UNIT 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Section 1



















- Slides: 19

UNIT 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Section 1 – Exponential Applications

Exponential Applications Supplemental Packet pg. 67 y = a(b)x is the parent function It is exponential because x is the exponent. i. a is the Starting value (this is y-intercept) ii. b is the growth or decay factor. This tells you the pattern for the function b > 1 is growth Ex. y = 4(1. 3)x iii. 0 < b < 1 is decay Ex. y = 4(. 7)x

Exponential Applications Example 1: y = 4(2)x a. a = 4 which means he started with 4, 000 b. It is growing since b = 2 is greater than 1 c. In 1 week, they double (b = 2), so 8, 000 In week 2, they double again, so 16, 000 We can use the equation to predict. y = 4(2)21 PRIZM: Table app (7) Y 1: 4(2)x Enter in 21, 28, and 42 into the table.

Exponential Applications Example 2: a. a = 8, and b = 3, since it triples y = 8(3)x b. PRIZM: Table app (7) Y 1: 4(3)x Enter in 24, 48, and 100 into the table.

Exponential Applications Example 3: 1024 is the starting value, this is how many grams we started with. ½ is the decay factor. After each week, there is half of what it previously was. After 1 week, there would be 512 grams. After week 2, there would be 256 grams. After week 3, there would be 128 grams.

Exponential Applications Example 4: A) a = 12, since we are starting with 12, 000 b = ? We are growing at a rate of 7% *rate is different than a growth factor B) b = 1 + rate b = 1 +. 07 b = 1. 07 *It is now > 1, indicating growth. Equation: y = 12(1. 07)x C) 2025 is x = 35 since the first year was 1990 y = 12(1. 07)35 = 128. 119 128, 119 people

Exponential Applications Example 5: A) a = 45, since we are starting with $45, 000 b = 1 + rate b = 1 + (-. 12) b =. 88 *It is now < 1, indicating decay. Equation: y = 45(. 88)x C) y = 45(. 88)9 = 14. 242 $14, 242

Exponential Applications PRACTICE supplemental packet pg. 68

Exponential Applications from data Example 1 A) Rate = 8700 – 9000 = -300 = -. 03 9000 This is 33% decrease for the RATE B) a = 9; b = 1+rate = 1 + -. 03 =. 97 y= 9(. 97)x C) y = 9(. 97)12 = 6. 245 = 6, 245 people

Exponential Applications Example 2 Rate = 20 – 60 = 40 = 2 20 20 This is 200% increase rate. So b = 1+r = 3 This is consistent throughout, it triples A) a = 20; b = 3 y= 20(3)x B) y = 20(3)9 = 393, 660. C) Not accurate, since it will start to decrease. This is called EXTRAPOLATION.

Exponential Applications Example 3 A) Rate = 34 – 32. 5 = -. 044 34 y= 34(. 956)x B) $21, 680 b=. 956

Exponential Applications Example 4 A) PRIZM: Go to the STATISTICS App (2) Enter in the data in List 1 and List 2 F 2, F 3, F 6, F 2 will calculate the regression equation for the EXPONENTIAL, this is different from the Linear in Unit 2 Equation: y = 42. 58(1. 04)x

Exponential Applications Example 4 B) F 6 will copy the equation into the Table App (7) so that you can make predictions. Go to the Table App (7) and enter in each year value (18, 20, 22, 24, 30, 50) C) Too many things change in population, 2050 is too far to predict accurately, so this probably extrapolation.

Exponential Applications PRACTICE: Supplemental packet pg. 70: 1, 3 and complete page 71

Exponential Applications Specific Applications (packet p. 72) Example 1: HALF LIFE y = a(b) t/h b is always ½ since we are halving the amount each cycle. h = half life time. The amount of time to complete a cycle. In the exponent, we divide t by h to find the number of cycles. Ex. If half life is h = 10 and we want to predict for 30 years from now, then 30/10 means there were 3 cycles. It become half of what it was 3 times.

Exponential Applications Specific Applications (packet p. 72) Practice p. 72: 7 – 9

Exponential Applications Specific Applications (packet p. 72) Example 2: Compound Interest Continuously A = P(e) rt A is the ending amount after so many years. P is the starting amount e is the growth factor e = 2. 71828 r is the rate and t is the time. .

Exponential Applications Specific Applications (packet p. 72) Example 2: Compound Interest Continuously A = P(e) rt 5) A = 5000(e). 069(30) (Use shift, ln on PRIZM) A = $39, 624. 12

Exponential Applications Practice: textbook p. 442: 15 – 17, 24 – 26, 37, 39