#set document(title: "3.8 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")) == 3.8#h(0.6em)Exponents #figure(figph[An illustration shows the solar system. The sun, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune are labeled.], alt: "An illustration shows the solar system. The sun, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune are labeled.", caption: [Astronomical distances are written using exponents.]) === Learning Objectives After completing this section, you should be able to: + Apply the rules of exponents to simplifying expressions. Sometimes, we look for shorthand when writing or expressing something that simply takes too long. The use of LOL and tl;dr. This shorthand only works if everyone reading the shorthand knows what it stands for. Using exponents is a similar instance. Writing out a long string of a number times itself over and over takes too much time, and eventually one would forget how many of the value has been written or read. For example, #math.equation(block: false, alt: "8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8 times 8")[$8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8$]. There has to be a shorter and more efficient way to write 8 times itself 1, 2, 3….hmmmm, 19 times. And that’s the role that exponents play in mathematics. They are shorthand for multiplying a number by itself a number of times. Without it, calculations would become a mess and we’d have to write a lot more. === Applying the Rules of Exponents to Simplify Expressions Squaring a number is multiplying it by itself, and has that name because it is the area of a square with that side length. Cubing a number is finding the volume of a cube with that length of sides. That’s why we refer to #math.equation(block: false, alt: "5 squared")[$5^(2)$] as five squared, or #math.equation(block: false, alt: "10 cubed")[$10^(3)$] as ten cubed. Exponents represent that multiplication. Let’s remind ourselves of the terminology associated with exponents and what exponents represent. Suppose you want to multiply a number, let’s label that number #math.equation(block: false, alt: "a")[$a$], by itself some number of times. Let’s label the number of times #math.equation(block: false, alt: "b")[$b$]. We denote that as #math.equation(block: false, alt: "a to the power b")[$a^(b)$]. We say #math.equation(block: false, alt: "a")[$a$] raised to the #math.equation(block: false, alt: "b")[$b$]th power. When we write or see #math.equation(block: false, alt: "7 to the power 5")[$7^(5)$], we call the 7 the #strong[base] and we call 5 the #strong[exponent]. What it represents is 7 multiplied by itself 5 times. This means exponents are used as a shorthand for repeated multiplications, where we write #math.equation(block: false, alt: "7 to the power 5 equals 7 times 7 times 7 times 7 times 7")[$7^(5) = 7 × 7 × 7 × 7 × 7$]. We would write #math.equation(block: false, alt: "7 to the power 5")[$7^(5)$] and say seven to the fifth power. #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Exponential_Notation")[Exponential Notation] ] The definitions of base and exponent make it possible to understand the exponent rules. ==== Product Rule for Exponents The first rule we examine is the product rule, #math.equation(block: false, alt: "a to the power n a to the power m equals a to the power n plus m")[$a^(n) a^(m) = a^(n + m)$]. This rule means that when we multiply a base raised to a power times the same base to another power, the result is the base raised to the sum of the powers. To demonstrate, consider #math.equation(block: false, alt: "9 cubed times 9 to the power 5")[$9^(3) × 9^(5)$]. If we apply the product rule to that we get #math.equation(block: false, alt: "9 cubed times 9 to the power 5 equals 9 to the power 3 plus 5 equals 9 to the power 8")[$9^(3) × 9^(5) = 9^(3 + 5) = 9^(8)$]. This can be tested by looking at the multiplications that are represented. The #math.equation(block: false, alt: "9 cubed")[$9^(3)$] is 9 times itself 3 times, while #math.equation(block: false, alt: "9 to the power 5")[$9^(5)$] is 9 times itself 5 times. Substituting those into #math.equation(block: false, alt: "9 cubed times 9 to the power 5")[$9^(3) × 9^(5)$] we see #math.equation(block: false, alt: "9 cubed times 9 to the power 5 equals open parenthesis 9 times 9 times 9 close parenthesis times open parenthesis 9 times 9 times 9 times 9 times 9 close parenthesis equals 9 to the power 8")[$9^(3) × 9^(5) = ( 9 × 9 × 9 ) × ( 9 × 9 × 9 × 9 × 9 ) = 9^(8)$], which is what the formula told us would happen. #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Caution: The product rule only applies when the bases are the same. If the bases are different, we do not apply this rule.] ] #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ If a number, #math.equation(block: false, alt: "a")[$a$], raised to a power, #math.equation(block: false, alt: "n")[$n$], is then multiplied by #math.equation(block: false, alt: "a")[$a$] raised to another power, #math.equation(block: false, alt: "m")[$m$], the result is #math.equation(block: false, alt: "a to the power n a to the power m equals a to the power n plus m")[$a^(n) a^(m) = a^(n + m)$]. ] #examplebox("Example 1")[Using the Product Rule for Exponents][ If possible, use the product rule to simplify the following: + #math.equation(block: false, alt: "21 to the power 9 times 21 to the power 15")[$21^(9) × 21^(15)$] + #math.equation(block: false, alt: "5 to the power 9 times 8 to the power 4")[$5^(9) × 8^(4)$] #solutionbox[ + We can apply the product rule to simplify the expression because the bases are the same and we are multiplying. #math.equation(block: true, alt: "21 to the power 9 times 21 to the power 15 equals 21 to the power open parenthesis 9 plus 15 close parenthesis equals 21 to the power 24")[$21^(9) × 21^(15) = 21^(( 9 + 15 )) = 21^(24)$] + Since the bases are not the same (one is 5, the other 8), this cannot be simplified using the product rule for exponents. ] ] These rules can be applied to unknowns too. #examplebox("Example 2")[Using the Product Rule for Exponents of Unknowns][ Use the product rule to simplify #math.equation(block: false, alt: "a to the power 4 times a to the power 10")[$a^(4) × a^(10)$]. #solutionbox[ The bases are the same, and we are multiplying, so we apply the multiplication rule to simplify the expression. #math.equation(block: true, alt: "a to the power 4 times a to the power 10 equals a to the power open parenthesis 4 plus 10 close parenthesis equals a to the power 14")[$a^(4) × a^(10) = a^(( 4 + 10 )) = a^(14)$] ] ] ==== Quotient Rule for Exponents The next rule we examine is the quotient, or division, rule. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ When a number, #math.equation(block: false, alt: "a")[$a$], raised to a power, #math.equation(block: false, alt: "n")[$n$], is divided by #math.equation(block: false, alt: "a")[$a$] raised to another power, #math.equation(block: false, alt: "m")[$m$], then the result is #math.equation(block: false, alt: "the fraction a to the power n over a to the power m equals a to the power open parenthesis n minus m close parenthesis")[$frac(a^(n), a^(m)) = a^(( n − m ))$]. ] This rule means that when we divide a base raised to a power by the same base to another power, the result is the base raised to the difference of the powers. To demonstrate, consider #math.equation(block: false, alt: "the fraction 14 to the power 13 over 14 to the power 6")[$frac(14^(13), 14^(6))$]. If we apply the quotient rule to that, we get #math.equation(block: false, alt: "the fraction 14 to the power 13 over 14 to the power 6 equals 14 to the power 13 minus 6 equals 14 to the power 7")[$frac(14^(13), 14^(6)) = 14^(13 − 6) = 14^(7)$]. This can be tested by looking at the division that is represented. Remember, #math.equation(block: false, alt: "14 to the power 13")[$14^(13)$] is 14 multiplied to itself 13 times, while #math.equation(block: false, alt: "14 to the power 6")[$14^(6)$] is 14 multiplied to itself 6 times. Substituting those into #math.equation(block: false, alt: "the fraction 14 to the power 13 over 14 to the power 6")[$frac(14^(13), 14^(6))$] gives the following: #math.equation(block: true, alt: "the fraction 4 to the power 13 over 4 to the power 6 equals the fraction 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 over 4 times 4 times 4 times 4 times 4 times 4")[$frac(4^(13), 4^(6)) = frac(4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4, 4 × 4 × 4 × 4 × 4 × 4)$] We see here that there are a LOT of fours to be divided out. #math.equation(block: true, alt: "equals the fraction 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 times 4 over 4 times 4 times 4 times 4 times 4 times 4 equals the fraction 4 times 4 times 4 times 4 times 4 times 4 times 4 over 1 equals 4 times 4 times 4 times 4 times 4 times 4 times 4")[$= frac(cancel(4) × cancel(4) × cancel(4) × cancel(4) × cancel(4) × cancel(4) × 4 × 4 × 4 × 4 × 4 × 4 × 4, cancel(4) × cancel(4) × cancel(4) × cancel(4) × cancel(4) × cancel(4)) = frac(4 × 4 × 4 × 4 × 4 × 4 × 4, 1) = 4 × 4 × 4 × 4 × 4 × 4 × 4$] What remains is 4 to the 7th power, #math.equation(block: false, alt: "4 times 4 times 4 times 4 times 4 times 4 times 4 equals 4 to the power 7")[$4 × 4 × 4 × 4 × 4 × 4 × 4 = 4^(7)$]. All of the work above confirmed what the formula told us would be the result. #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Caution: The quotient rule only applies when the bases are the same. If the bases are different, we do not apply this rule.] ] #examplebox("Example 3")[Using the Quotient Rule for Exponents][ Use the quotient rule to simplify #math.equation(block: false, alt: "the fraction 5 to the power 19 over 5 to the power 11")[$frac(5^(19), 5^(11))$]. #solutionbox[ We can apply the quotient rule to simplify the expression since the bases are the same and we are dividing. #math.equation(block: true, alt: "the fraction 5 to the power 19 over 5 to the power 11 equals 5 to the power open parenthesis 19 minus 11 close parenthesis equals 5 to the power 8")[$frac(5^(19), 5^(11)) = 5^(( 19 − 11 )) = 5^(8)$] ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Product_and_Quotient_Rule_for_Exponents")[Product and Quotient Rule for Exponents] ] A natural consequence of the quotient rule is what it means to raise a non-zero number to the zeroth power. Let’s look at the simplification when the exponents are equal. #math.equation(block: true, alt: "the fraction 3 to the power 6 over 3 to the power 6 equals 3 to the power open parenthesis 6 minus 6 close parenthesis equals 3 to the power 0")[$frac(3^(6), 3^(6)) = 3^(( 6 − 6 )) = 3^(0)$] We know that a number divided by itself is 1, so #math.equation(block: false, alt: "the fraction 3 to the power 6 over 3 to the power 6 equals 1")[$frac(3^(6), 3^(6)) = 1$]. From that is must be that #math.equation(block: false, alt: "the fraction 3 to the power 6 over 3 to the power 6 equals 3 to the power 0 equals 1")[$frac(3^(6), 3^(6)) = 3^(0) = 1$]. This provides the rule for a number raised to the power 0: #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ If you have a non-zero number #math.equation(block: false, alt: "a")[$a$], then #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$]. ] ==== Distributive Rule for Exponents The next rule we look to is a distributive rule for exponents. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ If you have a product, #math.equation(block: false, alt: "open parenthesis a times b close parenthesis")[$( a × b )$], and raise it to an exponent, #math.equation(block: false, alt: "n")[$n$], then #math.equation(block: false, alt: "open parenthesis a times b close parenthesis to the power n equals a to the power n times b to the power n")[$attach(( a × b ), t: n) = a^(n) × b^(n)$]. ] This means that when we have two numbers multiplied together, and that is raised to a power, it is the same as raising each of the numbers to the same power first, then multiplying. For example, #math.equation(block: false, alt: "open parenthesis 3 times 7 close parenthesis to the power 4 equals 3 to the power 4 times 7 to the power 4")[$attach(( 3 × 7 ), t: 4) = 3^(4) × 7^(4)$]. This can be explained using the definition of exponents and multiplying all the factors. #math.equation(block: true, alt: "open parenthesis 3 times 7 close parenthesis to the power 4 equals open parenthesis 3 times 7 close parenthesis times open parenthesis 3 times 7 close parenthesis times open parenthesis 3 times 7 close parenthesis times open parenthesis 3 times 7 close parenthesis")[$attach(( 3 × 7 ), t: 4) = ( 3 × 7 ) × ( 3 × 7 ) × ( 3 × 7 ) × ( 3 × 7 )$] We may change the order in which numbers are multiplied. This is the commutative property of the real numbers. This can be written as #math.equation(block: false, alt: "3 times 3 times 3 times 3 times 7 times 7 times 7 times 7")[$3 × 3 × 3 × 3 × 7 × 7 × 7 × 7$]. Using exponents, that shortens to #math.equation(block: false, alt: "3 to the power 4 times 7 to the power 4")[$3^(4) × 7^(4)$]. This also works in the other direction, #math.equation(block: false, alt: "a to the power n times b to the power n equals open parenthesis a times b close parenthesis to the power n")[$a^(n) × b^(n) = attach(( a × b ), t: n)$]. Read this way, if we have one base raised to an exponent, and another base raised to the same exponent, we can multiply the bases and raise that product to the shared exponent. For instance, #math.equation(block: false, alt: "7 to the power 8 times 11 to the power 8 equals open parenthesis 7 times 11 close parenthesis to the power 8 equals 77 to the power 8")[$7^(8) × 11^(8) = attach(( 7 × 11 ), t: 8) = 77^(8)$]. #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Caution: The exponent distributive rule], #math.equation(block: false, alt: "a to the power n times b to the power n equals open parenthesis a times b close parenthesis to the power n")[$a^(n) × b^(n) = attach(( a × b ), t: n)$], #emph[only works if the exponents are the same]. ] #examplebox("Example 4")[Using the Distributive Rule for Exponents][ Use the exponent distributive rule to expand #math.equation(block: false, alt: "open parenthesis 6 times 13 close parenthesis to the power 7")[$attach(( 6 × 13 ), t: 7)$]. #solutionbox[ Applying the distributive rule to the product, we get #math.equation(block: false, alt: "open parenthesis 6 times 13 close parenthesis to the power 7 equals 6 to the power 7 times 13 to the power 7")[$attach(( 6 × 13 ), t: 7) = 6^(7) × 13^(7)$]. ] ] #examplebox("Example 5")[Using the Distributive Rule for Exponents][ Use the exponent distributive rule to expand #math.equation(block: false, alt: "open parenthesis c times d close parenthesis to the power 10")[$attach(( c × d ), t: 10)$]. #solutionbox[ Applying the distributive rule to the product, we get #math.equation(block: false, alt: "open parenthesis c times d close parenthesis to the power 10 equals c to the power 10 times d to the power 10")[$attach(( c × d ), t: 10) = c^(10) × d^(10)$]. ] ] This distribution also works for quotients. A fraction raised to an exponent equals the numerator raised to the exponent divided by the denominator raised to the exponent. For example, #math.equation(block: false, alt: "open parenthesis the fraction 3 over 5 close parenthesis to the power 7 equals the fraction 3 to the power 7 over 5 to the power 7")[$attach(( frac(3, 5) ), t: 7) = frac(3^(7), 5^(7))$]. Demonstrating this is similar to the previous rule. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ When you have a fraction, #math.equation(block: false, alt: "the fraction a over b")[$frac(a, b)$], raised to an exponent, #math.equation(block: false, alt: "n")[$n$], then #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")[$attach(( frac(a, b) ), t: n) = frac(a^(n), b^(n))$]. ] #examplebox("Example 6")[Using the Distributive Rule for Exponents with Fractions][ Use the exponent distributive rule to expand the following: + #math.equation(block: false, alt: "open parenthesis the fraction 4 over 9 close parenthesis to the power 6")[$attach(( frac(4, 9) ), t: 6)$] + #math.equation(block: false, alt: "open parenthesis the fraction 3 over b close parenthesis to the power 11")[$attach(( frac(3, b) ), t: 11)$] #solutionbox[ + Applying the distributive rule to the quotient, we get #math.equation(block: false, alt: "open parenthesis the fraction 4 over 9 close parenthesis to the power 6 equals the fraction 4 to the power 6 over 9 to the power 6")[$attach(( frac(4, 9) ), t: 6) = frac(4^(6), 9^(6))$]. + Applying the distributive rule to the quotient, we get #math.equation(block: false, alt: "open parenthesis the fraction 3 over b close parenthesis to the power 11 equals the fraction 3 to the power 11 over b to the power 11")[$attach(( frac(3, b) ), t: 11) = frac(3^(11), b^(11))$]. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Fraction_Raised_to_a_Power")[Fraction Raised to a Power] ] ==== Power Rule In the previous two sets of rules, we’ve seen exponents applied to products and quotients. Now we look to exponents applied to other exponents. For example, #math.equation(block: false, alt: "open parenthesis 3 to the power 6 close parenthesis to the power 4 equals 3 to the power open parenthesis 6 times 4 close parenthesis equals 3 to the power 24")[$attach(( 3^(6) ), t: 4) = 3^(( 6 × 4 )) = 3^(24)$]. This can be explained by examining what the outer exponent does. We raise #math.equation(block: false, alt: "3 to the power 6")[$3^(6)$] to the fourth power, so we multiply #math.equation(block: false, alt: "3 to the power 6")[$3^(6)$] by itself 4 times, #math.equation(block: false, alt: "open parenthesis 3 to the power 6 close parenthesis to the power 4 equals 3 to the power 6 times 3 to the power 6 times 3 to the power 6 times 3 to the power 6")[$attach(( 3^(6) ), t: 4) = 3^(6) × 3^(6) × 3^(6) × 3^(6)$]. Now if we apply the product rule for exponents, this becomes #math.equation(block: false, alt: "3 to the power open parenthesis 6 plus 6 plus 6 plus 6 close parenthesis equals 3 to the power 24")[$3^(( 6 + 6 + 6 + 6 )) = 3^(24)$]. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ If you raise a non-zero base, say #math.equation(block: false, alt: "a")[$a$], to an exponent #math.equation(block: false, alt: "n")[$n$], and raise that to another exponent, #math.equation(block: false, alt: "m")[$m$], you get the base raised to the product of the exponents, which is #math.equation(block: false, alt: "open parenthesis a to the power n close parenthesis to the power m equals a to the power open parenthesis n times m close parenthesis")[$attach(( a^(n) ), t: m) = a^(( n × m ))$]. ] #examplebox("Example 7")[Raising an Exponent to an Exponent][ Expand the following: + #math.equation(block: false, alt: "open parenthesis 6 to the power 7 close parenthesis cubed")[$attach(( 6^(7) ), t: 3)$] + #math.equation(block: false, alt: "open parenthesis b to the power 12 close parenthesis to the power 4")[$attach(( b^(12) ), t: 4)$] #solutionbox[ + Using the power rule of exponents, #math.equation(block: false, alt: "open parenthesis 6 to the power 7 close parenthesis cubed equals 6 to the power open parenthesis 7 times 3 close parenthesis equals 6 to the power 21")[$attach(( 6^(7) ), t: 3) = 6^(( 7 × 3 )) = 6^(21)$]. + Using the power rule of exponents, #math.equation(block: false, alt: "open parenthesis b to the power 12 close parenthesis to the power 4 equals b to the power open parenthesis 12 times 4 close parenthesis equals b to the power 48")[$attach(( b^(12) ), t: 4) = b^(( 12 × 4 )) = b^(48)$]. ] ] ==== Negative Exponent Rule Up until now, we’ve only looked at positive exponents. The last exponent rule we look at is what negative exponents represent. Recall the quotient rule: #math.equation(block: false, alt: "the fraction a to the power n over a to the power m equals a to the power open parenthesis n plus m close parenthesis")[$frac(a^(n), a^(m)) = a^(( n + m ))$]. What would happen if the exponent in the denominator was larger than that in the numerator? For example, #math.equation(block: false, alt: "the fraction 4 to the power 5 over 4 to the power 7")[$frac(4^(5), 4^(7))$]. If we apply the quotient rule, we obtain #math.equation(block: false, alt: "the fraction 4 to the power 5 over 4 to the power 7 equals 4 to the power 5 minus 7 equals 4 to the power minus 2")[$frac(4^(5), 4^(7)) = 4^(5 − 7) = 4^(− 2)$]. We need to make sense of that negative exponent. To do so, we can expand the quotient and see what happens: #math.equation(block: false, alt: "the fraction 4 to the power 5 over 4 to the power 7 equals the fraction 4 times 4 times 4 times 4 times 4 over 4 times 4 times 4 times 4 times 4 times 4 times 4")[$frac(4^(5), 4^(7)) = frac(4 × 4 × 4 × 4 × 4, 4 × 4 × 4 × 4 × 4 × 4 × 4)$]. When we divide out common factors, only two factors of 4 are left in the denominator, as we see here:#math.equation(block: false, alt: "the fraction 1 over 4 times 4")[$frac(1, 4 × 4)$]. Using exponent notation, this is #math.equation(block: false, alt: "the fraction 1 over 4 squared")[$frac(1, 4^(2))$]. Since #math.equation(block: false, alt: "4 to the power minus squared")[$4^(−) 2$] and #math.equation(block: false, alt: "the fraction 1 over 4 squared")[$frac(1, 4^(2))$] represent the same number, #math.equation(block: false, alt: "the fraction 4 to the power 5 over 4 to the power 7")[$frac(4^(5), 4^(7))$], they are equal. This demonstrates how negative exponents are defined. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #math.equation(block: false, alt: "a to the power minus n equals the fraction 1 over a to the power n")[$a^(− n) = frac(1, a^(n))$] provided that #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. Similarly, #math.equation(block: false, alt: "the fraction 1 over a to the power minus n equals a to the power n")[$frac(1, a^(− n)) = a^(n)$]. ] #examplebox("Example 8")[Eliminating Negative Exponents][ Convert the following to expressions with no negative exponent: + #math.equation(block: false, alt: "3 to the power 4 times 5 to the power minus 8")[$3^(4) × 5^(− 8)$] + #math.equation(block: false, alt: "a to the power minus 9 times b to the power 5")[$a^(− 9) × b^(5)$] + #math.equation(block: false, alt: "the fraction 7 over c to the power minus 2")[$frac(7, c^(− 2))$] #solutionbox[ + Using the negative exponent rule on the #math.equation(block: false, alt: "5 to the power minus 8")[$5^(− 8)$] and multiplying, #math.equation(block: false, alt: "3 to the power 4 times 5 to the power minus 8 equals 3 to the power 4 times the fraction 1 over 5 to the power 8 equals the fraction 3 to the power 4 over 5 to the power 8")[$3^(4) × 5^(− 8) = 3^(4) × frac(1, 5^(8)) = frac(3^(4), 5^(8))$]. + Using the negative exponent rule on the #math.equation(block: false, alt: "a to the power minus 9")[$a^(− 9)$] and multiplying, #math.equation(block: false, alt: "a to the power minus 9 times b to the power 5 equals the fraction 1 over a to the power 9 times b to the power 5 equals the fraction b to the power 5 over a to the power 9")[$a^(− 9) × b^(5) = frac(1, a^(9)) × b^(5) = frac(b^(5), a^(9))$]. + Begin by rewriting the expression as #math.equation(block: false, alt: "the fraction 7 over c to the power minus 2 equals the fraction 7 over 1 times the fraction 1 over c to the power minus 2")[$frac(7, c^(− 2)) = frac(7, 1) × frac(1, c^(− 2))$]. Apply the negative exponent rule to #math.equation(block: false, alt: "the fraction 1 over c to the power minus 2")[$frac(1, c^(− 2))$] in the expression, which becomes #math.equation(block: false, alt: "the fraction 7 over 1 times the fraction 1 over c to the power minus 2 equals 7 times c squared")[$frac(7, 1) × frac(1, c^(− 2)) = 7 × c^(2)$], which has no negative exponents. ] ] #examplebox("Example 9")[Eliminating Denominators by Using Negative Exponents][ Use negative exponents to rewrite the following expressions with no denominator: + #math.equation(block: false, alt: "the fraction 7 cubed over 13 to the power 9")[$frac(7^(3), 13^(9))$] + #math.equation(block: false, alt: "the fraction c to the power 4 over d to the power 8")[$frac(c^(4), d^(8))$] #solutionbox[ + Rewrite the expression #math.equation(block: false, alt: "the fraction 7 cubed over 13 to the power 9")[$frac(7^(3), 13^(9))$] as #math.equation(block: false, alt: "the fraction 7 cubed over 1 times the fraction 1 over 13 to the power 9")[$frac(7^(3), 1) × frac(1, 13^(9))$]. Then use the definition of negative exponents to rewrite the #math.equation(block: false, alt: "the fraction 1 over 13 to the power 9")[$frac(1, 13^(9))$] as #math.equation(block: false, alt: "13 to the power minus 9")[$13^(− 9)$]. Last, multiply, yielding #math.equation(block: false, alt: "the fraction 7 cubed over 1 times the fraction 1 over 13 to the power 9 equals 7 cubed times 13 to the power minus 9")[$frac(7^(3), 1) × frac(1, 13^(9)) = 7^(3) × 13^(− 9)$]. + Rewrite the expression #math.equation(block: false, alt: "the fraction c to the power 4 over d to the power 8")[$frac(c^(4), d^(8))$] as #math.equation(block: false, alt: "the fraction c to the power 4 over 1 times the fraction 1 over d to the power 8")[$frac(c^(4), 1) × frac(1, d^(8))$]. Then use the definition of negative exponents to rewrite the #math.equation(block: false, alt: "the fraction 1 over d to the power 8")[$frac(1, d^(8))$] as #math.equation(block: false, alt: "d to the power minus 8")[$d^(− 8)$]. Last, multiply, yielding #math.equation(block: false, alt: "the fraction c to the power 4 over 1 times the fraction 1 over d to the power 8 equals c to the power 4 times d to the power −8")[$frac(c^(4), 1) × frac(1, d^(8)) = c^(4) × d^(−8)$]. ] ] The table below shows a summary of the exponent rules from this section. #figure(table( columns: 3, align: left, inset: 6pt, table.header([Rule], [Example], [In Words]), [Product Rule #math.equation(block: false, alt: "a to the power n a to the power m equals a to the power n plus m")[$a^(n) a^(m) = a^(n + m)$]], [#math.equation(block: false, alt: "8 squared times 8 to the power 5 equals 8 to the power 7")[$8^(2) × 8^(5) = 8^(7)$]], [A base raised to a power, times the same based raised to another power, is the base raised to the sum of the powers.], [Quotient Rule #math.equation(block: false, alt: "the fraction a to the power n over a to the power m equals a to the power open parenthesis n minus m close parenthesis")[$frac(a^(n), a^(m)) = a^(( n − m ))$]], [#math.equation(block: false, alt: "the fraction 11 to the power 14 over 11 to the power 12 equals 11 to the power 12")[$frac(11^(14), 11^(12)) = 11^(12)$]], [A base raised to a power, divided by the same based raised to another power, is the base raised to the difference of the powers.], [Zero Power Rule #linebreak() #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$] provided that #math.equation(block: false, alt: "a not equal to 1")[$a ≠ 1$]], [#math.equation(block: false, alt: "412 to the power 0 equals 1")[$412^(0) = 1$]], [Any non-zero number raised to the zeroth power equals 1.], [Distributive Rule, Multiplication #math.equation(block: false, alt: "open parenthesis a times b close parenthesis to the power n equals a to the power n times b to the power n")[$attach(( a × b ), t: n) = a^(n) × b^(n)$]], [#math.equation(block: false, alt: "open parenthesis 14 times 31 close parenthesis to the power 9 equals 14 to the power 9 times 31 to the power 9")[$attach(( 14 × 31 ), t: 9) = 14^(9) × 31^(9)$]], [Exponents distribute across multiplication.], [Distributive Rule, Division #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")[$attach(( frac(a, b) ), t: n) = frac(a^(n), b^(n))$]], [#math.equation(block: false, alt: "open parenthesis the fraction 62 over 91 close parenthesis to the power 8 equals the fraction 62 to the power 8 over 91 to the power 8")[$attach(( frac(62, 91) ), t: 8) = frac(62^(8), 91^(8))$]], [Exponents distribute across division.], [Power Rule #math.equation(block: false, alt: "open parenthesis a to the power n close parenthesis to the power m equals a to the power open parenthesis n times m close parenthesis")[$attach(( a^(n) ), t: m) = a^(( n × m ))$]], [#math.equation(block: false, alt: "open parenthesis 5 to the power 9 close parenthesis to the power 15 equals 5 to the power 135")[$attach(( 5^(9) ), t: 15) = 5^(135)$]], [A base raised to a power, raised to another power, is the base raised to the first power times the second power.], [Negative Exponent Rule #math.equation(block: false, alt: "a to the power minus n equals the fraction 1 over a to the power n")[$a^(− n) = frac(1, a^(n))$] #linebreak() provided that #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]], [#math.equation(block: false, alt: "6 to the power minus 8 equals the fraction 1 over 6 to the power 8")[$6^(− 8) = frac(1, 6^(8))$] #linebreak() #math.equation(block: false, alt: "the fraction 1 over 12 to the power 7 equals 12 to the power minus 7")[$frac(1, 12^(7)) = 12^(− 7)$]], [A base raised to a negative exponent is 1 divided by the base raised to the positive power, and vice versa.], )) These rules often occur in tandem with each other, but it requires that you carefully apply the rules. #examplebox("Example 10")[Simplifying Expressions Using Exponent Rules][ Simplify the following: + #math.equation(block: false, alt: "open parenthesis the fraction 4 squared times 7 over 9 cubed close parenthesis to the power 5")[$attach(( frac(4^(2) × 7, 9^(3)) ), t: 5)$] + #math.equation(block: false, alt: "open parenthesis the fraction 5 a to the power 4 over b to the power 9 close parenthesis to the power 6")[$attach(( frac(5 a^(4), b^(9)) ), t: 6)$] #solutionbox[ + #strong[Step 1:] To simplify this, we start by distributing the power 5 across the quotient: #math.equation(block: true, alt: "open parenthesis the fraction 4 squared times 7 over 9 cubed close parenthesis to the power 5 equals the fraction open parenthesis 4 squared times 7 close parenthesis to the power 5 over open parenthesis 9 cubed close parenthesis to the power 5")[$attach(( frac(4^(2) × 7, 9^(3)) ), t: 5) = frac(attach(( 4^(2) × 7 ), t: 5), attach(( 9^(3) ), t: 5))$]#strong[Step 2:] We distribute the power 5 in the numerator across that multiplication: #math.equation(block: true, alt: "open parenthesis the fraction 4 squared times 7 over 9 cubed close parenthesis to the power 5 equals the fraction open parenthesis 4 squared times 7 close parenthesis over open parenthesis 9 cubed close parenthesis to the power 5 to the power 5 equals the fraction open parenthesis 4 squared close parenthesis to the power 5 times 7 to the power 5 over open parenthesis 9 cubed close parenthesis to the power 5")[$attach(( frac(4^(2) × 7, 9^(3)) ), t: 5) = attach(frac(( 4^(2) × 7 ), attach(( 9^(3) ), t: 5)), t: 5) = frac(attach(( 4^(2) ), t: 5) × 7^(5), attach(( 9^(3) ), t: 5))$]#strong[Step 3:] We apply the power rule where indicated: #math.equation(block: true, alt: "open parenthesis the fraction 4 squared times 7 over 9 cubed close parenthesis to the power 5 equals the fraction open parenthesis 4 squared times 7 close parenthesis over open parenthesis 9 cubed close parenthesis to the power 5 to the power 5 equals the fraction open parenthesis 4 squared close parenthesis to the power 5 times 7 to the power 5 over open parenthesis 9 cubed close parenthesis to the power 5 equals the fraction 4 to the power open parenthesis 2 times 5 close parenthesis times 7 to the power 5 over 9 to the power open parenthesis 3 times 5 close parenthesis equals the fraction 4 to the power 10 times 7 to the power 5 over 9 to the power 15")[$attach(( frac(4^(2) × 7, 9^(3)) ), t: 5) = attach(frac(( 4^(2) × 7 ), attach(( 9^(3) ), t: 5)), t: 5) = frac(attach(( 4^(2) ), t: 5) × 7^(5), attach(( 9^(3) ), t: 5)) = frac(4^(( 2 × 5 )) × 7^(5), 9^(( 3 × 5 ))) = frac(4^(10) × 7^(5), 9^(15))$] + #strong[Step 1:] To simplify this, we start by distributing the power 6 across the quotient: #math.equation(block: true, alt: "open parenthesis the fraction 5 a to the power 4 over b to the power 9 close parenthesis to the power 9 equals the fraction open parenthesis 5 times a to the power 4 close parenthesis to the power 6 over open parenthesis b to the power 9 close parenthesis to the power 6")[$attach(( frac(5 a^(4), b^(9)) ), t: 9) = frac(attach(( 5 × a^(4) ), t: 6), attach(( b^(9) ), t: 6))$]#strong[Step 2:] We distribute the power 5 in the numerator across that multiplication: #math.equation(block: true, alt: "the fraction open parenthesis 5 times a to the power 4 close parenthesis to the power 6 over open parenthesis b to the power 9 close parenthesis to the power 6 equals the fraction open parenthesis 5 close parenthesis to the power 6 times open parenthesis a to the power 4 close parenthesis to the power 6 over open parenthesis b to the power 9 close parenthesis to the power 6")[$frac(attach(( 5 × a^(4) ), t: 6), attach(( b^(9) ), t: 6)) = frac(attach(( 5 ), t: 6) × attach(( a^(4) ), t: 6), attach(( b^(9) ), t: 6))$]#strong[Step 3:] We apply the power rule where indicated: #math.equation(block: true, alt: "the fraction open parenthesis 5 close parenthesis to the power 6 times open parenthesis a to the power 4 close parenthesis to the power 6 over open parenthesis b to the power 9 close parenthesis to the power 6 equals the fraction 5 to the power 6 a to the power 24 over b to the power 54")[$frac(attach(( 5 ), t: 6) × attach(( a^(4) ), t: 6), attach(( b^(9) ), t: 6)) = frac(5^(6) a^(24), b^(54))$] ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Simplifying_Expressions_with_Exponents")[Simplifying Expressions with Exponents] ] === Key Terms - base - exponent === Key Concepts - Exponents are used to express multiplying a number by itself a number of times. The number being multiplied by itself is the base. The number of times it is multiplied by itself is the exponent, which is often referred to as the power. - Understanding that exponents represent repeated multiplication of a base makes it possible to establish some rules for combining exponential expressions, using the product rule, the quotient rule, and the power rule. Additionally, it allows us to formulate distributive rules for exponents. - Any non-zero number raised to the 0th power is 1. This makes the definition of the 0th power consistent with the division rule for exponents. - For consistency, negative exponents represent the reciprocal of the base raised to the power, so that #math.equation(block: false, alt: "a to the power minus n equals the fraction 1 over a to the power n")[$a^(− n) = frac(1, a^(n))$], provided that #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. === Formulas - #math.equation(block: true, alt: "a to the power n a to the power m equals a to the power n plus m")[$a^(n) a^(m) = a^(n + m)$] - #math.equation(block: true, alt: "the fraction a to the power n over a to the power m equals a to the power open parenthesis n minus m close parenthesis")[$frac(a^(n), a^(m)) = a^(( n − m ))$] - #math.equation(block: false, alt: "a to the power 0 equals 1")[$a^(0) = 1$], provided that #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$] - #math.equation(block: true, alt: "open parenthesis a times b close parenthesis to the power n equals a to the power n times b to the power n")[$attach(( a × b ), t: n) = a^(n) × b^(n)$] - #math.equation(block: true, 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")[$attach(( frac(a, b) ), t: n) = frac(a^(n), b^(n))$] - #math.equation(block: true, alt: "open parenthesis a to the power n close parenthesis to the power m equals a to the power open parenthesis n times m close parenthesis")[$attach(( a^(n) ), t: m) = a^(( n × m ))$] - #math.equation(block: false, alt: "a to the power minus n equals the fraction 1 over a to the power n")[$a^(− n) = frac(1, a^(n))$], provided that #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$] === Videos - #link("https://openstax.org/r/Exponential_Notation")[Exponential Notation] - #link("https://openstax.org/r/Product_and_Quotient_Rule_for_Exponents")[Product and Quotient Rule for Exponents] - #link("https://openstax.org/r/Fraction_Raised_to_a_Power")[Fraction Raised to a Power] - #link("https://openstax.org/r/Simplifying_Expressions_with_Exponents")[Simplifying Expressions with Exponents]