6.6 The Natural Base, e

# 6.6 The Natural Base, e

## 6.6 The Natural Base, e

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1. 6.6 The Natural Base, e Objectives: Evaluate natural exponential and natural logarithmic functions.

2. As n becomes very large, the value of approaches the number 2.71828…, named e The natural base, e, is used to estimate the ages of artifacts and to calculate interest that is compounded continuously.

3. The Natural Exponential Function • The exponential function with base e, f(x) = ex is called the natural exponential function and e is called the natural base. • The function ex is graphed. • Notice that the domain is all real numbers • The range is all positive numbers.

4. Ex 1. Evaluate f(x) = ex to the nearest thousandth for each value of x below. c. x = -1 e-1 = .368 • x= 2 • e2 = 7.389 b. x= ½ e1/2 = 1.649 d. x = 6 e6 = 403.429 e. x = 1/3 e1/3 = 1.396 f. x = -2 e-2 = .135

5. Continuous Compounding Formula

6. Ex 2: An investment of \$1000 earns an annual interest rate of 7.6%. Compare the final amounts after 8 years for interest compounded quarterly and for interest compounded continuously. Continuously A = Pert A = 1000e .076 * 8 A = 1836.75 Quarterly A = P(1+ r/n)nt A = 1000(1+ .076/4)4*8 A = 1826.31

7. Ex 3: Find the value of \$500 after 4 years invested at an annual interest rate of 9% compounded continuously. P = 500 t = 4 r = .09 A = 500e.36 = \$716.66

8. The Natural Logarithmic Function The natural logarithmic function y = loge x, abbreviated y = In x, is the inverse of the natural exponential function, y = ex. The function y = In x is graphed along with y = ex. y=ex y=x y = Inx

9. Ex 4 Evaluate f(x) = ln x to the nearest thousandth for each value of x below. • x = 2 • ln 2 = .693 b. x = ½ In ½ = -.693 c. x = -1 In -1 = undefined d. x = 5 In 5 = 1.609 e. x= 0.85 In.85 = -.163 f. x = 1 In 1 = 0

10. The natural logarithmic function can be used to solve an equation of the form A = Pert for the exponent t in order to find the time it takes for an investment that is compounded continuously to reach a specific amount. **** In e = 1 ****

11. Ex 5 How long does it take for an investment to double at an annual interest rate of 8.5% compounded continuously? A = Pert 2 P = Pert 2 = e0.085t ln2 = ln e0.085t ln 2 = 0.085t t = ln 2/0.085 t = 8.15

12. Ex 5 How long does it take for an investment to triple at an annual interest rate of 7.2% compounded continuously?

13. N(t) = Noe-0.00012t 0.78 No = Noe-0.00012t 0.78 = e-0.00012t ln 0.78 = -0.00012t -0.00012t = ln 0.78 t = ln 0.78/-0.00012 t = 2070.5 • Ex 7 Radiocarbon Dating Suppose that archaeologists find scrolls and claim that they are 2000 years old. Tests indicate that the scrolls contain 78% of their original carbon-14.

14. Homework Pg. 397-398 #12-30 even #72-75