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Solve the equations. 4 2π₯ = 8 π₯+7 πππ 125 25 8 5π₯ = 64 5π₯β5 πππ 7 49
Warm Up Solve the equations. 4 2π₯ = 8 π₯+7 πππ 8 5π₯ = 64 5π₯β5 πππ 7 49
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Solving Logarithmic Equations
Lessons Solving Logarithmic Equations (log on one side) (log on both sides) Product Rule Quotient Rule Power Rule Solving Logs
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Solving Logarithm Equations with a logarithm on one side and a number on the other
πππ π π= π π
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7.4 Logarithm Equations Letβs begin this lesson with a puzzle.
See if you can figure out a way to solve the following questions based on the previous lesson.
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πππ 9 π₯= 3 2 7.4 Logarithm Equations
Convert the logarithm equation into an exponential equation to solve NO MATTER where the π₯ is! πππ 9 π₯= 3 2
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Solve the following logarithmic equations. πππ 16 π₯= 5 2 πππ 81 π₯= 3 4
Practice Solve the following logarithmic equations. πππ 16 π₯= 5 2 πππ 81 π₯= 3 4
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Solving Logarithm Equations with a logarithm both sides
πππ π π+ πππ π π= πππ π π
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The bases must be the same!
7.5 Properties Where have we talked about the βProperty of Equalityβ before? If πππ π π₯= πππ π π¦, then π₯=π¦. The bases must be the same! Example: πππ 10 π₯= πππ 10 (5π₯β20) βdelete the logsβ
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Practice Solve the following logarithmic equations.
πππ 4 π₯ 2 = πππ 4 (β6π₯β8) πππ 3 π₯ 2 β15 = πππ 3 2π₯
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You Try!!! Solve the following logarithmic equations.
πππ 2 π₯β4 = πππ 2 3π₯ πππ 5 π₯ 2 β10 = πππ 5 3π₯
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Product Property of Logarithms πππ π π₯π¦ = πππ π π₯+ πππ π π¦
Logarithm of a Product Product Property of Logarithms πππ π π₯π¦ = πππ π π₯+ πππ π π¦
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Express as a sum of logarithms. πππ π 16β32 πππ π (8β16) πππ π§ (3ππ)
Practice βthe log of a product is the sum of the logsβ Express as a sum of logarithms. πππ π 16β32 πππ π (8β16) πππ π§ (3ππ)
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Practice Express as a single logarithms. πππ π¦ 65+ πππ π¦ 2
βthe sum of the logs is the log of the productsβ Express as a single logarithms. πππ π¦ 65+ πππ π¦ 2 πππ π π»+ πππ π K πππ π 8+ πππ π π
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Logarithm of a Quotient
Quotient Property of Logarithms πππ π π₯ π¦ = πππ π π₯β πππ π π¦
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Express as a difference of logarithms. πππ π 6 5 πππ π π¦ π₯ πππ π§ 8 3
Practice βthe log of a quotient is the difference of the logsβ Express as a difference of logarithms. πππ π 6 5 πππ π π¦ π₯ πππ π§ 8 3
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Practice Express as a single logarithm. πππ π 42β πππ π 7
βthe difference of the logs is the log of the quotientβ Express as a single logarithm. πππ π 42β πππ π 7 πππ π π΄β πππ π πΆ πππ π‘ 5β πππ π‘ 13
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Power Property of Logarithms πππ π π₯ π = πβπππ π π₯
Logarithm of a Power Power Property of Logarithms πππ π π₯ π = πβπππ π π₯
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Express as a product. πππ π π‘ 2 πππ 3 π β2
Practice Express as a product. πππ π π‘ 2 πππ 3 π β2
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Express as a single logarithm. 5 πππ π 2 β2πππ 3 10
Practice Express as a single logarithm. 5 πππ π 2 β2πππ 3 10
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Solving Logarithms Use a property of logarithms to combine the left side of the equation. Use the equality property of logarithms to write a new equation. Solve the equation for π₯.
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Practice Solve each equation. πππ 6 π₯+ πππ 6 9= πππ 6 54
πππ 9 3π₯+14 β πππ 9 5= πππ 9 2π₯
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Practice Solve each equation. 4πππ 2 π₯+ πππ 2 5= πππ 2 405
πππ 3 π¦=β πππ πππ 3 64
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YOU TRY!!! πππ 8 48β πππ 8 π€= πππ 8 4 Solve each equation.
πππ 2 π₯=5 πππ 2 2β πππ 2 8
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HOMEWORK Lesson Pg. 480 #βs 11-15 Pg. 488 #βs 23-26
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