#set document(title: "10.2 Evaluate and Graph Exponential Functions", 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")) == 10.2#h(0.6em)Evaluate and Graph Exponential Functions #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Simplify: #math.equation(block: false, alt: "open parenthesis the fraction x cubed over x squared close parenthesis .")[$( frac(x^(3), x^(2)) ) .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x")[$x$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Evaluate: ⓐ #math.equation(block: false, alt: "2 to the power 0")[$2^(0)$] ⓑ #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power 0 .")[$attach(( frac(1, 3) ), t: 0) .$] #linebreak() If you missed this problem, review . #solutionbox[ ⓐ #math.equation(block: false, alt: "1")[$1$]; ⓑ #math.equation(block: false, alt: "1")[$1$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Evaluate: ⓐ #math.equation(block: false, alt: "2 to the power −1")[$2^(−1)$] ⓑ #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power −1 .")[$attach(( frac(1, 3) ), t: −1) .$] #linebreak() If you missed this problem, review . #solutionbox[ ⓐ #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]; ⓑ #math.equation(block: false, alt: "3")[$3$] ] ] === Graph Exponential Functions The functions we have studied so far do not give us a model for many naturally occurring phenomena. From the growth of populations and the spread of viruses to radioactive decay and compounding interest, the models are very different from what we have studied so far. These models involve exponential functions. An #strong[exponential function] is a function of the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] where #math.equation(block: false, alt: "a greater than 0")[$a > 0$] and #math.equation(block: false, alt: "a not equal to 1 .")[$a ≠ 1 .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Function] An exponential function, where #math.equation(block: false, alt: "a greater than 0")[$a > 0$] and #math.equation(block: false, alt: "a not equal to 1 ,")[$a ≠ 1 ,$] is a function of the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] ] Notice that in this function, the variable is the exponent. In our functions so far, the variables were the base. #figure(figph[This figure shows three functions: f of x equals negative 3x plus 4, which is marked as linear; f of x equals 2x squared plus 5x minus 3, which is marked as quadratic; and f of x equals 6 to the x power, which is marked exponential. For the functions marked linear and quadratic, x is the base. For the function marked exponential, x is the exponent for the base 6.], alt: "This figure shows three functions: f of x equals negative 3x plus 4, which is marked as linear; f of x equals 2x squared plus 5x minus 3, which is marked as quadratic; and f of x equals 6 to the x power, which is marked exponential. For the functions marked linear and quadratic, x is the base. For the function marked exponential, x is the exponent for the base 6.", caption: none) Our definition says #math.equation(block: false, alt: "a not equal to 1 .")[$a ≠ 1 .$] If we let #math.equation(block: false, alt: "a equals 1 ,")[$a = 1 ,$] then #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] becomes #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1 to the power x .")[$f ( x ) = 1^(x) .$] Since #math.equation(block: false, alt: "1 to the power x equals 1")[$1^(x) = 1$] for all real numbers, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1 to the power .")[$f ( x ) = 1^ .$] This is the constant function. Our definition also says #math.equation(block: false, alt: "a greater than 0 .")[$a > 0 .$] If we let a base be negative, say #math.equation(block: false, alt: "−4 ,")[$−4 ,$] then #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis −4 close parenthesis to the power x")[$f ( x ) = attach(( −4 ), t: x)$] is not a real number when #math.equation(block: false, alt: "x equals the fraction 1 over 2 .")[$x = frac(1, 2) .$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis, equals, open parenthesis −4 close parenthesis to the power x; f open parenthesis the fraction 1 over 2 close parenthesis, equals, open parenthesis −4 close parenthesis to the power the fraction 1 over 2; f open parenthesis the fraction 1 over 2 close parenthesis, equals, the square root of −4 not a real number")[$f ( x ) & = & attach(( −4 ), t: x) \ f ( frac(1, 2) ) & = & attach(( −4 ), t: frac(1, 2)) \ f ( frac(1, 2) ) & = & sqrt(−4) #h(1em) "not a real number"$]In fact, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis −4 close parenthesis to the power x")[$f ( x ) = attach(( −4 ), t: x)$] would not be a real number any time #math.equation(block: false, alt: "x")[$x$] is a fraction with an even denominator. So our definition requires #math.equation(block: false, alt: "a greater than 0 .")[$a > 0 .$] By graphing a few exponential functions, we will be able to see their unique properties. #examplebox("Example 1")[][ On the same coordinate system graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x .")[$g ( x ) = 3^(x) .$] #solutionbox[ We will use point plotting to graph the functions. #figure(figph[This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 3 to the x power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 3 to the negative 2 power equals 1 divided by 3 squared which equals 1 over 9, (negative 2, 1 over 9). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 3 to the negative 1 power equals 1 divided by 3 to the first power which equals 1 over 3, (negative 1, 1 over 3). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 3 to the 1 power equals 9, (1, 3). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 3 to the 2 power equals 9, (2, 9). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 3 to the 3 power equals 27, (3, 27).], alt: "This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 3 to the x power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 3 to the negative 2 power equals 1 divided by 3 squared which equals 1 over 9, (negative 2, 1 over 9). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 3 to the negative 1 power equals 1 divided by 3 to the first power which equals 1 over 3, (negative 1, 1 over 3). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 3 to the 1 power equals 9, (1, 3). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 3 to the 2 power equals 9, (2, 9). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 3 to the 3 power equals 27, (3, 27).", caption: none) #figure(figph[This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3).], alt: "This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Graph: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 to the power x .")[$f ( x ) = 4^(x) .$] #solutionbox[ #figure(figph[This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 4).], alt: "This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 4).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Graph: #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 to the power x .")[$g ( x ) = 5^(x) .$] #solutionbox[ #figure(figph[This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 5).], alt: "This figure shows a curve that slopes swiftly upward from just above (negative 3, 0) through (0, 1) up to (1, 5).", caption: none) ] ] If we look at the graphs from the previous Example and Try Its, we can identify some of the properties of exponential functions. The graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x ,")[$g ( x ) = 3^(x) ,$] as well as the graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 to the power x")[$f ( x ) = 4^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 to the power x ,")[$g ( x ) = 5^(x) ,$] all have the same basic shape. This is the shape we expect from an exponential function where #math.equation(block: false, alt: "a greater than 1 .")[$a > 1 .$] We notice, that for each function, the graph contains the point #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis .")[$( 0 , 1 ) .$] This make sense because #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$] for any #emph[a]. The graph of each function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] also contains the point #math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis .")[$( 1 , a ) .$] The graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] contained #math.equation(block: false, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$] and the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x")[$g ( x ) = 3^(x)$] contained #math.equation(block: false, alt: "open parenthesis 1 , 3 close parenthesis .")[$( 1 , 3 ) .$] This makes sense as #math.equation(block: false, alt: "a to the power 1 equals a .")[$a^(1) = a .$] Notice too, the graph of each function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] also contains the point #math.equation(block: false, alt: "open parenthesis −1 , the fraction 1 over a close parenthesis .")[$( −1 , frac(1, a) ) .$] The graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] contained #math.equation(block: false, alt: "open parenthesis −1 , the fraction 1 over 2 close parenthesis")[$( −1 , frac(1, 2) )$] and the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x")[$g ( x ) = 3^(x)$] contained #math.equation(block: false, alt: "open parenthesis −1 , the fraction 1 over 3 close parenthesis .")[$( −1 , frac(1, 3) ) .$] This makes sense as #math.equation(block: false, alt: "a to the power −1 equals the fraction 1 over a .")[$a^(−1) = frac(1, a) .$] What is the domain for each function? From the graphs we can see that the domain is the set of all real numbers. There is no restriction on the domain. We write the domain in interval notation as #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$] Look at each graph. What is the range of the function? The graph never hits the #math.equation(block: false, alt: "x")[$x$]-axis. The range is all positive numbers. We write the range in interval notation as #math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis .")[$( 0 , ∞ ) .$] Whenever a graph of a function approaches a line but never touches it, we call that line an #strong[asymptote]. For the exponential functions we are looking at, the graph approaches the #math.equation(block: false, alt: "x")[$x$]-axis very closely but will never cross it, we call the line #math.equation(block: false, alt: "y equals 0 ,")[$y = 0 ,$] the #emph[x]-axis, a horizontal asymptote. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of the Graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] when #math.equation(block: false, alt: "a greater than 1")[$a > 1$]] #figure(table( columns: 2, align: left, inset: 6pt, table.header([Domain], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]]), [Range], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [#emph[x]-intercept], [None], [#emph[y]-intercept], [#math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis , open parenthesis −1 , the fraction 1 over a close parenthesis")[$( 1 , a ) , #h(0.2em) ( −1 , frac(1, a) )$]], [Asymptote], [#math.equation(block: false, alt: "x")[$x$]-axis, the line #math.equation(block: false, alt: "y equals 0")[$y = 0$]], )) #figure(figph[This figure shows a curve that slopes upward from (negative 1, 1 over a) through (0, 1), up to (1, a).], alt: "This figure shows a curve that slopes upward from (negative 1, 1 over a) through (0, 1), up to (1, a).", caption: none) ] Our definition of an exponential function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$] says #math.equation(block: false, alt: "a greater than 0 ,")[$a > 0 ,$] but the examples and discussion so far has been about functions where #math.equation(block: false, alt: "a greater than 1 .")[$a > 1 .$] What happens when #math.equation(block: false, alt: "0 less than a less than 1")[$0 < a < 1$]? The next example will explore this possibility. #examplebox("Example 2")[][ On the same coordinate system, graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 2) ), t: x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x .")[$g ( x ) = attach(( frac(1, 3) ), t: x) .$] #solutionbox[ We will use point plotting to graph the functions. #figure(figph[This table has seven rows and five columns. The first row is header row and reads x, f of x, equals 1 over 2 to the x power, (x, f of x), g of x equals 1 over 3 to the x power, and (x, g of x). The second row reads negative 2, 1 over 2 to the negative 2 power equals 2 squared which equals 4, (negative 2, 4), 3 to the negative 2 power equals 3 squared which equals 9, (negative 2, 9). The third row reads negative 1, 1 over 2 to the negative 1 power equals 2 to the first power which equals 2, (negative 1, 2), 1 over 3 to the negative 1 power equals 3 to the first power which equals 3, (negative 1, 3). The fourth row reads 0, 1 over 2 to the 0 power equals 1, (0, 1), 1 over 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 1 over 2 to the 1 power equals 1 over 2, (1, 1 over 2), 1 over 3 to the 1 power equals 1 over 3, (1, 1 over 3). The sixth row reads 2, 1 over 2 to the 2 power equals 1 over 4, (2, 1 over 4), 1 over 3 to the 2 power equals 1 over 9, (2, 1 over 9). The seventh row reads 3, 1 over 2 to the 3 power equals 1 over 8, (3, 1 over 8), 1 over 3 to the 3 power equals 1 over 27, (3, 1 over 27).], alt: "This table has seven rows and five columns. The first row is header row and reads x, f of x, equals 1 over 2 to the x power, (x, f of x), g of x equals 1 over 3 to the x power, and (x, g of x). The second row reads negative 2, 1 over 2 to the negative 2 power equals 2 squared which equals 4, (negative 2, 4), 3 to the negative 2 power equals 3 squared which equals 9, (negative 2, 9). The third row reads negative 1, 1 over 2 to the negative 1 power equals 2 to the first power which equals 2, (negative 1, 2), 1 over 3 to the negative 1 power equals 3 to the first power which equals 3, (negative 1, 3). The fourth row reads 0, 1 over 2 to the 0 power equals 1, (0, 1), 1 over 3 to the 0 power equals 1, (0, 1). The fifth row reads 1, 1 over 2 to the 1 power equals 1 over 2, (1, 1 over 2), 1 over 3 to the 1 power equals 1 over 3, (1, 1 over 3). The sixth row reads 2, 1 over 2 to the 2 power equals 1 over 4, (2, 1 over 4), 1 over 3 to the 2 power equals 1 over 9, (2, 1 over 9). The seventh row reads 3, 1 over 2 to the 3 power equals 1 over 8, (3, 1 over 8), 1 over 3 to the 3 power equals 1 over 27, (3, 1 over 27).", caption: none) #figure(figph[This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 2), (0, 1), and (1, 1 over 2). The second curve is marked in red and passes through the points (negative 1, 3), (0, 1), and (1, 1 over 3).], alt: "This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 2), (0, 1), and (1, 1 over 2). The second curve is marked in red and passes through the points (negative 1, 3), (0, 1), and (1, 1 over 3).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Graph: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 4 close parenthesis to the power x .")[$f ( x ) = attach(( frac(1, 4) ), t: x) .$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 4), (0, 1) to a point just above (3, 0).], alt: "This figure shows a curve that passes through (negative 1, 4), (0, 1) to a point just above (3, 0).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Graph: #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 5 close parenthesis to the power x .")[$g ( x ) = attach(( frac(1, 5) ), t: x) .$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 5), (0, 1) to a point just above (3, 0).], alt: "This figure shows a curve that passes through (negative 1, 5), (0, 1) to a point just above (3, 0).", caption: none) ] ] Now let’s look at the graphs from the previous Example and Try Its so we can now identify some of the properties of exponential functions where #math.equation(block: false, alt: "0 less than a less than 1 .")[$0 < a < 1 .$] The graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 2) ), t: x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$g ( x ) = attach(( frac(1, 3) ), t: x)$] as well as the graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 4 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 4) ), t: x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 5 close parenthesis to the power x")[$g ( x ) = attach(( frac(1, 5) ), t: x)$] all have the same basic shape. While this is the shape we expect from an exponential function where #math.equation(block: false, alt: "0 less than a less than 1 ,")[$0 < a < 1 ,$] the graphs go down from left to right while the previous graphs, when #math.equation(block: false, alt: "a greater than 1 ,")[$a > 1 ,$] went from up from left to right. We notice that for each function, the graph still contains the point (0, 1). This make sense because #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$] for any #emph[a]. As before, the graph of each function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x ,")[$f ( x ) = a^(x) ,$] also contains the point #math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis .")[$( 1 , a ) .$] The graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 2) ), t: x)$] contained #math.equation(block: false, alt: "open parenthesis 1 , the fraction 1 over 2 close parenthesis")[$( 1 , frac(1, 2) )$] and the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$g ( x ) = attach(( frac(1, 3) ), t: x)$] contained #math.equation(block: false, alt: "open parenthesis 1 , the fraction 1 over 3 close parenthesis .")[$( 1 , frac(1, 3) ) .$] This makes sense as #math.equation(block: false, alt: "a to the power 1 equals a .")[$a^(1) = a .$] Notice too that the graph of each function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x ,")[$f ( x ) = a^(x) ,$] also contains the point #math.equation(block: false, alt: "open parenthesis −1 , the fraction 1 over a close parenthesis .")[$( −1 , frac(1, a) ) .$] The graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 2) ), t: x)$] contained #math.equation(block: false, alt: "open parenthesis −1 , 2 close parenthesis")[$( −1 , 2 )$] and the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$g ( x ) = attach(( frac(1, 3) ), t: x)$] contained #math.equation(block: false, alt: "open parenthesis −1 , 3 close parenthesis .")[$( −1 , 3 ) .$] This makes sense as #math.equation(block: false, alt: "a to the power −1 equals the fraction 1 over a .")[$a^(−1) = frac(1, a) .$] What is the domain and range for each function? From the graphs we can see that the domain is the set of all real numbers and we write the domain in interval notation as #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$] Again, the graph never hits the #math.equation(block: false, alt: "x")[$x$]-axis. The range is all positive numbers. We write the range in interval notation as #math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis .")[$( 0 , ∞ ) .$] We will summarize these properties in the chart below. Which also include when #math.equation(block: false, alt: "a greater than 1 .")[$a > 1 .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of the Graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x")[$f ( x ) = a^(x)$]] #figure(table( columns: 4, align: left, inset: 6pt, [when #math.equation(block: false, alt: "a greater than 1")[$a > 1$]], [when #math.equation(block: false, alt: "0 less than a less than 1")[$0 < a < 1$]], [], [], [Domain], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [Domain], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [Range], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [Range], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [#math.equation(block: false, alt: "x")[$x$]-intercept], [none], [#math.equation(block: false, alt: "x")[$x$]-intercept], [none], [#math.equation(block: false, alt: "y")[$y$]-intercept], [#math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]], [#math.equation(block: false, alt: "y")[$y$]-intercept], [#math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis , open parenthesis −1 , the fraction 1 over a close parenthesis")[$( 1 , a ) , #h(0.2em) ( −1 , frac(1, a) )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis , open parenthesis −1 , the fraction 1 over a close parenthesis")[$( 1 , a ) , #h(0.2em) ( −1 , frac(1, a) )$]], [Asymptote], [#math.equation(block: false, alt: "x")[$x$]-axis, the line #math.equation(block: false, alt: "y equals 0")[$y = 0$]], [Asymptote], [#math.equation(block: false, alt: "x")[$x$]-axis, the line #math.equation(block: false, alt: "y equals 0")[$y = 0$]], [Basic shape], [increasing], [Basic shape], [decreasing], )) #figure(figph[This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).], alt: "This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).", caption: none) ] It is important for us to notice that both of these graphs are one-to-one, as they both pass the horizontal line test. This means the exponential function will have an inverse. We will look at this later. When we graphed quadratic functions, we were able to graph using translation rather than just plotting points. Will that work in graphing exponential functions? #examplebox("Example 3")[][ On the same coordinate system graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x plus 1 .")[$g ( x ) = 2^(x + 1) .$] #solutionbox[ We will use point plotting to graph the functions. #figure(figph[This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 2 to the x plus 1 power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 2 to the negative 2 plus 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 2, 1 over 2). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 2 to the negative 1 plus 1 power equals 2 to the 0 power which equals 1, (negative 1, 1). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 2 to the 0 plus 1 power equals 2 to the 1 power which equals 2, (0, 2). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 2 to the 1 plus 1 power equals 2 to the second power which equals 4, (1, 4). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 2 to the 2 plus 1 power equals 2 to the third power which equals 8, (2, 8). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 2 to the 3 plus 1 power equals 2 to the fourth power which equals 16, (3, 16).], alt: "This table has seven rows and five columns. The first row is header row and reads x, f of x equals 2 to the x power, (x, f of x), g of x equals 2 to the x plus 1 power, and (x, g of x). The second row reads negative 2, 2 to the negative 2 power equals 1 divided by 2 squared which equals 1 over 4, (negative 2, 1 over 4), 2 to the negative 2 plus 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 2, 1 over 2). The third row reads negative 1, 2 to the negative 1 power equals 1 divided by 2 to the first power which equals 1 over 2, (negative 1, 1 over 2), 2 to the negative 1 plus 1 power equals 2 to the 0 power which equals 1, (negative 1, 1). The fourth row reads 0, 2 to the 0 power equals 1, (0, 1), 2 to the 0 plus 1 power equals 2 to the 1 power which equals 2, (0, 2). The fifth row reads 1, 2 to the 1 power equals 2, (1, 2), 2 to the 1 plus 1 power equals 2 to the second power which equals 4, (1, 4). The sixth row reads 2, 2 to the 2 power equals 4, (2, 4), 2 to the 2 plus 1 power equals 2 to the third power which equals 8, (2, 8). The seventh row reads 3, 2 to the 3 power equals 8, (3, 8), 2 to the 3 plus 1 power equals 2 to the fourth power which equals 16, (3, 16).", caption: none) #figure(figph[This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1), (0, 2) and (1, 4).], alt: "This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second curve is marked in red and passes through the points (negative 1, 1), (0, 2) and (1, 4).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On the same coordinate system, graph: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x minus 1 .")[$g ( x ) = 2^(x − 1) .$] #solutionbox[ #figure(figph[This figure shows the graphs of two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 1 power is marked in red and passes through the points (0, 1 over 2), (1, 1), and (2, 2).], alt: "This figure shows the graphs of two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 1 power is marked in red and passes through the points (0, 1 over 2), (1, 1), and (2, 2).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On the same coordinate system, graph: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 to the power x")[$f ( x ) = 3^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x plus 1 .")[$g ( x ) = 3^(x + 1) .$] #solutionbox[ #figure(figph[This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x plus 1 power is marked in red and passes through the points (negative 2, 1 over 3), (negative 1, 1), and (0, 3).], alt: "This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x plus 1 power is marked in red and passes through the points (negative 2, 1 over 3), (negative 1, 1), and (0, 3).", caption: none) ] ] Looking at the graphs of the functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x plus 1")[$g ( x ) = 2^(x + 1)$] in the last example, we see that adding one in the exponent caused a horizontal shift of one unit to the left. Recognizing this pattern allows us to graph other functions with the same pattern by translation. Let’s now consider another situation that might be graphed more easily by translation, once we recognize the pattern. #examplebox("Example 4")[][ On the same coordinate system graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 to the power x")[$f ( x ) = 3^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x minus 2 .")[$g ( x ) = 3^(x) − 2 .$] #solutionbox[ We will use point plotting to graph the functions. #figure(figph[This table has five rows and six columns. The first row is header row and reads x, f of x equals 3 to the x power, (x, f of x), g of x equals 3 to the x power minus 2, and (x, g of x). The second row reads negative 2, 3 to the negative 2 power equals 1 over 9, (negative 2, 1 over 9), 3 to the negative 2 power minus 2 equals 1 over 9 minus 2 which equals negative 17 over 9, (negative 2, negative 17 over 9). The third row reads negative 1, 3 to the negative 1 power equals 1 over 3, (negative 1, 1 over 3), 3 to the negative 1 power minus 2 equals 1 over 3 minus 2 which equals negative 5 over 3, (negative 1, negative 5 over 3). The fourth row reads 0, 3 to the 0 power equals 1, (0, 1), 3 to the 0 power minus 2 equals 1 minus 2 which equals negative 1, (0, negative 1). The fifth row reads 1, 3 to the 1 power equals 3, (1, 3), 3 to the 1 power minus 2 equals 3 minus 2 which equals 1, (1, 1). The sixth row reads 2, 3 squared equals 9, (2, 9), 3 squared minus 2 equals 9 minus 2 which equals 7, (2, 7).], alt: "This table has five rows and six columns. The first row is header row and reads x, f of x equals 3 to the x power, (x, f of x), g of x equals 3 to the x power minus 2, and (x, g of x). The second row reads negative 2, 3 to the negative 2 power equals 1 over 9, (negative 2, 1 over 9), 3 to the negative 2 power minus 2 equals 1 over 9 minus 2 which equals negative 17 over 9, (negative 2, negative 17 over 9). The third row reads negative 1, 3 to the negative 1 power equals 1 over 3, (negative 1, 1 over 3), 3 to the negative 1 power minus 2 equals 1 over 3 minus 2 which equals negative 5 over 3, (negative 1, negative 5 over 3). The fourth row reads 0, 3 to the 0 power equals 1, (0, 1), 3 to the 0 power minus 2 equals 1 minus 2 which equals negative 1, (0, negative 1). The fifth row reads 1, 3 to the 1 power equals 3, (1, 3), 3 to the 1 power minus 2 equals 3 minus 2 which equals 1, (1, 1). The sixth row reads 2, 3 squared equals 9, (2, 9), 3 squared minus 2 equals 9 minus 2 which equals 7, (2, 7).", caption: none) #figure(figph[This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second curve is marked in red and passes through the points (negative 1, negative 5 over 3), (0, negative 1), and (1, 1).], alt: "This figure shows two curves. The first curve is marked in blue and passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second curve is marked in red and passes through the points (negative 1, negative 5 over 3), (0, negative 1), and (1, 1).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On the same coordinate system, graph: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 to the power x")[$f ( x ) = 3^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x plus 2 .")[$g ( x ) = 3^(x) + 2 .$] #solutionbox[ #figure(figph[This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 1, 7 over 3), (0, 3) and (1, 5).], alt: "This figure shows the graphs of two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 1, 7 over 3), (0, 3) and (1, 5).", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ On the same coordinate system, graph: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 to the power x")[$f ( x ) = 4^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 4 to the power x minus 2 .")[$g ( x ) = 4^(x) − 2 .$] #solutionbox[ #figure(figph[This figure shows the graphs of two functions. The first function f of x equals 4 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 4), (0, 1) and (1, 4). The second function g of x equals 4 to the x power minus 2 is marked in red and passes through the points (negative 1, negative 7 over 4), (0, negative 1), and (1, 2).], alt: "This figure shows the graphs of two functions. The first function f of x equals 4 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 4), (0, 1) and (1, 4). The second function g of x equals 4 to the x power minus 2 is marked in red and passes through the points (negative 1, negative 7 over 4), (0, negative 1), and (1, 2).", caption: none) ] ] Looking at the graphs of the functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 to the power x")[$f ( x ) = 3^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 to the power x minus 2")[$g ( x ) = 3^(x) − 2$] in the last example, we see that subtracting 2 caused a vertical shift of down two units. Notice that the horizontal asymptote also shifted down 2 units. Recognizing this pattern allows us to graph other functions with the same pattern by translation. All of our exponential functions have had either an integer or a rational number as the base. We will now look at an exponential function with an irrational number as the base. Before we can look at this exponential function, we need to define the irrational number, #emph[e]. This number is used as a base in many applications in the sciences and business that are modeled by exponential functions. The number is defined as the value of #math.equation(block: false, alt: "open parenthesis 1 plus the fraction 1 over n close parenthesis to the power n")[$attach(( 1 + frac(1, n) ), t: n)$] as #emph[n] gets larger and larger. We say, as #emph[n] approaches infinity, or increases without bound. The table shows the value of #math.equation(block: false, alt: "open parenthesis 1 plus the fraction 1 over n close parenthesis to the power n")[$attach(( 1 + frac(1, n) ), t: n)$] for several values of #math.equation(block: false, alt: "n .")[$n .$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "open parenthesis 1 plus the fraction 1 over n close parenthesis to the power n")[$attach(( 1 + frac(1, n) ), t: n)$]]), [1], [2], [2], [2.25], [5], [2.48832], [10], [2.59374246], [100], [2.704813829…], [1,000], [2.716923932…], [10,000], [2.718145927…], [100,000], [2.718268237…], [1,000,000], [2.718280469…], [1,000,000,000], [2.718281827…], )) If carried out to even larger values of #emph[n], we get #math.equation(block: true, alt: "e approximately equals 2.718281828")[$e ≈ 2.718281828$]The number #emph[e] is like the number #math.equation(block: false, alt: "π")[$π$] in that we use a symbol to represent it because its decimal representation never stops or repeats. The irrational number #emph[e] is called the #strong[natural base]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Natural Base #math.equation(block: false, alt: "e")[$e$]] The number #emph[e] is defined as the value of #math.equation(block: false, alt: "open parenthesis 1 plus the fraction 1 over n close parenthesis to the power n ,")[$attach(( 1 + frac(1, n) ), t: n) ,$] as #emph[n] increases without bound. We say, as #emph[n] approaches infinity, #math.equation(block: true, alt: "e approximately equals 2.718281828.. .")[$e ≈ 2.718281828.. .$] ] The exponential function whose base is #math.equation(block: false, alt: "e ,")[$e ,$] #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$] is called the #strong[natural exponential function]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Natural Exponential Function] The natural exponential function is an exponential function whose base is #math.equation(block: false, alt: "e")[$e$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$]The domain is #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$] and the range is #math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis .")[$( 0 , ∞ ) .$] ] Let’s graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$] on the same coordinate system as #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x")[$g ( x ) = 2^(x)$] and #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals 3 to the power x .")[$h ( x ) = 3^(x) .$] #figure(figph[This figure shows the graphs of three functions. The first function, f of x equals 2 to the x, is marked in red and passes through the points (negative 1, negative 1 over 2), (0, negative 1), and (2, 1). The second function, f of x equals 3 to the x power, is marked in green and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The third function, f of x equals e to the x power, is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over e), (0, 1) and (0, e).], alt: "This figure shows the graphs of three functions. The first function, f of x equals 2 to the x, is marked in red and passes through the points (negative 1, negative 1 over 2), (0, negative 1), and (2, 1). The second function, f of x equals 3 to the x power, is marked in green and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1) and (1, 3). The third function, f of x equals e to the x power, is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over e), (0, 1) and (0, e).", caption: none) Notice that the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$] is “between” the graphs of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x")[$g ( x ) = 2^(x)$] and #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals 3 to the power x .")[$h ( x ) = 3^(x) .$] Does this make sense as #math.equation(block: false, alt: "2 less than e less than 3")[$2 < e < 3$]? === Solve Exponential Equations Equations that include an exponential expression #math.equation(block: false, alt: "a to the power x")[$a^(x)$] are called exponential equations. To solve them we use a property that says as long as #math.equation(block: false, alt: "a greater than 0")[$a > 0$] and #math.equation(block: false, alt: "a not equal to 1 ,")[$a ≠ 1 ,$] if #math.equation(block: false, alt: "a to the power x equals a to the power y")[$a^(x) = a^(y)$] then it is true that #math.equation(block: false, alt: "x equals y .")[$x = y .$] In other words, in an exponential equation, if the bases are equal then the exponents are equal. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[One-to-One Property of Exponential Equations] For #math.equation(block: false, alt: "a greater than 0")[$a > 0$] and #math.equation(block: false, alt: "a not equal to 1 ,")[$a ≠ 1 ,$] #math.equation(block: true, alt: "If a to the power x equals a to the power y , then x equals y .")[$"If" #h(0.2em) a^(x) = a^(y) , #h(0.2em) "then" #h(0.2em) x = y .$] ] To use this property, we must be certain that both sides of the equation are written with the same base. #examplebox("Example 5")[How to Solve an Exponential Equation][ Solve: #math.equation(block: false, alt: "3 to the power 2 x minus 5 equals 27 .")[$3^(2 x − 5) = 27 .$] #solutionbox[ #figure(figph[Step 1 is to write both sides of the equation with the same base. This means that, since the left side has base 3, we write the right side with base 3. Hence, 27 equals 3 to the third power. We have 3 to the 2x minus 5 power equals 27, which we write as 3 to the 2x minus 5 power equals 3 cubed.], alt: "Step 1 is to write both sides of the equation with the same base. This means that, since the left side has base 3, we write the right side with base 3. Hence, 27 equals 3 to the third power. We have 3 to the 2x minus 5 power equals 27, which we write as 3 to the 2x minus 5 power equals 3 cubed.", caption: none) #figure(figph[Step 2 is to write a new equation by setting the exponents equal. This means that, since the bases are the same, the exponents must be equal. Hence, 2x minus 5 equals 3.], alt: "Step 2 is to write a new equation by setting the exponents equal. This means that, since the bases are the same, the exponents must be equal. Hence, 2x minus 5 equals 3.", caption: none) #figure(figph[Step 3 is to solve the equation. This means that we add 5 to each side and divide by 2. Hence, 2x equals 8, which means that x equals 4.], alt: "Step 3 is to solve the equation. This means that we add 5 to each side and divide by 2. Hence, 2x equals 8, which means that x equals 4.", caption: none) #figure(figph[Step 3 is to check the solution. This means that we substitute x equals 4 into the original equation. We start with 3 to the 2 x minus 5 power equals 27. We want to know whether 3 to the 2 times 4 minus 5 power equals 27. This becomes a question of whether 3 cubed equals 27, which of course is true.], alt: "Step 3 is to check the solution. This means that we substitute x equals 4 into the original equation. We start with 3 to the 2 x minus 5 power equals 27. We want to know whether 3 to the 2 times 4 minus 5 power equals 27. This becomes a question of whether 3 cubed equals 27, which of course is true.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "3 to the power 3 x minus 2 equals 81 .")[$3^(3 x − 2) = 81 .$] #solutionbox[ #math.equation(block: true, alt: "x equals 2")[$x = 2$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "7 to the power x minus 3 equals 7 .")[$7^(x − 3) = 7 .$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] ] The steps are summarized below. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[How to Solve an Exponential Equation] + Write both sides of the equation with the same base, if possible. + Write a new equation by setting the exponents equal. + Solve the equation. + Check the solution. ] In the next example, we will use our properties on exponents. #examplebox("Example 6")[][ Solve #math.equation(block: false, alt: "the fraction e to the power x squared over e cubed equals e to the power 2 x")[$frac(e^(x^(2)), e^(3)) = e^(2 x)$]. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "the fraction e to the power x squared over e cubed equals e to the power 2 x")[$#h(2.4em) frac(e^(x^(2)), e^(3)) = e^(2 x)$]]), [Use the Property of Exponents: #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 .")[$frac(a^(m), a^(n)) = a^(m − n) .$]], [#math.equation(block: false, alt: "e to the power x squared minus 3 equals e to the power 2 x")[$#h(2.45em) e^(x^(2) − 3) = e^(2 x)$]], [Write a new equation by setting the exponents #linebreak() equal.], [#math.equation(block: false, alt: "x squared minus 3 equals 2 x")[$#h(1.1em) x^(2) − 3 = 2 x #h(0.2em)$]], [Solve the equation.], [#math.equation(block: false, alt: "x squared minus 2 x minus 3 equals 0")[$x^(2) − 2 x − 3 = 0 #h(0.65em)$]], [], [#math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 1 close parenthesis equals 0")[$( x − 3 ) ( x + 1 ) = 0 #h(0.7em)$]], [], [#math.equation(block: false, alt: "x equals 3 , x equals minus 1")[$x = 3 , #h(0.4em) x = − 1$]], [Check the solutions.], [], [#figure(figph[Step-by-step verification that e^(x^2) / e^3 = e^(2x) for x=3 and x=-1, demonstrating equality through algebraic simplification of exponents.], alt: "Step-by-step verification that e^(x^2) / e^3 = e^(2x) for x=3 and x=-1, demonstrating equality through algebraic simplification of exponents.", caption: none)], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "the fraction e to the power x squared over e to the power x equals e squared .")[$frac(e^(x^(2)), e^(x)) = e^(2) .$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals 2")[$x = −1 , x = 2$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "the fraction e to the power x squared over e to the power x equals e to the power 6 .")[$frac(e^(x^(2)), e^(x)) = e^(6) .$] #solutionbox[ #math.equation(block: true, alt: "x equals −2 , x equals 3")[$x = −2 , x = 3$] ] ] === Use Exponential Models in Applications Exponential functions model many situations. If you own a bank account, you have experienced the use of an exponential function. There are two formulas that are used to determine the balance in the account when interest is earned. If a principal, #emph[P], is invested at an interest rate, #emph[r], for #emph[t] years, the new balance, #emph[A], will depend on how often the interest is compounded. If the interest is compounded #emph[n] times a year we use the formula #math.equation(block: false, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t .")[$A = P attach(( 1 + frac(r, n) ), t: n t) .$] If the interest is compounded continuously, we use the formula #math.equation(block: false, alt: "A equals P e to the power r t .")[$A = P e^(r t) .$] These are the formulas for #strong[compound interest]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Interest] For a principal, #emph[P], invested at an interest rate, #emph[r], for #emph[t] years, the new balance, #emph[A], is: #math.equation(block: true, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t, when compounded n times a year.; A equals P e to the power r t, when compounded continuously.")[$A = P attach(( 1 + frac(r, n) ), t: n t) & & & & & "when compounded" #h(0.2em) n #h(0.2em) "times a year." \ A = P e^(r t) & & & & & "when compounded continuously."$] ] As you work with the Interest formulas, it is often helpful to identify the values of the variables first and then substitute them into the formula. #examplebox("Example 7")[][ A total of #math.equation(block: false, alt: "$ 10,000")[$"$" 10,000$] was invested in a college fund for a new grandchild. If the interest rate is #math.equation(block: false, alt: "5 % ,")[$5 "%" ,$] how much will be in the account in 18 years by each method of compounding? ⓐ compound quarterly ⓑ compound monthly ⓒ compound continuously #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "A equals ?")[$#h(3em) A = ?$]]), [Identify the values of each variable in the formulas.], [#math.equation(block: false, alt: "P equals $ 10,000")[$#h(3em) P = "$" 10,000$]], [Remember to express the percent as a decimal.], [#math.equation(block: false, alt: "r equals 0.05")[$#h(3em) r = 0.05$]], [], [#math.equation(block: false, alt: "t equals 18 years")[$#h(3em) t = 18 #h(0.2em) "years"$]], )) ⓐ #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([For quarterly compounding, #math.equation(block: false, alt: "n equals 4")[$n = 4$]. There are 4 quarters in a year.], [#math.equation(block: false, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$#h(3em) A = P attach(( 1 + frac(r, n) ), t: n t)$]]), [Substitute the values in the formula.], [#math.equation(block: false, alt: "A equals 10,000 open parenthesis 1 plus the fraction 0.05 over 4 close parenthesis to the power 4 times 18")[$#h(3em) A = 10,000 attach(( 1 + frac(0.05, 4) ), t: 4 · 18)$]], [Compute the amount. Be careful to consider the order of operations as you enter the expression into your calculator.], [#math.equation(block: false, alt: "A equals $ 24,459.20")[$#h(3em) A = "$" 24,459.20$]], )) ⓑ #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([For monthly compounding, #math.equation(block: false, alt: "n equals 12")[$n = 12$]. There are 12 months in a year.], [#math.equation(block: false, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$#h(3em) A = P attach(( 1 + frac(r, n) ), t: n t)$]]), [Substitute the values in the formula.], [#math.equation(block: false, alt: "A equals 10,000 open parenthesis 1 plus the fraction 0.05 over 12 close parenthesis to the power 12 times 18")[$#h(3em) A = 10,000 attach(( 1 + frac(0.05, 12) ), t: 12 · 18)$]], [Compute the amount.], [#math.equation(block: false, alt: "A equals $ 24,550.08")[$#h(3em) A = "$" 24,550.08$]], )) ⓒ #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([For compounding continuously,], [#math.equation(block: false, alt: "A equals P e to the power r t")[$#h(3em) A = P e^(r t)$]]), [Substitute the values in the formula.], [#math.equation(block: false, alt: "A equals 10,000 e to the power 0.05 times 18")[$#h(3em) A = 10,000 e^(0.05 · 18)$]], [Compute the amount.], [#math.equation(block: false, alt: "A equals $ 24,596.03")[$#h(3em) A = "$" 24,596.03$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Angela invested #math.equation(block: false, alt: "$ 15,000")[$"$" 15,000$] in a savings account. If the interest rate is #math.equation(block: false, alt: "4 % ,")[$4 "%" ,$] how much will be in the account in 10 years by each method of compounding? ⓐ compound quarterly ⓑ compound monthly ⓒ compound continuously #solutionbox[ ⓐ #math.equation(block: false, alt: "$ 22,332.96")[$"$" 22,332.96$] #linebreak() ⓑ #math.equation(block: false, alt: "$ 22,362.49")[$"$" 22,362.49$] ⓒ #math.equation(block: false, alt: "$ 22,377.37")[$"$" 22,377.37$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Allan invested \$10,000 in a mutual fund. If the interest rate is #math.equation(block: false, alt: "5 % ,")[$5 "%" ,$] how much will be in the account in 15 years by each method of compounding? ⓐ compound quarterly ⓑ compound monthly ⓒ compound continuously #solutionbox[ ⓐ \$21,071.81 ⓑ \$21,137.04 #linebreak() ⓒ \$21,170.00 ] ] Other topics that are modeled by exponential functions involve growth and decay. Both also use the formula #math.equation(block: false, alt: "A equals P e to the power r t")[$A = P e^(r t)$] we used for the growth of money. For growth and decay, generally we use#math.equation(block: false, alt: "A sub 0 ,")[$A_(0) ,$] as the original amount instead of calling it #math.equation(block: false, alt: "P ,")[$P ,$] the principal. We see that #strong[exponential growth] has a positive rate of growth and #strong[exponential decay] has a negative rate of growth. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Growth and Decay] For an original amount, #math.equation(block: false, alt: "A sub 0 ,")[$A_(0) ,$] that grows or decays at a rate, #emph[r], for a certain time, #emph[t], the final amount, #emph[A], is: #math.equation(block: true, alt: "A equals A sub 0 e to the power r t")[$A = A_(0) e^(r t)$] ] Exponential growth is typically seen in the growth of populations of humans or animals or bacteria. Our next example looks at the growth of a virus. #examplebox("Example 8")[][ Chris is a researcher at the Center for Disease Control and Prevention and he is trying to understand the behavior of a new and dangerous virus. He starts his experiment with 100 of the virus that grows continously at a rate of 25% per hour. He will check on the virus in 24 hours. How many viruses will he find? #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Identify the values of each variable in the formulas.], [#math.equation(block: false, alt: "A equals ?")[$#h(0.5em) A = ?$]]), [Be sure to put the percent in decimal form.], [#math.equation(block: false, alt: "A sub 0 equals 100")[$A_(0) = 100$]], [Be sure the units match—the rate is per hour and the time is in hours.], [#math.equation(block: false, alt: "r equals 0.25 /hour")[$#h(0.8em) r = 0.25 "/hour"$]], [], [#math.equation(block: false, alt: "t equals 24 hours")[$#h(0.8em) t = 24 #h(0.2em) "hours"$]], [], [], [Substitute the values in the formula: #math.equation(block: false, alt: "A equals A sub 0 e to the power r t")[$A = A_(0) e^(r t)$].], [#math.equation(block: false, alt: "A equals 100 e to the power 0.25 times 24")[$A = 100 e^(0.25 · 24)$]], [Compute the amount.], [#math.equation(block: false, alt: "A equals 40,342.88")[$A = 40,342.88$]], [Round to the nearest whole virus.], [#math.equation(block: false, alt: "A equals 40,343")[$A = 40,343$]], [], [The researcher will find 40,343 viruses.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Another researcher at the Center for Disease Control and Prevention, Lisa, is studying the growth of a bacteria. She starts her experiment with 50 of the bacteria that grows at a rate of #math.equation(block: false, alt: "15 %")[$15 "%"$] per hour. She will check on the bacteria every 8 hours. How many bacteria will she find in 8 hours? #solutionbox[ She will find 166 bacteria. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Milan, a biologist is observing the growth pattern of a virus. They start with 100 of the virus that grows at a rate of #math.equation(block: false, alt: "10 %")[$10 "%"$] per hour. They will check on the virus in 24 hours. How many viruses will they find? #solutionbox[ They will find 1,102 viruses. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with evaluating and graphing exponential functions. - #link("https://openstax.org/l/37Graphexponent")[Graphing Exponential Functions] - #link("https://openstax.org/l/37Solvelikebase")[Solving Exponential Equations] - #link("https://openstax.org/l/37Exponentapp")[Applications of Exponential Functions] - #link("https://openstax.org/l/37Compoundint")[Continuously Compound Interest] - #link("https://openstax.org/l/37Exponentdecay")[Radioactive Decay and Exponential Growth] ] === Key Concepts - #strong[Properties of the Graph of] #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a to the power x :")[$f ( x ) = a^(x) :$] #linebreak() #figure(table( columns: 4, align: left, inset: 6pt, [when #math.equation(block: false, alt: "a greater than 1")[$a > 1$]], [when #math.equation(block: false, alt: "0 less than a less than 1")[$0 < a < 1$]], [], [], [Domain], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [Domain], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [Range], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [Range], [#math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , ∞ )$]], [#math.equation(block: false, alt: "x")[$x$]-intercept], [none], [#math.equation(block: false, alt: "x")[$x$]-intercept], [none], [#math.equation(block: false, alt: "y")[$y$]-intercept], [#math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]], [#math.equation(block: false, alt: "y")[$y$]-intercept], [#math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis , open parenthesis −1 , the fraction 1 over a close parenthesis")[$( 1 , a ) , #h(0.2em) ( −1 , frac(1, a) )$]], [Contains], [#math.equation(block: false, alt: "open parenthesis 1 , a close parenthesis , open parenthesis −1 , the fraction 1 over a close parenthesis")[$( 1 , a ) , #h(0.2em) ( −1 , frac(1, a) )$]], [Asymptote], [#math.equation(block: false, alt: "x")[$x$]-axis, the line #math.equation(block: false, alt: "y equals 0")[$y = 0$]], [Asymptote], [#math.equation(block: false, alt: "x")[$x$]-axis, the line #math.equation(block: false, alt: "y equals 0")[$y = 0$]], [Basic shape], [increasing], [Basic shape], [decreasing], )) #linebreak() #figure(figph[This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).], alt: "This figure has two parts. On the left, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a). On the right, where a is noted to be less than 1, we have a curve that passes through (negative 1, 1 over a) through (0, 1) to (1, a).", caption: none) - #strong[One-to-One Property of Exponential Functions:] #linebreak() For #math.equation(block: false, alt: "a greater than 0")[$a > 0$] and #math.equation(block: false, alt: "a not equal to 1 ,")[$a ≠ 1 ,$] #linebreak() #math.equation(block: true, alt: "If a to the power x equals a to the power y , then x equals y .")[$"If" #h(0.2em) a^(x) = a^(y) , #h(0.2em) "then" #h(0.2em) x = y .$] - #strong[How to Solve an Exponential Equation] + Write both sides of the equation with the same base, if possible. + Write a new equation by setting the exponents equal. + Solve the equation. + Check the solution. - #strong[Compound Interest:] For a principal, #math.equation(block: false, alt: "P ,")[$P ,$] invested at an interest rate, #math.equation(block: false, alt: "r ,")[$r ,$] for #math.equation(block: false, alt: "t")[$t$] years, the new balance, #math.equation(block: false, alt: "A ,")[$A ,$] is #linebreak() #math.equation(block: false, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t, when compounded n times a year.; A equals P e to the power r t, when compounded continuously.")[$A = P attach(( 1 + frac(r, n) ), t: n t) & & & & & "when compounded" #h(0.2em) n #h(0.2em) "times a year." \ A = P e^(r t) & & & & & "when compounded continuously."$] - #strong[Exponential Growth and Decay:] For an original amount, #math.equation(block: false, alt: "A sub 0")[$A_(0)$] that grows or decays at a rate, #math.equation(block: false, alt: "r ,")[$r ,$] for a certain time #math.equation(block: false, alt: "t ,")[$t ,$] the final amount,#math.equation(block: false, alt: "A ,")[$A ,$] is #math.equation(block: false, alt: "A equals A sub 0 e to the power r t .")[$A = A_(0) e^(r t) .$] ==== Practice Makes Perfect #strong[Graph Exponential Functions] In the following exercises, graph each exponential function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 1 over 2) through (0, 1) to (1, 2).], alt: "This figure shows a curve that passes through (negative 1, 1 over 2) through (0, 1) to (1, 2).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 3 to the power x")[$g ( x ) = 3^(x)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 6 to the power x")[$f ( x ) = 6^(x)$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 1 over 6) through (0, 1) to (1, 6).], alt: "This figure shows a curve that passes through (negative 1, 1 over 6) through (0, 1) to (1, 6).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 7 to the power x")[$g ( x ) = 7^(x)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis 1.5 close parenthesis to the power x")[$f ( x ) = attach(( 1.5 ), t: x)$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 2 over 3) through (0, 1) to (1, 3 over 2).], alt: "This figure shows a curve that passes through (negative 1, 2 over 3) through (0, 1) to (1, 3 over 2).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis 2.5 close parenthesis to the power x")[$g ( x ) = attach(( 2.5 ), t: x)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 2) ), t: x)$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 2) through (0, 1) to (1, 1 over 2).], alt: "This figure shows a curve that passes through (negative 1, 2) through (0, 1) to (1, 1 over 2).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$g ( x ) = attach(( frac(1, 3) ), t: x)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 6 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 6) ), t: x)$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative1, 6) through (0, 1) to (1, 1 over 6).], alt: "This figure shows a curve that passes through (negative1, 6) through (0, 1) to (1, 1 over 6).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 7 close parenthesis to the power x")[$g ( x ) = attach(( frac(1, 7) ), t: x)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis 0.4 close parenthesis to the power x")[$f ( x ) = attach(( 0.4 ), t: x)$] #solutionbox[ #figure(figph[This figure shows a curve that passes through (negative 1, 5 over 2) through (0, 1) to (1, 2 over 5).], alt: "This figure shows a curve that passes through (negative 1, 5 over 2) through (0, 1) to (1, 2 over 5).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis 0.6 close parenthesis to the power x")[$g ( x ) = attach(( 0.6 ), t: x)$] In the following exercises, graph each function in the same coordinate system. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 to the power x , g open parenthesis x close parenthesis equals 4 to the power x minus 1")[$f ( x ) = 4^(x) , #h(0.2em) g ( x ) = 4^(x − 1)$] #solutionbox[ #figure(figph[This figure shows two functions. The first function f of x equals 4 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 4), (0, 1) and (1, 4). The second function g of x equals 4 to the x minus 1 power is marked in red and passes through the points (0, 1 over 4), (1, 1) and (2, 4).], alt: "This figure shows two functions. The first function f of x equals 4 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 4), (0, 1) and (1, 4). The second function g of x equals 4 to the x minus 1 power is marked in red and passes through the points (0, 1 over 4), (1, 1) and (2, 4).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 to the power x , g open parenthesis x close parenthesis equals 3 to the power x minus 1")[$f ( x ) = 3^(x) , #h(0.2em) g ( x ) = 3^(x − 1)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x , g open parenthesis x close parenthesis equals 2 to the power x minus 2")[$f ( x ) = 2^(x) , #h(0.2em) g ( x ) = 2^(x − 2)$] #solutionbox[ #figure(figph[This figure shows two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 2 power is marked in red and passes through the points (0, 1 over 4), (1, 1 over 2), and (2, 1).], alt: "This figure shows two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1) and (1, 2). The second function g of x equals 2 to the x minus 2 power is marked in red and passes through the points (0, 1 over 4), (1, 1 over 2), and (2, 1).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x , g open parenthesis x close parenthesis equals 2 to the power x plus 2")[$f ( x ) = 2^(x) , #h(0.2em) g ( x ) = 2^(x + 2)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 to the power x , g open parenthesis x close parenthesis equals 3 to the power x plus 2")[$f ( x ) = 3^(x) , #h(0.2em) g ( x ) = 3^(x) + 2$] #solutionbox[ #figure(figph[This figure shows two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 2, 1), (negative 1, 3), and (0, 5).], alt: "This figure shows two functions. The first function f of x equals 3 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 3), (0, 1), and (1, 3). The second function g of x equals 3 to the x power plus 2 is marked in red and passes through the points (negative 2, 1), (negative 1, 3), and (0, 5).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 to the power x , g open parenthesis x close parenthesis equals 4 to the power x plus 2")[$f ( x ) = 4^(x) , #h(0.2em) g ( x ) = 4^(x) + 2$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x , g open parenthesis x close parenthesis equals 2 to the power x plus 1")[$f ( x ) = 2^(x) , #h(0.2em) g ( x ) = 2^(x) + 1$] #solutionbox[ #figure(figph[This figure shows two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second function g of x equals 2 to the x power plus 1 is marked in red and passes through the points (negative 1, 1), (0, 2), and (1, 4).], alt: "This figure shows two functions. The first function f of x equals 2 to the x power is marked in blue and corresponds to a curve that passes through the points (negative 1, 1 over 2), (0, 1), and (1, 2). The second function g of x equals 2 to the x power plus 1 is marked in red and passes through the points (negative 1, 1), (0, 2), and (1, 4).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x , g open parenthesis x close parenthesis equals 2 to the power x minus 1")[$f ( x ) = 2^(x) , #h(0.2em) g ( x ) = 2^(x) − 1$] In the following exercises, graph each exponential function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 to the power x plus 2")[$f ( x ) = 3^(x + 2)$] #solutionbox[ #figure(figph[This figure shows an exponential curve that passes through (negative 3, 1 over 3), (negative 2, 1), and (0, 9).], alt: "This figure shows an exponential curve that passes through (negative 3, 1 over 3), (negative 2, 1), and (0, 9).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 to the power x minus 2")[$f ( x ) = 3^(x − 2)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x plus 3")[$f ( x ) = 2^(x) + 3$] #solutionbox[ #figure(figph[This figure shows an exponential that passes through (negative 1, 7 over 2), (0, 4), and (1, 5).], alt: "This figure shows an exponential that passes through (negative 1, 7 over 2), (0, 4), and (1, 5).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x minus 3")[$f ( x ) = 2^(x) − 3$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x minus 4")[$f ( x ) = attach(( frac(1, 2) ), t: x − 4)$] #solutionbox[ #figure(figph[This figure shows an exponential that passes through (2, 4), (3, 2), and (4, 1).], alt: "This figure shows an exponential that passes through (2, 4), (3, 2), and (4, 1).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x minus 3")[$f ( x ) = attach(( frac(1, 2) ), t: x) − 3$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power x plus 1")[$f ( x ) = e^(x) + 1$] #solutionbox[ #figure(figph[This figure shows an exponential that passes through (1, 1 plus 1 over e), (0, 2), and (1, e).], alt: "This figure shows an exponential that passes through (1, 1 plus 1 over e), (0, 2), and (1, e).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power x minus 2")[$f ( x ) = e^(x − 2)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals − 2 to the power x")[$f ( x ) = "−" 2^(x)$] #solutionbox[ #figure(figph[This figure shows an exponential that passes through (negative 1, negative 1 over 2), (0, negative 1), and (1, 2).], alt: "This figure shows an exponential that passes through (negative 1, negative 1 over 2), (0, negative 1), and (1, 2).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power minus x minus 1 −1")[$f ( x ) = 2^(− x − 1) −1$] #strong[Solve Exponential Equations] In the following exercises, solve each equation. #math.equation(block: true, alt: "2 to the power 3 x minus 8 equals 16")[$2^(3 x − 8) = 16$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] #math.equation(block: true, alt: "2 to the power 2 x minus 3 equals 32")[$2^(2 x − 3) = 32$] #math.equation(block: true, alt: "3 to the power x plus 3 equals 9")[$3^(x + 3) = 9$] #solutionbox[ #math.equation(block: true, alt: "x equals −1")[$x = −1$] ] #math.equation(block: true, alt: "3 to the power x squared equals 81")[$3^(x^(2)) = 81$] #math.equation(block: true, alt: "4 to the power x squared equals 4")[$4^(x^(2)) = 4$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals 1")[$x = −1 , x = 1$] ] #math.equation(block: true, alt: "4 to the power x equals 32")[$4^(x) = 32$] #math.equation(block: true, alt: "4 to the power x plus 2 equals 64")[$4^(x + 2) = 64$] #solutionbox[ #math.equation(block: true, alt: "x equals 1")[$x = 1$] ] #math.equation(block: true, alt: "4 to the power x plus 3 equals 16")[$4^(x + 3) = 16$] #math.equation(block: true, alt: "2 to the power x squared plus 2 x equals the fraction 1 over 2")[$2^(x^(2) + 2 x) = frac(1, 2)$] #solutionbox[ #math.equation(block: true, alt: "x equals −1")[$x = −1$] ] #math.equation(block: true, alt: "3 to the power x squared minus 2 x equals the fraction 1 over 3")[$3^(x^(2) − 2 x) = frac(1, 3)$] #math.equation(block: true, alt: "e to the power 3 x times e to the power 4 equals e to the power 10")[$e^(3 x) · e^(4) = e^(10)$] #solutionbox[ #math.equation(block: true, alt: "x equals 2")[$x = 2$] ] #math.equation(block: true, alt: "e to the power 2 x times e cubed equals e to the power 9")[$e^(2 x) · e^(3) = e^(9)$] #math.equation(block: true, alt: "the fraction e to the power x squared over e squared equals e to the power x")[$frac(e^(x^(2)), e^(2)) = e^(x)$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals 2")[$x = −1 , x = 2$] ] #math.equation(block: true, alt: "the fraction e to the power x squared over e cubed equals e to the power 2 x")[$frac(e^(x^(2)), e^(3)) = e^(2 x)$] In the following exercises, match the graphs to one of the following functions: ⓐ #math.equation(block: false, alt: "2 to the power x")[$2^(x)$] ⓑ #math.equation(block: false, alt: "2 to the power x plus 1")[$2^(x + 1)$] ⓒ #math.equation(block: false, alt: "2 to the power x minus 1")[$2^(x − 1)$] ⓓ #math.equation(block: false, alt: "2 to the power x plus 2")[$2^(x) + 2$] ⓔ #math.equation(block: false, alt: "2 to the power x minus 2")[$2^(x) − 2$] ⓕ #math.equation(block: false, alt: "3 to the power x")[$3^(x)$] #figure(figph[This figure shows an exponential that passes through (1, 1 over 3), (0, 1), and (1, 3).], alt: "This figure shows an exponential that passes through (1, 1 over 3), (0, 1), and (1, 3).", caption: none) #solutionbox[ ⓕ ] #figure(figph[This figure shows an exponential that passes through (negative 2, 1 over 2), (negative 1, 1), and (0, 2).], alt: "This figure shows an exponential that passes through (negative 2, 1 over 2), (negative 1, 1), and (0, 2).", caption: none) #figure(figph[This figure shows an exponential that passes through (1, 1 over 2), (0, 1), and (1, 2).], alt: "This figure shows an exponential that passes through (1, 1 over 2), (0, 1), and (1, 2).", caption: none) #solutionbox[ ⓐ ] #figure(figph[This figure shows an exponential that passes through (0, 1 over 2), (1, 1), and (2, 2).], alt: "This figure shows an exponential that passes through (0, 1 over 2), (1, 1), and (2, 2).", caption: none) #figure(figph[This figure shows an exponential that passes through (negative 1, 3 over 2), (0, negative 1), and (1, 0).], alt: "This figure shows an exponential that passes through (negative 1, 3 over 2), (0, negative 1), and (1, 0).", caption: none) #solutionbox[ ⓔ ] #figure(figph[This figure shows an exponential that passes through (negative 1, 5 over 2), (0, 3), and (1, 4).], alt: "This figure shows an exponential that passes through (negative 1, 5 over 2), (0, 3), and (1, 4).", caption: none) #strong[Use exponential models in applications] In the following exercises, use an exponential model to solve. Edgar accumulated #math.equation(block: false, alt: "$ 5,000")[$"$" 5,000$] in credit card debt. If the interest rate is #math.equation(block: false, alt: "20 %")[$20 "%"$] per year, and he does not make any payments for 2 years, how much will he owe on this debt in 2 years by each method of compounding? ⓐ compound quarterly ⓑ compound monthly ⓒ compound continuously #solutionbox[ ⓐ #math.equation(block: false, alt: "$ 7,387.28")[$"$" 7,387.28$] ⓑ #math.equation(block: false, alt: "$ 7,434.57")[$"$" 7,434.57$] ⓒ #math.equation(block: false, alt: "$ 7,459.12")[$"$" 7,459.12$] ] Cynthia invested #math.equation(block: false, alt: "$ 12,000")[$"$" 12,000$] in a savings account. If the interest rate is #math.equation(block: false, alt: "6 % ,")[$6 "%" ,$] how much will be in the account in 10 years by each method of compounding? ⓐ compound quarterly #linebreak() ⓑ compound monthly ⓒ compound continuously Rochelle deposits #math.equation(block: false, alt: "$ 5,000")[$"$" 5,000$] in an IRA. What will be the value of her investment in 25 years if the investment is earning #math.equation(block: false, alt: "8 %")[$8 "%"$] per year and is compounded continuously? #solutionbox[ #math.equation(block: true, alt: "$ 36,945.28")[$"$" 36,945.28$] ] Nazerhy deposits #math.equation(block: false, alt: "$ 8,000")[$"$" 8,000$] in a certificate of deposit. The annual interest rate is #math.equation(block: false, alt: "6 %")[$6 "%"$] and the interest will be compounded quarterly. How much will the certificate be worth in 10 years? A researcher at the Center for Disease Control and Prevention is studying the growth of a bacteria. He starts his experiment with 100 of the bacteria that grows at a rate of #math.equation(block: false, alt: "6 %")[$6 "%"$] per hour. He will check on the bacteria every 8 hours. How many bacteria will he find in 8 hours? #solutionbox[ 162 bacteria ] A biologist is observing the growth pattern of a virus. She starts with 50 of the virus that grows at a rate of #math.equation(block: false, alt: "20 %")[$20 "%"$] per hour. She will check on the virus in 24 hours. How many viruses will she find? In the last ten years the population of Indonesia has grown at a rate of #math.equation(block: false, alt: "1.12 %")[$1.12 "%"$] per year to 258,316,051. If this rate continues, what will be the population in 10 more years? #solutionbox[ 288,929,825 ] In the last ten years the population of Brazil has grown at a rate of #math.equation(block: false, alt: "0.9 %")[$0.9 "%"$] per year to 205,823,665. If this rate continues, what will be the population in 10 more years? ==== Writing Exercises Explain how you can distinguish between exponential functions and polynomial functions. #solutionbox[ Answers will vary. ] Compare and contrast the graphs of #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$] and #math.equation(block: false, alt: "y equals 2 to the power x")[$y = 2^(x)$]. What happens to an exponential function as the values of #math.equation(block: false, alt: "x")[$x$] decreases? Will the graph ever cross the #linebreak() #math.equation(block: false, alt: "x")[$x$]-axis? Explain. #solutionbox[ Answers will vary. ] ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Graph exponential functions, solve exponential equations, and use exponential models in applications.], alt: "This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Graph exponential functions, solve exponential equations, and use exponential models in applications.", caption: none) ⓑ After reviewing this checklist, what will you do to become confident for all objectives?