Glencoe Algebra 1: Multiplication and Division Properties of Exponents, Rational Exponents, and Exponential Functions Study Guide

Lesson 7-1: Multiplication Properties of Exponents

  • Learning Objectives and Goals:

    • Multiply monomials using the Product of Powers property.

    • Simplify complex expressions involving monomials using Power of a Power and Power of a Product properties.

    • Perform operations on expressions with exponents.

  • New Vocabulary Definitions:

    • Monomial: A number, a variable, or a product of a number and one or more variables with nonnegative integer exponents. It has only one term.

    • Constant: A monomial that is a real number (e.g., 34\frac{3}{4}).

  • Identifying Monomials:

    • Monomial Example: 17−c17 - c is NOT a monomial because the expression involves subtraction, meaning it has more than one term.

    • Monomial Example: 8f2g8f^2g IS a monomial because it is the product of a number and two variables.

    • Monomial Example: 34\frac{3}{4} IS a monomial because it is a constant.

    • Monomial Example: 5t\frac{5}{t} is NOT a monomial because the expression involves division by a variable.

  • Key Concept: Product of Powers:

    • Word Definition: To multiply two powers that have the same base, add their exponents.

    • Mathematical Symbols: For any real number aa and any integers mm and pp, am⋅ap=am+pa^m \cdot a^p = a^{m+p}.

    • Example 1: b3⋅b5=b3+5b^3 \cdot b^5 = b^{3+5} or b8b^8.

    • Example 2: g4⋅g6=g4+6g^4 \cdot g^6 = g^{4+6} or g10g^{10}.

  • Product of Powers Examples:

    • Simplifying (r4)(−12r7)(r^4)(-12r^7).

      • Group coefficients and variables: [1⋅(−12)](r4)(r7)[1 \cdot (-12)](r^4)(r^7).

      • Apply Product of Powers: [1⋅(−12)](r4+7)[1 \cdot (-12)](r^{4+7}).

      • Final Answer: −12r11-12r^{11}.

    • Simplifying (6cd5)(5c5d2)(6cd^5)(5c^5d^2).

      • Group coefficients and variables: (6⋅5)(c⋅c5)(d5⋅d2)(6 \cdot 5)(c \cdot c^5)(d^5 \cdot d^2).

      • Apply Product of Powers: (6⋅5)(c1+5)(d5+2)(6 \cdot 5)(c^{1+5})(d^{5+2}).

      • Final Answer: 30c6d730c^6d^7.

  • Key Concept: Power of a Power:

    • Word Definition: To find the power of a power, multiply the exponents.

    • Mathematical Symbols: For any real number aa and any integers mm and pp, (am)p=am⋅p(a^m)^p = a^{m \cdot p}.

    • Example 1: (63)5=63⋅5(6^3)^5 = 6^{3 \cdot 5} or 6156^{15}.

    • Example 2: (96)7=96⋅7(9^6)^7 = 9^{6 \cdot 7} or 9429^{42}.

  • Power of a Power Simplification:

    • Simplifying [(23)3]2[(2^3)^3]^2.

      • Inner power: (23)3=23⋅3=29(2^3)^3 = 2^{3 \cdot 3} = 2^9.

      • Outer power: (29)2=29⋅2=218(2^9)^2 = 2^{9 \cdot 2} = 2^{18}.

      • Final Answer: 2182^{18} or 262,144262,144.

  • Key Concept: Power of a Product:

    • Word Definition: To find the power of a product, find the power of each factor and multiply.

    • Mathematical Symbols: For any real numbers aa and bb and any integer mm, (ab)m=ambm(ab)^m = a^m b^m.

    • Example 1: (−2xy3)5=(−2)5x5(y3)5(-2xy^3)^5 = (-2)^5 x^5 (y^3)^5 or −32x5y15-32x^5y^{15}.

  • Application in Geometry:

    • Finding the Volume of a Cube with side length 5xyz5xyz.

      • Formula for volume of a cube: V=s3V = s^3.

      • Substitution: V=(5xyz)3V = (5xyz)^3.

      • Power of a Product: 53x3y3z35^3 x^3 y^3 z^3.

      • Final Answer: 125x3y3z3125x^3y^3z^3.

  • Criteria for a Simplified Monomial Expression:

    • Each variable base appears exactly once.

    • There are no powers of powers.

    • All fractions are in simplest form.

  • Exhaustive Simplification Example:

    • Simplifying [(8g3h4)2]2(2gh5)4[(8g^3h^4)^2]^2(2gh^5)^4.

      • Power of a Power (first term): (8g3h4)4(2gh5)4(8g^3h^4)^4(2gh^5)^4.

      • Power of a Product: (8)4(g3)4(h4)4(2)4g4(h5)4(8)^4(g^3)^4(h^4)^4 (2)^4 g^4 (h^5)^4.

      • Evaluate and apply Power of a Power: 4096g12h16(16)g4h204096g^{12}h^{16}(16)g^4h^{20}.

      • Commutative Property: 4096(16)g12⋅g4⋅h16⋅h204096(16) g^{12} \cdot g^4 \cdot h^{16} \cdot h^{20}.

      • Product of Powers: 65,536g16h3665,536g^{16}h^{36}.

Lesson 7-2: Division Properties of Exponents

  • Learning Objectives and Goals:

    • Find the quotient of two monomials.

    • Simplify expressions containing negative and zero exponents.

  • New Vocabulary Definitions:

    • Zero Exponent: Any nonzero number raised to the power of zero.

    • Negative Exponent: For any nonzero number aa and integer nn, a−na^{-n} is the reciprocal of ana^n.

    • Order of Magnitude: Used to compare large or small numbers by powers of 10.

  • Key Concept: Quotient of Powers:

    • Word Definition: To divide two powers with the same base, subtract the exponents.

    • Mathematical Symbols: For any nonzero number aa, and any integers mm and pp, amap=am−p\frac{a^m}{a^p} = a^{m-p}.

    • Example 1: c11c8=c11−8\frac{c^{11}}{c^8} = c^{11-8} or c3c^3.

    • Example 2: k14k2=k12\frac{k^{14}}{k^2} = k^{12}.

  • Quotient of Powers Example:

    • Simplifying x3y12x2y3\frac{x^3y^{12}}{x^2y^3}.

      • Group same bases: (x3x2)(y12y3)(\frac{x^3}{x^2})(\frac{y^{12}}{y^3}).

      • Product of Powers (Subtraction): x3−2y12−3=xy9x^{3-2}y^{12-3} = xy^9.

  • Key Concept: Power of a Quotient:

    • Word Definition: To find the power of a quotient, find the power of the numerator and the power of the denominator.

    • Mathematical Symbols: For any real numbers aa and b≠0b \neq 0, and any integer mm, (ab)m=ambm(\frac{a}{b})^m = \frac{a^m}{b^m}.

    • Example 1: (35)4=3454(\frac{3}{5})^4 = \frac{3^4}{5^4}.

  • Power of a Quotient Example:

    • Simplifying (4c3d25)3(\frac{4c^3d^2}{5})^3.

      • Apply Power of a Quotient: (4c3d2)353\frac{(4c^3d^2)^3}{5^3}.

      • Apply Power of a Product (Numerator): 43(c3)3(d2)3125\frac{4^3(c^3)^3(d^2)^3}{125}.

      • Final Answer: 64c9d6125\frac{64c^9d^6}{125}.

  • Key Concept: Zero Exponent Property:

    • Word Definition: Any nonzero number raised to the zero power is equal to 1.

    • Mathematical Symbols: For any nonzero number aa, a0=1a^0 = 1.

    • Examples: 150=115^0 = 1; (b20)0=1(\frac{b}{20})^0 = 1.

  • Zero Exponent Example:

    • Simplifying 12m8n78m5n100\frac{12m^8n^7}{8m^5n^{10}}^0.

      • Since the entire bracket is raised to the 0 power, the answer is 11.

    • Simplifying m0n3n2\frac{m^0n^3}{n^2}.

      • Substitute m0=1m^0 = 1: 1⋅n3n2\frac{1 \cdot n^3}{n^2}.

      • Subtract exponents (3−23-2): n1n^1 or nn.

  • Key Concept: Negative Exponent Property:

    • Word Definition: For any nonzero number aa and any integer nn, a−na^{-n} is the reciprocal of ana^n. Also, the reciprocal of a−na^{-n} is ana^n.

    • Mathematical Symbols: For any nonzero number aa and any integer nn, a−n=1ana^{-n} = \frac{1}{a^n} and 1a−n=an\frac{1}{a^{-n}} = a^n.

    • Example 1: 2−4=1242^{-4} = \frac{1}{2^4} or 116\frac{1}{16}.

  • Negative Exponent Example:

    • Simplifying x−2y3zx\frac{x^{-2}y^3z}{x}.

      • Regroup: (x−2x)(y3)(z)(\frac{x^{-2}}{x})(y^3)(z).

      • Combine exponents: x−2−1y3z=x−3y3zx^{-2-1}y^3z = x^{-3}y^3z.

      • Remove negatives: y3zx3\frac{y^3z}{x^3}.

    • Complex Negative Exponent: 5p−2q−1r−3\frac{5p^{-2}}{q^{-1}r^{-3}}.

      • Final Answer by moving negative powers to the opposite side of the fraction: 5qr3p2\frac{5qr^3}{p^2}.

  • Real World Example: Order of Magnitude:

    • Scenario: Darin has $123,456\$123,456 in savings. Tabo has $156\$156.

    • Standard Calculation: Darin is close to $100,000\$100,000 (10510^5). Tabo is close to $100\$100 (10210^2).

    • Ratio Calculation: 105102=103\frac{10^5}{10^2} = 10^3.

    • Conclusion: Darin has about 1000 times (10310^3) as much as Tabo, meaning Darin has 3 orders of magnitude as much money.

Lesson 7-3: Rational Exponents

  • Learning Objectives and Goals:

    • Evaluate and rewrite expressions involving rational exponents.

    • Solve equations involving expressions with rational exponents.

  • New Vocabulary Definitions:

    • Rational Exponent: An exponent that is a fraction.

    • Cube Root: If a3=ba^3 = b, then aa is the cube root of bb.

    • nth Root: For any real numbers aa and bb and any positive integer nn, if an=ba^n = b, then aa is an nth root of bb.

    • Exponential Equation: An equation in which variables occur as exponents.

  • Key Concept: b12b^{\frac{1}{2}}:

    • Word Definition: For any nonnegative real number bb, b12=bb^{\frac{1}{2}} = \sqrt{b}.

    • Examples: 1612=1616^{\frac{1}{2}} = \sqrt{16} or 44.

  • Key Concept: nth Root:

    • Word Definition: If an=ba^n = b, then bn=a\sqrt[n]{b} = a.

    • Examples: 2564\sqrt[4]{256}. Since 4⋅4⋅4⋅4=2564 \cdot 4 \cdot 4 \cdot 4 = 256, 2564=4\sqrt[4]{256} = 4.

    • Examples: 15,6256\sqrt[6]{15,625}. Since 56=15,6255^6 = 15,625, the answer is 55.

  • Key Concept: b1nb^{\frac{1}{n}}:

    • Word Definition: For any positive real number bb and any integer n>1n > 1, b1n=bnb^{\frac{1}{n}} = \sqrt[n]{b}.

    • Example: 813=83=28^{\frac{1}{3}} = \sqrt[3]{8} = 2.

    • Example: 133113=13313=111331^{\frac{1}{3}} = \sqrt[3]{1331} = 11.

  • Key Concept: bmnb^{\frac{m}{n}}:

    • Power Formulation: For any positive real number bb and any integers mm and n>1n > 1, bmn=(bn)mb^{\frac{m}{n}} = (\sqrt[n]{b})^m or bmn\sqrt[n]{b^m}.

    • Example: 823=(83)2=228^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 or 44.

    • Evaluation Example: 3225=(325)2=2232^{\frac{2}{5}} = (\sqrt[5]{32})^2 = 2^2 or 44.

    • Evaluation Example: 8152=(81)5=9581^{\frac{5}{2}} = (\sqrt{81})^5 = 9^5 or 59,04959,049.

  • Key Concept: Power Property of Equality:

    • Word Definition: For any real number b>0b > 0 and b≠1b \neq 1, bx=byb^x = b^y if and only if x=yx = y.

    • Example: If 5x=535^x = 5^3, then x=3x = 3.

  • Solving Exponential Equations:

    • Basic Example: Solve 9x=7299^x = 729.

      • Rewrite 729729 as a power of 9: 9x=939^x = 9^3.

      • By Power Property of Equality: x=3x = 3.

    • Intermediate Example: Solve 162x−1=816^{2x-1} = 8.

      • Rewrite with common base 2: (24)2x−1=23(2^4)^{2x-1} = 2^3.

      • Power of a Power: 28x−4=232^{8x-4} = 2^3.

      • Equate exponents: 8x−4=38x - 4 = 3.

      • Solve: 8x=7→x=788x = 7 \rightarrow x = \frac{7}{8}.

  • Real-World Application: Biology Populations:

    • Formula: p=40(2)t8p = 40(2)^{\frac{t}{8}}, where pp is population and tt is time in hours.

    • Problem: Find tt if p=20,480p = 20,480.

    • Step 1: 20,480=40(2)t820,480 = 40(2)^{\frac{t}{8}}.

    • Step 2: Divide by 40: 512=2t8512 = 2^{\frac{t}{8}}.

    • Step 3: Rewrite 512 as 292^9: 29=2t82^9 = 2^{\frac{t}{8}}.

    • Step 4: Equate exponents: 9=t89 = \frac{t}{8}.

    • Result: t=72t = 72 hours.

Lesson 7-4: Scientific Notation

  • Learning Objectives and Goals:

    • Express numbers in scientific notation.

    • Find products and quotients of numbers expressed in scientific notation.

  • New Vocabulary Definitions:

    • Scientific Notation: A way of expressing numbers that are too large or too small to be conveniently written in decimal form. It is written in the form a×10na \times 10^n, where 1≤a<101 \le a < 10 and nn is an integer.

  • Standard Form to Scientific Notation Procedure:

    • Step 1: Move the decimal point until it is to the right of the first nonzero digit. This results in the number aa.

    • Step 2: Note the number of places nn and the direction moved.

    • Step 3: If moved left, nn is positive (a×10na \times 10^n). If moved right, nn is negative (a×10−na \times 10^{-n}).

    • Step 4: Remove unnecessary trailing or leading zeros.

    • Example A (Large Number): 4,062,000,000,000→4.062×10124,062,000,000,000 \rightarrow 4.062 \times 10^{12}.

    • Example B (Small Number): 0.000000823→8.23×10−70.000000823 \rightarrow 8.23 \times 10^{-7}.

  • Scientific Notation to Standard Form Procedure:

    • Step 1: Note whether n>0n > 0 (positive) or n<0n < 0 (negative).

    • Step 2: If n>0n > 0, move the decimal point nn places right. If n<0n < 0, move it −n-n places left.

    • Step 3: Insert placeholder zeros and commas.

    • Example A: 6.49×105=649,0006.49 \times 10^5 = 649,000.

    • Example B: 1.8×10−3=0.00181.8 \times 10^{-3} = 0.0018.

  • Multiplication with Scientific Notation:

    • Example: Evaluate (5×10−6)(2.3×1012)(5 \times 10^{-6})(2.3 \times 10^{12}).

      • Group coefficients and group powers: (5×2.3)×(10−6×1012)(5 \times 2.3) \times (10^{-6} \times 10^{12}).

      • Multiply: 11.5×10611.5 \times 10^6.

      • Adjust to proper scientific notation (1.15×101×1061.15 \times 10^1 \times 10^6): 1.15×1071.15 \times 10^7.

      • Standard form: 11,500,00011,500,000.

  • Division with Scientific Notation:

    • Example: Evaluate 4.5×10−21.5×10−3\frac{4.5 \times 10^{-2}}{1.5 \times 10^{-3}}.

      • Quotient of coefficients: 4.51.5=3\frac{4.5}{1.5} = 3.

      • Quotient of powers: 10−2−(−3)=10110^{-2 - (-3)} = 10^1.

      • Result: 3×101=303 \times 10^1 = 30.

  • Real World Example: Business Revenue:

    • Newspaper circulation: 150150 thousand = 150,000150,000 or 1.5×1051.5 \times 10^5.

    • Advertising revenue: $1.2\$1.2 million = $1,200,000\$1,200,000 or $1.2×106\$1.2 \times 10^6.

Lesson 7-5: Exponential Functions

  • Learning Objectives and Goals:

    • Graph exponential functions.

    • Identify data that display exponential behavior.

  • New Vocabulary Definitions:

    • Exponential Function: A function that can be described by an equation of the form y=abxy = ab^x, where a≠0a \neq 0, b>0b > 0, and b≠1b \neq 1.

    • Exponential Growth Function: A function where a>0a > 0 and b>1b > 1.

    • Exponential Decay Function: A function where a>0a > 0 and 0<b<10 < b < 1.

  • Characteristics of Exponential Growth Graphs (f(x)=abx,b>1f(x) = ab^x, b > 1):

    • Equation: f(x)=abxf(x) = ab^x.

    • Domain: All real numbers.

    • Range: All positive real numbers.

    • Intercepts: One y-intercept (at aa), no x-intercepts.

    • End Behavior: As xx increases, f(x)f(x) increases; as xx decreases, f(x)f(x) approaches 00.

  • Characteristics of Exponential Decay Graphs (f(x)=abx,0<b<1f(x) = ab^x, 0 < b < 1):

    • Equation: f(x)=abxf(x) = ab^x.

    • Domain: All real numbers.

    • Range: All positive real numbers (transcript incorrectly notes range as all negative reals, but the graph confirms all positive reals).

    • Intercepts: One y-intercept, no x-intercepts.

    • End Behavior: As xx increases, f(x)f(x) approaches 00; as xx decreases, f(x)f(x) increases.

  • Real-World Application: Car Depreciation:

    • Formula: V=25,000⋅0.82tV = 25,000 \cdot 0.82^t.

    • VV is value, tt is time in years.

    • Initial cost: $25,000\$25,000.

    • Question: Value after 5 years?

    • Calculation: V=25,000⋅0.825≈9268V = 25,000 \cdot 0.82^5 \approx 9268.

    • Meaningful Values: t≥0t \ge 0 and 0≤V≤25,0000 \le V \le 25,000.

  • Identifying Exponential Behavior from Data:

    • Look for a pattern in the domain and range.

    • If domain values are at regular intervals and range values have a common factor (not common difference), the behavior is exponential.

    • Example Data: (0,10),(10,25),(20,62.5)(0, 10), (10, 25), (20, 62.5).

      • Common factor in range: 2510=2.5\frac{25}{10} = 2.5; 62.525=2.5\frac{62.5}{25} = 2.5.

      • This indicates exponential behavior: y=a(2.5)xy = a(2.5)^x.

Lesson 7-6: Growth and Decay

  • Learning Objectives and Goals:

    • Solve problems involving exponential growth (e.g., populations, investments).

    • Solve problems involving exponential decay (e.g., charity trends, depreciation).

  • New Vocabulary Definitions:

    • Compound Interest: Interest earned or paid on both the initial principal and previously earned interest.

  • Key Concept: Equation for Exponential Growth:

    • Equation: y=a(1+r)ty = a(1 + r)^t.

    • aa: Initial amount.

    • yy: Final amount.

    • tt: Time.

    • rr: Rate of growth as a decimal (r>0r > 0).

  • Exponential Growth Application: Population:

    • Problem: Town population was 280,000280,000 in 2015; growth rate is 0.85%0.85\%.

    • Equation: y=280,000(1+0.0085)ty = 280,000(1 + 0.0085)^t or y=280,000(1.0085)ty = 280,000(1.0085)^t.

    • To find population in 2028: Set t=13t = 13 (2028−20152028 - 2015).

    • Calculation: y=280,000(1.0085)13≈312,216y = 280,000(1.0085)^{13} \approx 312,216 (Note: page 170 uses a 2018-2028 window for t=10t=10, resulting in 304,731304,731).

  • Key Concept: Equation for Compound Interest:

    • Equation: A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt}.

    • AA: Current amount.

    • PP: Principal (initial amount).

    • rr: Annual interest rate (r>0r > 0).

    • nn: Number of times interest is compounded each year.

    • tt: Time in years.

  • Compound Interest Example:

    • Investment: $1000\$1000 at 7%7\% compounded annually (n=1n=1) for 1818 years.

    • Calculation: A=1000(1+0.07)1⋅18A = 1000(1 + 0.07)^{1 \cdot 18}.

    • A=1000(1.07)18≈$3379.93A = 1000(1.07)^{18} \approx \$3379.93.

  • Key Concept: Equation for Exponential Decay:

    • Equation: y=a(1−r)ty = a(1 - r)^t.

    • aa: Initial amount.

    • yy: Final amount.

    • rr: Rate of decay as a decimal (0<r<10 < r < 1).

    • tt: Time.

  • Exponential Decay Application: Charity Donations:

    • Initial donations: $390,000\$390,000.

    • Drop rate: 1.1%1.1\% per year (r=0.011r = 0.011).

    • Equation: y=390,000(1−0.011)t=390,000(0.989)ty = 390,000(1 - 0.011)^t = 390,000(0.989)^t.

    • Estimate after 5 years: y=390,000(0.989)5≈$369,017y = 390,000(0.989)^5 \approx \$369,017.

Questions & Discussion

  • Substitution/Elimination Question: Solve system r−t=−5r - t = -5, r+t=25r + t = 25. (Answer: (10,15)(10, 15)).

  • Substitution/Elimination Question: Solve system 2x+y=72x + y = 7, y=0.5x+2y = 0.5x + 2. (Answer: (2,3)(2, 3)).

  • Word Problem: Tens digit of a two-digit number is 5 more than twice the ones digit. Sum of digits is 8. What is the number? (Calculation: Let t=2o+5t = 2o + 5. t+o=8→2o+5+o=8→3o=3→o=1t + o = 8 \rightarrow 2o + 5 + o = 8 \rightarrow 3o = 3 \rightarrow o = 1. Number is 71).

  • Division Checkpoint: Simplify a3b9ab2\frac{a^3b^9}{ab^2}. Answer: a3−1b9−2=a2b7a^{3-1}b^{9-2} = a^2b^7.

  • Power Checkpoint: Simplify [(42)2]3[(4^2)^2]^3. Answer: 42⋅2⋅3=4124^{2 \cdot 2 \cdot 3} = 4^{12}.

  • Scientific Notation Checkpoint: Express 0.00004520.0000452 in scientific notation. Answer: 4.52×10−54.52 \times 10^{-5}.