1. Holt McDougal Algebra 2 7-6 The Natural Base, e Warm Up Simplify. x 1. log 10 x 2. log b b 3w 3. 10 log z 3w3w z 4. b log b (x – 1 ) x – 1 Slide 1 of 16

2. Holt McDougal Algebra 2 7-6 The Natural Base, e e is from euler’s number and e = 2.71828… e is a number just like ∏ is a number that is equal to 3.14159… Evaluate e 2 Evaluate e -3 The decimal value of e looks like it repeats: 2.718281828… The value is actually 2.71828182890… There is no repeating portion. Caution Slide 2 of 16

3. Holt McDougal Algebra 2 7-6 The Natural Base, e Recall graphing the exponential function: y = 2 x Now graph y = e x Now what is the domain, range, asymptote, EB and inc/dec? Slide 3 of 16

4. Holt McDougal Algebra 2 7-6 The Natural Base, e Graph f(x) = e x–2 + 1. Example 1: Graphing Exponential Functions Slide 4 of 16

5. Holt McDougal Algebra 2 7-6 The Natural Base, e A logarithm with a base of e is called a natural logarithm and is abbreviated as “ln” (rather than as log e ). Natural logarithms have the same properties as log base 10 and logarithms with other bases. Slide 5 of 16

6. Holt McDougal Algebra 2 7-6 The Natural Base, e Simplify. A. ln e 0.15t B. e 3ln(x +1) ln e 0.15t = 0.15te 3ln(x +1) = (x + 1) 3 ln e 2x + ln e x = 2x + x = 3x Example 2: Simplifying Expression with e or ln C. ln e 2x + ln e x Slide 6 of 16

7. Holt McDougal Algebra 2 7-6 The Natural Base, e Simplify. a. ln e 3.2 b. e 2lnx c. ln e x +4y ln e 3.2 = 3.2 e 2lnx = x 2 ln e x + 4y = x + 4y Check It Out! Example 2 Slide 7 of 16

8. Holt McDougal Algebra 2 7-6 The Natural Base, e Recall the compound interest formula A = P(1 + ) nt, where A is the amount, P is the principal, r is the annual interest, n is the number of times the interest is compounded per year and t is the time in years. n r The formula for continuously compounded interest is A = Pe rt, where A is the total amount, P is the principal, r is the annual interest rate, and t is the time in years. Slide 8 of 16

9. Holt McDougal Algebra 2 7-6 The Natural Base, e What is the total amount for an investment of $500 invested at 5.25% for 40 years and compounded continuously? Example 3: Economics Application The total amount is $4083.08. A = Pe rt Substitute 500 for P, 0.0525 for r, and 40 for t. A = 500e 0.0525(40) Use the e x key on a calculator. A ≈ 4083.08 Slide 9 of 16

10. Holt McDougal Algebra 2 7-6 The Natural Base, e What is the total amount for an investment of $100 invested at 3.5% for 8 years and compounded continuously? The total amount is $132.31. A = Pe rt Substitute 100 for P, 0.035 for r, and 8 for t. A = 100e 0.035(8) Use the e x key on a calculator. A ≈ 132.31 Check It Out! Example 3 Slide 10 of 16

11. Holt McDougal Algebra 2 7-6 The Natural Base, e The half-life of a substance is the time it takes for half of the substance to breakdown or convert to another substance during the process of decay. Natural decay is modeled by the function below. Slide 11 of 16

12. Holt McDougal Algebra 2 7-6 The Natural Base, e Pluonium-239 (Pu-239) has a half-life of 24,110 years. How long does it take for a 13 g sample of Pu-239 to decay to 4 g? Example 4: Science Application It takes approximately 40,998 years to decay. Slide 12 of 16

13. Holt McDougal Algebra 2 7-6 The Natural Base, e Determine how long it will take for 650 mg of a sample of chromium-51 which has a half-life of about 28 days to decay to 200 mg. Check It Out! Example 4 It takes approximately 47.7 days to decay. Slide 13 of 16

14. Holt McDougal Algebra 2 7-6 The Natural Base, e Doubling Your Investment. How long does it take for an investment to double at an annual interest rate of 8.5% compounded continuously? How long does it take for an investment to triple at an annual interest rate of 7.2% compounded continuously? Slide 14 of 16

15. Holt McDougal Algebra 2 7-6 The Natural Base, e Lesson Quiz Simplify. 1. ln e –10t t 0.25 –10t –x–x2x22x2 2. e 0.25 lnt 3. –ln e x 4. 2ln e x 2 5. What is the total amount for an investment of $1000 invested at 7.25% for 15 years and compounded continuously? ≈ 21,000 yr 6. The half-life of carbon-14 is 5730 years. What is the age of a fossil that only has 8% of its original carbon-14? ≈ $2966.85 Slide 15 of 16

16. Holt McDougal Algebra 2 7-6 The Natural Base, e HW: PG 534 2, 6-14, 17-21, 31-36, 54-58 evens Slide 16 of 16 Remember on 12 we use a different formula so you can’t find the decay constant