4-6

1. The Natural Base, e 4-6 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

2. x=7 • x=12.5 • No solution • x=2 • x=2 • x=-4.5 • x=29 • x=-1/3 • x=7.5 • x=4 • x=17/16 • x=4 • x=6 • x=-2 • x=53 • x=7 • No solution • X=3.5 • x=11/4 • x=-2 • x=28 • x=9 • No solution • x=2 • No solution • No solution 27. x=-4 cannot be a solution

3. Warm Up Simplify. x 1. log10x 3w 2. logbb3w z 3. 10log z x –1 4. blogb(x –1) 3x – 2 5.

4. Objectives Use the number e to write and graph exponential functions representing real-world situations. Solve equations and problems involving e or natural logarithms.

5. Vocabulary natural logarithm natural logarithmic function

6. Caution The decimal value of e looks like it repeats: 2.718281828… The value is actually 2.71828182890… There is no repeating portion.

7. Exponential functions with e as a base have the same properties as the functions you have studied. The graph of f(x) = ex is like other graphs of exponential functions, such as f(x) = 3x. The domain of f(x) = ex is all real numbers. The range is {y|y > 0}.

8. Example 1: Graphing Exponential Functions Graph f(x) = ex–2 + 1. Make a table. Because e is irrational, the table values are rounded to the nearest tenth.

9. Check It Out! Example 1 Graph f(x) = ex – 3. Make a table. Because e is irrational, the table values are rounded to the nearest tenth.

10. A logarithm with a base of e is called a natural logarithmand is abbreviated as “ln” (rather than as loge). Natural logarithms have the same properties as log base 10 and logarithms with other bases. The natural logarithmic functionf(x) = ln x is the inverse of the natural exponential function f(x) = ex.

11. The domain of f(x) = ln x is {x|x > 0}. The range of f(x) = ln x is all real numbers.

12. Example 2: Simplifying Expression with e or ln Simplify. A. ln e0.15t B. e3ln(x +1) ln e0.15t= 0.15t eln(x +1)3= (x + 1)3 C. ln e2x + ln ex ln e3x = 3x

13. You Try! Example 2 Simplify. a. ln e3.2 b. e2lnx ln e3.2= 3.2 elnx2= x2 c.ln ex +4y ln ex + 4y= x + 4y

14. Recall the compound interest formulaA = P(1 + )nt, A =amount P = principal r =annual interest as a decimal n =# of times the interest is compounded per year t =time in years. r n 1 n Suppose that $1 is invested at 100% interest (r = 1) compounded n times for one year as represented by the function f(n) = P(1 + )n.

15. 1 n As n gets very large, interest is continuouslycompounded. Examine the graph of f(n)= (1 + )n. The function has a horizontal asymptote. As n becomes infinitely large, the value of the function approaches approximately 2.7182818…. This number is called e. Like , the constant e is an irrational number.

16. The formula for continuously compounded interest is A = Pert where A = total amount P =principal r =annual interest rate as a decimal t =time in years.

17. Example 3: Economics Application What is the total amount for an investment of $500 invested at 5.25% for 40 years and compounded continuously? A = Pert A = 500e0.0525(40) Substitute 500 for P, 0.0525 for r, and 40 for t. A ≈ 4083.08 Use the ex key on a calculator. The total amount is $4083.08.

18. You Try! Example 3 What is the total amount for an investment of $100 invested at 3.5% for 8 years and compounded continuously? A = Pert A = 100e0.035(8) Substitute 100 for P, 0.035 for r, and 8 for t. A ≈ 132.31 Use the ex key on a calculator. The total amount is $132.31.

19. Assignment: 7.6 Worksheet evens

20. 10. 0.61 12. 1.58 14. -0.11 16. 2.12 18. 1.84 20. 2 22. 1.83 24. -0.93 26. 151.48 28. 2.71 30. 1.91 32. 4.69 34. 0 36. 2.60 38. -0.02 40. 1.61 42. 21 44. 40.2 days 46. 31.8 days 48. No, max velocity of 6.11 km/s is less than 7.7 km/s 50. About 28.91 52.About 1553 years 54. 25 56. 100

21. Warm Up How much money would you have at the end of 6 years if you invest $4500 in an account that earns 5% interest compounded continuously? $6074.37

22. 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.

23. 1 2 1 Substitute 1 for N0 ,24,110 for t, and for N(t) because half of the initial quantity will remain. 2 = 1e–k(24,110) Example 4: Science Application Plutonium-239 (Pu-239) has a half-life of 24,110 years. How long does it take for a 1 g sample of Pu-239 to decay to 0.1 g? Step 1 Find the decay constant for plutonium-239. N(t) = N0e–kt Use the natural decay function.

24. ln = lne–24,110k 1 2 1 2 Example 4 Continued Simplify and take ln of both sides. ln = –24,110k k ≈ 0.000029

25. t = – ≈ 80,000 ln 0.1 0.000029 Example 4 Continued Step 2 Write the decay function and solve for t. Substitute 0.000029 for k. N(t) = N0e–0.000029t Substitute 1 for N0 and 0.01 for N(t). 0.1 = 1e–0.000029t Take ln of both sides. ln 0.1 = lne–0.000029t ln 0.1 = –0.000029t Simplify. It takes approximately 80,000 years to decay.

26. 1 2 Substitute 1 for N0 ,28 for t, and for N(t) because half of the initial quantity will remain. = 1e–k(28) 1 2 You Try! Example 4 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. Step 1 Find the decay constant for Chromium-51. N(t) = N0e–kt Use the natural decay function. t.

27. ln = lne–28k 1 2 1 k≈ 0.0247 2 You Try! Example 4 Continued Simplify and take ln of both sides. ln= –28k

28. ln = ln e–0.0247t ln = –0.0247t 200 200 650 650 650 200 ln t = ≈ 47.7 –0.0247 Check It Out! Example 4 Continued Step 2 Write the decay function and solve for t. Substitute 0.0247 for k. N(t) = N0e–0.0247t Substitute 650 for N0 and 200 for N(t). 200 = 650e–0.0247t Take ln of both sides. Simplify. It takes approximately 47.7 days to decay.

29. Assignment: Worksheet Natural Base e