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Section 7.3 Rational Exponents
Algebra 1
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Learning Targets Define a rational exponent Define radical form
Define and apply β n th β root Evaluate β π π‘β β root expressions Define and apply Power Property of Equality Solve exponential equations
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Recall: Exponent Definition
3 4 =3β3β3β3=81 Remember, this is saying that I want 4 pieces of the base multiplied together
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Multiplicative VS Additive Half
12 = 6+6 18=9+9 16=8+8 Multiplicative Half: 16=4β4 36 = 6β6 25= 5β5 Thus, the additive half of 16 is 8 and the multiplicative half of 16 is 4.
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Multiplicative VS Additive Thirds
9= 3+3+3 12 = 4+4+4 30= Multiplicative Third: 8= 2 β2 β2 27= 3 β3 β3 125= 5β5β5 Thus, the additive third of 30 is 10 and the multiplicative third of 125 is 5.
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Explore: Exponent Definition
What if I have something likeβ¦ This is saying that I know 81 is made of 4 equal pieces and I only want 1 of those pieces. Or I want the multiplicative fourth of 81 Thus, 81=3β3β3β3 and one of the four pieces is just 3. So, =3
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Concept 1: Basic Rational Exponent
The most common rational exponent is π₯ = π₯ π₯ is also known as the square root. This representation is also known as radical form. π₯ is asking for the multiplicative half of a number π₯
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Concept 1: Basic Rational Exponent
Practice 1: What is ? 16=4β4 Thus, =4 Practice 2: Find 100 100=10β10 Thus, =10
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Concept 2: β π ππ β Roots If π π =π, then π 1 π = π π =π
Ex: 2 4 =16, then = 4 16 =2 4 16 = This is saying, I know 16 has 4 equal multiplicative pieces. Itβs then asking, what is 1 of those 4 pieces. 2β2β2β2=16 Thus, =2
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Concept 2: β π ππ β Roots Practice 1: Practice 2: What is 27 1 3 ?
27=3β3β3 Thus, =3 Practice 2: Find 32=2β2β2β2β2 Thus, =2
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Concept 2: β π ππ β Roots Practice 3: Practice 4: What is 64 1 3 ?
64=4β4β4 Thus, =4 Practice 4: Find 125=5β5β5 Thus, =5
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Concept 3: Advanced β π ππ β Roots
π π π = π π π Ex: = =8 = This is saying, I know 16 has 4 equal multiplicative pieces. Itβs then asking, what is 3 of those 4 pieces. 2β2β2β2=16 Thus, =8
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Concept 3: Advanced β π ππ β Roots
Practice 1: What is ? 27=3β3β3 Thus, =9 Practice 2: Find 36=6β6 Thus, =216
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Concept 3: Advanced β π ππ β Roots
Practice 3: What is ? 64=4β4β4 Thus, =16 Practice 4: Find 32=2β2β2β2β2 Thus, =4
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Concept 4: Solving Exponential Equations
Power Property of Equality For any real number π>0 and πβ 1, then π π₯ = π π¦ if and only if π₯=π¦. Example 1: If 5 π₯ = 5 3 , then π₯=3 Example 2: If 2 π₯+1 = 2 7 , then π₯+1=7
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Concept 4: Solving Exponential Equations
Practice 2: Solve 25 π₯β1 =5 5 2 π₯β1 = 5 1 2 π₯β1 =1 2π₯β2=1 Thus, π₯= 3 2 Practice 1: Solve 6 π₯ =216 6 π₯ = 6 3 Thus, π₯=3
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Concept 4: Solving Exponential Equations
Practice 4: Solve 12 2π₯+3 =144 12 2π₯+3 = 12 2 2π₯+3=2 Thus, π₯=β 1 2 Practice 3: Solve 5 π₯ =125 5 π₯ = 5 3 Thus, π₯=3
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Exit Ticket for Feedback
1. Solve 4 2π₯β1 = 2 3 2. Find
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