1. 5-Minute Check on Activity 5-14 Click the mouse button or press the Space Bar to display the answers. Use the properties of logarithms or your calculator to solve the following equations: 1.14 = 3e x 2.3 = (1.04) x 3.6 = 4(2.2) x 4.5 = 1.3e 3x ln (14/3) = ln e x = x ln 3 = x ln 1.04 x = ln 3  ln 1.04 y1 = 3e x y2 = 14 x = 1.54 y1 = (1.04) x y2 = 3 x = 28.01 ln 6/4 = x ln 2.2 x = ln 1.5  ln 2.2 y1 = 4(2.2) x y2 = 6 x = 0.51 ln 5/1.3 = 3x ln e x = (ln 5/1.3) / 3 y1 = 1.3e 3x y2 = 5 x = 0.45

2. Activity 5 - 15 Frequency and Pitch

3. Objectives Solve logarithmic equations both graphically and algebraically

4. Vocabulary None new

5. Activity Raising a musical note one octave has the effect of doubling the pitch, or frequency, of the sound. However, you do not perceive the note to sound “twice as high,” as you might predict. Perceived pitch is given by the function P(f) = 2410 log (0.0016f + 1) where P is the perceived pitch in mels (units of pitch) and f is the frequency in hertz. Graph the function What is the perceived pitch, P, for an input of 10,000 hertz? P(10000) = 2410 log (0.0016f(10000) + 1) ≈ 2965.38 mels

6. Activity cont Write an equation that can be used to determine what frequency, f, gives an output of 2000 mels. Solve it using the graphing approach 2410 log (0.0016f + 1) = 2000 Y1 = 2410 log (0.0016x + 1) Y2 = 2000 Find the intersection: 3599.31 hertz

7. Algebraic Approach 1.Rewrite equation into form: log b (f(x)) = c (all positive) 2.Rewrite step 1 in exponential form: f(x) = b c 3.Solve the resulting equation from step 2 algebraically 4.Check solution in the original equation

8. Activity cont Solve the equation 2410 log (0.0016f + 1) = 2000 using an algebraic approach Solve the equation 2410 log (0.0016f + 1) = 2000 Divide both sides: log (0.0016f + 1) = 2000/2410 Exponential Form: (0.0016f + 1) = 10 2000/2410 Solve: f = ( 10 2000/2410 - 1) / 0.0016 f ≈ 3,599 Hz

9. Activity cont Use an algebraic approach to determine the frequency, f, that produces a perceived pitch of 3000 mels. Solve the equation 2410 log (0.0016f + 1) = 3000 Divide both sides: log (0.0016f + 1) = 3000/2410 Exponential Form: (0.0016f + 1) = 10 3000/2410 Solve: f = ( 10 3000/2410 - 1) / 0.0016 f ≈ 10,357.30 Hz

10. Summary and Homework Summary –Graphical Solution 1.Y1 = log function and Y2 = constant value 2.Graph and find intersection –Algebraic Solution 1.Rewrite equation into form: log b (f(x)) = c (all positive) 2.Rewrite step 1 in exponential form: f(x) = b c 3.Solve the resulting equation from step 2 algebraically 4.Check solution in the original equation Homework –pg 675 – 76; problems 1-6, 8