 2 equals open parenthesis 1 plus 2 close parenthesis to the power minus 2 equals 3 to the power minus 2 equals the fraction 1 over 9")[$( x + y )^(− 2) = ( 1 + 2 )^(− 2) = 3^(− 2) = frac(1, 9)$], but #math.equation(block: false, alt: "x to the power minus 2 plus y to the power minus 2 equals 1 to the power minus 2 plus 2 to the power minus 2 equals 1 plus the fraction 1 over 4 equals the fraction 5 over 4")[$x^(− 2) + y^(− 2) = 1^(− 2) + 2^(− 2) = 1 + frac(1, 4) = frac(5, 4)$] + Is it true that #math.equation(block: false, alt: "open parenthesis a minus b close parenthesis to the power minus 1 equals a to the power minus 1 minus b to the power minus 1")[$( a − b )^(− 1) = a^(− 1) − b^(− 1)$]? Explain why or why not. + Give a numerical example to support your answer. + Show that #math.equation(block: false, alt: "x plus x to the power minus 1 equals the fraction x squared plus 1 over x")[$x + x^(− 1) = display(frac(x^(2) + 1, x))$]. + Show that #math.equation(block: false, alt: "x cubed plus x to the power minus 3 equals the fraction x to the power 6 plus 1 over x cubed")[$x^(3) + x^(− 3) = display(frac(x^(6) + 1, x^(3)))$]. + Write #math.equation(block: false, alt: "x to the power n plus x to the power minus n")[$x^(n) + x^(− n)$] as an algebraic fraction. Justify your answer. + #math.equation(block: false, alt: "x plus x to the power minus 1 equals x plus the fraction 1 over x equals the fraction x squared over x plus the fraction 1 over x equals the fraction x squared plus 1 over x")[$x + x^(− 1) = x + display(frac(1, x)) = display(frac(x^(2), x)) + display(frac(1, x)) = display(frac(x^(2) + 1, x))$] + #math.equation(block: false, alt: "x cubed plus x to the power minus 3 equals x cubed plus the fraction 1 over x cubed equals the fraction x to the power 6 over x cubed plus the fraction 1 over x cubed equals the fraction x to the power 6 plus 1 over x cubed")[$x^(3) + x^(− 3) = x^(3) + display(frac(1, x^(3))) = display(frac(x^(6), x^(3))) + display(frac(1, x^(3))) = display(frac(x^(6) + 1, x^(3)))$] + #math.equation(block: false, alt: "x to the power n plus x to the power minus n equals x to the power n plus the fraction 1 over x to the power n equals the fraction x to the power 2 n over x to the power n plus the fraction 1 over x to the power n equals the fraction x to the power 2 n plus 1 over x squared")[$x^(n) + x^(− n) = x^(n) + display(frac(1, x^(n))) = display(frac(x^(2 n), x^(n))) + display(frac(1, x^(n))) = display(frac(x^(2 n) + 1, x^(2)))$] + Show that #math.equation(block: false, alt: "x to the power minus m plus x to the power minus n equals the fraction x to the power n plus x to the power m over x to the power n plus m")[$x^(− m) + x^(− n) = display(frac(x^(n) + x^(m), x^(n + m)))$]. + If #math.equation(block: false, alt: "m less than n")[$m < n$], show that #math.equation(block: false, alt: "x to the power minus m plus x to the power minus n equals the fraction x to the power n minus m plus 1 over x to the power n")[$x^(− m) + x^(− n) = display(frac(x^(n − m) + 1, x^(n)))$]. By rewriting the expressions in Problems 59–62 as fractions, verify that the laws of exponents hold for negative exponents. Show where you apply the corresponding law for positive exponents. Here is the fourth law as an example: #math.equation(block: true, alt: "open parenthesis a b close parenthesis to the power minus 3, equals the fraction 1 over open parenthesis a b close parenthesis cubed equals the fraction 1 over a cubed b cubed, By the fourth law of exponents.; equals the fraction 1 over a cubed times the fraction 1 over b cubed equals a to the power minus 3 b to the power minus 3")[$( a b )^(− 3) & = frac(1, ( a b )^(3)) = frac(1, a^(3) b^(3)) & & "By the fourth law of exponents." \ & = frac(1, a^(3)) ⋅ frac(1, b^(3)) = a^(− 3) b^(− 3)$] #math.equation(block: true, alt: "a to the power minus 2 a to the power minus 3 equals a to the power minus 5")[$a^(− 2) a^(− 3) = a^(− 5)$] #math.equation(block: true, alt: "a to the power minus 2 a to the power minus 3, equals the fraction 1 over a squared times the fraction 1 over a cubed equals the fraction 1 over a squared times a cubed; equals the fraction 1 over a to the power 2 plus 3, By the first law of exponents.; equals the fraction 1 over a to the power 5 equals a to the power minus 5")[$a^(− 2) a^(− 3) & = frac(1, a^(2)) ⋅ frac(1, a^(3)) = frac(1, a^(2) ⋅ a^(3)) \ & = frac(1, a^(2 + 3)) & & "By the first law of exponents." \ & = frac(1, a^(5)) = a^(− 5)$] #math.equation(block: true, alt: "the fraction a to the power minus 6 over a to the power minus 2 equals a to the power minus 4")[$display(frac(a^(− 6), a^(− 2))) = a^(− 4)$] #math.equation(block: true, alt: "the fraction a to the power minus 2 over a to the power minus 6 equals a to the power 4")[$display(frac(a^(− 2), a^(− 6))) = a^(4)$] #math.equation(block: true, alt: "the fraction a to the power minus 2 over a to the power minus 6, equals a to the power minus 2 divided by a to the power minus 6 equals the fraction 1 over a squared divided by the fraction 1 over a to the power 6; equals the fraction 1 over a squared times the fraction a to the power 6 over 1 equals the fraction a to the power 6 over a squared; equals a to the power 6 minus 2, By the second law of exponents.; equals a to the power 4")[$frac(a^(− 2), a^(− 6)) & = a^(− 2) ÷ a^(− 6) = frac(1, a^(2)) ÷ frac(1, a^(6)) \ & = frac(1, a^(2)) ⋅ frac(a^(6), 1) = frac(a^(6), a^(2)) \ & = a^(6 − 2) & & "By the second law of exponents." \ & = a^(4)$] #math.equation(block: true, alt: "open parenthesis a to the power minus 2 close parenthesis to the power minus 3 equals a to the power 6")[$attach(( a^(− 2) ), t: − 3) = a^(6)$]