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Mastering Exponential and Logarithmic Equations with Change of Base Formula

In this lesson, students will learn to solve exponential equations using the Change of Base Formula. They will gain proficiency in evaluating and solving logarithms, including tricky logarithmic equations. Key concepts include natural logarithms and variable exponents. Through practical examples, such as Newton's Law of Cooling, students will apply these principles to real-world scenarios. By the end of the session, participants will be prepared to tackle homework problems and deepen their understanding of exponential and logarithmic relationships.

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Mastering Exponential and Logarithmic Equations with Change of Base Formula

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  1. 5.7 – Exponential Equations

  2. 5.7 Exponential Equations Objectives: • Solve Exponential Equations using the Change of Base Formula • Evaluate logarithms • Solve logarithms Vocabulary: logarithms, natural logarithms

  3. Daily Objectives • Perform the Change of Base formula. • Master solving tricky logarithm equations. • Exponent variables

  4. Topic One: Change of Base Formula The BASE goes to the BOTTOM

  5. Example 1: Using Change of Base Formula On the other hand, why is problem #2 not the type of problem that you should use the change of base formula on? • = 3 • log 1000 This is a great problems to use the change of base formula on – Why? #3 can work with either – Why?

  6. Example 2: Solve for a variable exponent

  7. Example 3: Solve for a variable exponent

  8. Example 4: Solve for a variable exponent

  9. Example 5: Solve for a variable exponent

  10. Solving Logarithmic Equations Example 14: Newton’s Law of Cooling • The temperature T of a cooling substance at time t (in minutes) is: • T = (T0 – TR) e-rt + TR • T0= initial temperature • TR= room temperature • r = constant cooling rate of the substance

  11. Solving Logarithmic Equations Example 14: You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r = 0.046. How long will it take to cool the stew to a serving temp. of 100°? T = (T0 – TR) e -rt + TR T0 = 212, TR = 70, T = 100 r = 0.046 So solve: 100 = (212 – 70)e -0.046t + 70

  12. Solving Logarithmic Equations 30 = 142e -0.046t(subtract 70) 15/71 = e -0.046t(divide by 142) • How do you get the variable out of the exponent? ln(15/71) = lne-.046t (take the ln of both sides) ln(15/71) = – 0.046t ln(15/71)/(– 0.046) = t t =(ln15 – ln71)/(– 0.046) = t ≈ – 1.556/(– 0.046) t ≈ 33.8 about 34 minutes to cool!

  13. Homework • TB p. 205 #9-14, 23

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