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 This is a great problems to use the change of base formula on – Why? On the other hand, why is problem #2 not the type of problem that you should use the change of base formula on? #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. Example 14: Newton’s Law of Cooling The temperature T of a cooling substance at time t (in minutes) is: T = (T 0 – T R ) e -rt + T R T 0 = initial temperature T R = room temperature r = constant cooling rate of the substance Solving Logarithmic Equations

11. 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°? Solving Logarithmic Equations T = (T 0 – T R ) e -rt + T R T 0 = 212, T R = 70, T = 100 r = 0.046 So solve: 100 = (212 – 70)e -0.046t + 70

12. 30 = 142e -0.046t (subtract 70) 15/71 = e -0.046t (divide by 142) How do you get the variable out of the exponent? Solving Logarithmic Equations 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