1. Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic function f(x) = log b x: If log b u = log b v, then u = v

2. Solve: Use one-to-one property

3. Solve: Use one-to-one property -OR- rewrite exponentially

4. Solve: rewrite exponentially Check:

5. log 2 x(x - 7) = 3 use product rule & rewrite 2 3 = x(x - 7) rewrite exponentially 8 = x 2 - 7x simplify 0 = x 2 - 7x - 8 set equal to zero 0 = (x-8)(x+1) factor Solve: log 2 x + log 2 (x - 7) = 3

6. x - 8 = 0 or x + 1 =0 set each factor = 0 x = 8 or x = -1 solve Check log 2 x + log 2 (x - 7) = 3 log 2 8 + log 2 (8 - 7) = 3 3 + log 2 1 = 3 3 + 0 = 3 3 = 3 Check log 2 x + log 2 (x - 7) = 3 log 2 -1 + log 2 (-1 - 7) = 3 The number -1 does not check. Negative numbers do not have logarithms. ? continued: 0 = (x-8)(x+1)

7. +7 +7 add 7 to both sides 3 x+2 = 34 ln 3 x+2 = ln 34 take ln of both sides (x+2) ln 3 = ln 34 rewrite exponent ln 3 ln 3 divide both sides by ln 3 x+2 = 3.21 simplify -2 -2 subtract 2 to both sides x = 1.21 Solve: 3 x+2 -7 = 27

8. Solve:

10. .

11. Orders of Magnitude The common logarithm of a positive quantity is its’ order of magnitude. (the difference in their powers of ten) DAY 2 Determine the order of magnitude: A kilometer and a meter. A dollar and a penny A horse weighing 400 kg and a mouse weighing 40 g. New York City with 7 million people and Earmuff Junction with a population of 7. A kilometer is 3 orders of magnitude longer than a meter. A dollar is 2 orders of magnitude greater than a penny. The horse is 4 orders of magnitude heavier than the mouse. New York is 6 orders of magnitude bigger than Earmuff Junction.

12. Earthquake Intensities The Richter scale magnitude R of an earthquake is a is the amplitude in micrometers (μm) of the vertical ground motion at the receiving station, T is the period of the associated seismic wave in seconds, and B accounts for the weakening of seismic wave with increasing distance from the epicenter of the earthquake. Measure of Acidity The acidity of a water-based solution is measured by the concentration of hydrogen ion [H + ] in the solution (in moles per liter). The measure of acidity used is pH, the opposite of the common log of the hydrogen-ion concentration:

13. Newton’s Law of Cooling The temperature, T, of a heated object at time t is given by T(t) = T m + (T 0 - T m )e -kt Where T m is the constant temperature of the surrounding medium, T 0 is the initial temperature of the heated object, and k is a constant that is associated with the cooling object.

14. Example A cake removed from the oven has a temperature of 210 0 F. It is left to cool in a room that has a temperature of 70 0 F. After 30 minutes, the temperature of the cake is 140 0 F. a.Use Newton’s Law of Cooling to find a model for the temperature of the cake, T after t minutes. b.What is the temperature of the cake after 40 minutes? c.When will the temperature of the cake be 90 0 F?

15. T = 70 + (210 - 70)e -kt T 0 =70 0 T m =210 0 T = 70 + 140e kt After 30 minutes the cake is 140 0 F 140 = 70 + 140e -k30 T=140 0 t=30 -70 70 = 140e -30k 140 1/2 = e -30k ln1/2 = ln e -30k take the ln of both sides ln(1/2) = -30k a. T = T m + (T 0 - T m )e -kt Newton’s Law of Cooling

16. ln(1/2) = -30k -30 -30 0.0231 = k The temperature of the cake is modeled by: T = 70 + 140e -0.0231t b. Find the temperature after 40 minutes T = 70 + 140e -0.0231(40) T = 126 0 F c. Find when the temperature of the cake will be 90 0 F. 90 = 70 + 140e -0.0231t

17. -70 -70 20 = 140e -0.0231t 140 140 1/7 = e -0.0231t ln1/7 = ln e -0.0231t take the ln of both sides ln(1/7) = -0.0231t -0.0231 84 = t The temperature of the cake will be 90 0 F after 84 minutes.

18. Complete Student Checkpoint An object is heated to 100C. It is left to cool in a room that has a temperature of 30C. After 5 minutes, the temperature of the object is 80C. a. Use Newton’s Law of Cooling to find a model for the temperature of the object, T, after t minutes.

19. Complete Student Checkpoint An object is heated to 100C. It is left to cool in a room that has a temperature of 30C. After 5 minutes, the temperature of the object is 80C. b. What is the temperature of the object after 20 minutes? c. When will the temperature of the object be 35C? The temperature of the object will be 35C after 39 minutes.

20. Equation Solving and Modeling