1. Introduction Logarithms can be used to solve exponential equations that have a variable as an exponent. In compound interest problems that use the formula, logarithms can be used to solve for t, which indicates the amount of time an account or investment takes to mature. 1 4.3.2: Natural Logarithms

2. Introduction, continued A financial advisor at a business can use this information to forecast and plan for the company’s future. In natural exponential equations (those with the variable e), a natural logarithm can be used to solve for a variable that exists as an exponent, since ln x = log e x. So, natural logarithms can be used to model situations, graphed as functions, and written as exponents. Additionally, they can be evaluated by hand using the properties of logarithms (e.g., ln 1 = 0 because e 0 = 1), and with a calculator when properties of logarithms cannot be applied. For example, ln 8 ≈ 2.07 because e 2.07 ≈ 8. 2 4.3.2: Natural Logarithms

3. Key Concepts Recall that natural logarithms consist of a number with a base of e. As with common logarithms, natural logarithms have an inverse relationship with exponential functions. That is, the natural logarithm function ln b = a can also be rewritten as the exponential function, e a = b, where e is an irrational number with an approximate value of 2.71828. 3 4.3.2: Natural Logarithms

4. Key Concepts, continued Natural logarithms differ from common logarithms because of their base. Common logarithms have a base of 10 whereas natural logarithms have a base of e. For convenience, natural logarithms can be converted to base 10 by using the change of base formula,. The same properties that apply to common logarithms also apply to natural logarithms. 4 4.3.2: Natural Logarithms

5. Key Concepts, continued One application for natural logarithms involves calculating interest on money that is compounded continuously, or at every instant. The continuously compounded interest formula is A = Pe rt, where A is the ending amount, P is the principal or initial amount, e is a constant, r is the annual interest rate expressed as a decimal, and t is the time in years. Note this formula’s similarity to the compound interest formula,. 5 4.3.2: Natural Logarithms

6. Key Concepts, continued The continuously compounded interest formula, A = Pe rt, is a natural exponential function because it is in the form f(x) = e x and involves a base of e. Other uses of natural logarithms include situations of exponential growth or decay, which involve rapid increase (growth) or decrease (decay). One formula for modeling the exponential growth or decay of a substance or population is P = P 0 e kt. In this formula, P represents the amount of a substance or population after the time, t, has elapsed, P 0 is the initial amount, e is a constant, and k is the rate of growth or decay. 6 4.3.2: Natural Logarithms

7. Key Concepts, continued Notice this formula is similar to the continuously compounded interest formula, but instead of modeling an amount of money over time, this formula models the amount of a substance or population over time. When k > 0, the formula models growth; when k < 0, the formula models decay. Since this formula uses the base e, it represents a natural exponential function. A common application of exponential decay is finding the half-life of a substance. Half-life is the time it takes for a substance that is decaying exponentially to decrease to 50% of its original amount. 7 4.3.2: Natural Logarithms

8. Common Errors/Misconceptions confusing the formula for continuously compounded interest with that for compound interest and vice versa making substitution errors when converting exponential functions to logarithmic functions and vice versa misinterpreting the parts of an exponential function 8 4.3.2: Natural Logarithms

9. Guided Practice Example 1 Cheyenne deposited $200 in a bank account earning continuously compounded interest. After 10 years, she closed the account and withdrew the entire balance, which totaled $364.42. What was her annual interest rate? Rounded to the nearest dollar, how much would Cheyenne have received if she had left the money in the account for 15 years? 20 years? Use the continuously compounded interest formula, A = Pe rt, where A is the ending amount, P is the principal or initial amount, e is a constant, r is the annual interest rate expressed as a decimal, and t is the time in years. 9 4.3.2: Natural Logarithms

10. Guided Practice: Example 1, continued 1.Determine values for the continuously compounded interest formula. The continuously compounded interest formula is A = Pe rt. The ending amount, A, is the account balance when Cheyenne withdrew the money, $364.42. The principal, P, is $200. The time, t, is 10 years. Therefore, let A = 364.42, P = 200, and t = 10. 10 4.3.2: Natural Logarithms

11. Guided Practice: Example 1, continued 2.Substitute the known values into the formula and solve for the annual interest rate, r. A = Pe rt Continuously compounded interest formula (364.42) = (200)e r(10) Substitute 364.42 for A, 200 for P, and 10 for t. 1.82 = e 10r Divide both sides by 200 and simplify the exponent. ln 1.82 = ln e 10r Rewrite each side as a natural logarithm. 11 4.3.2: Natural Logarithms

12. Guided Practice: Example 1, continued ln 1.82 = 10r ln e Apply the Power Property for natural logarithms. Divide to isolate r and then apply the Symmetric Property of Equality. r ≈ 0.06Evaluate each natural logarithm using a calculator. The value of r is about 0.06; therefore, the annual interest rate was 6%. 12 4.3.2: Natural Logarithms

13. Guided Practice: Example 1, continued 3.Determine how much money Cheyenne would have received if she had left the money in the account for 15 years. Use the continuously compounded interest formula, A = Pe rt. The values of P and r remain the same. The value of t is 15. Substitute these values into the formula and solve for the ending amount, A. 13 4.3.2: Natural Logarithms

14. Guided Practice: Example 1, continued A = Pe rt Continuously compounded interest formula A = (200)e (0.06)(15) Substitute 200 for P, 0.06 for r, and 15 for t. A = 200e 0.9 Simplify the exponent. A ≈ 491.92Evaluate using a calculator. If Cheyenne had left her money in the account for 15 years, she would have received about $491.92. 14 4.3.2: Natural Logarithms

15. Guided Practice: Example 1, continued 4.Determine how much money Cheyenne would have received if she had left the money in the account for 20 years. Once again, use the continuously compounded interest formula, A = Pe rt. The values of P and r remain the same. The value of t is 20. Substitute these values into the formula and solve for the ending amount, A. 15 4.3.2: Natural Logarithms

16. Guided Practice: Example 1, continued A = Pe rt Continuously compounded interest formula A = (200)e (0.06)(20) Substitute 200 for P, 0.06 for r, and 20 for t. A = 200e 1.2 Simplify the exponent. A ≈ 664.02Evaluate using a calculator. If Cheyenne had left her money in the account for 20 years, she would have received about $664.02. 16 4.3.2: Natural Logarithms ✔

17. Guided Practice: Example 1, continued 17 4.3.2: Natural Logarithms

18. Guided Practice Example 3 Studies have shown that the number of bacteria on a public restroom sink can grow exponentially from 5,000 bacteria to 12,000 bacteria in 10 hours. Write the natural logarithmic equation that would represent how the number of bacteria would grow over the given time period. Use the exponential growth/decay formula, P = P 0 e kt, where P represents the number of bacteria after t hours, P 0 is the initial number of bacteria, and k is the rate of growth or decay. Given the rate of bacterial growth, how many bacteria would there be after 24 hours? How many bacteria would you expect to be present after 48 hours? Verify your answers algebraically. 18 4.3.2: Natural Logarithms

19. Guided Practice: Example 3, continued 1.Determine the growth rate, k, using properties of logarithms. Begin by identifying the known values. Let P, the final population of the bacteria, be 12,000. Let P 0, the initial population of the bacteria, be 5,000. Let t, the time in hours, be 10. Substitute these values into the given formula and solve for k. 19 4.3.2: Natural Logarithms

20. Guided Practice: Example 3, continued P = P 0 e kt Exponential growth/decay formula (12,000) = (5000)e k(10) Substitute 12,000 for P, 5,000 for P 0, and 10 for t. 2.4 = e 10k Divide both sides by 5,000 and simplify the exponent. ln 2.4 = ln e 10k Rewrite each side as a natural logarithm. ln 2.4 = 10k ln eApply the Power Property of natural logarithms. 20 4.3.2: Natural Logarithms

21. Guided Practice: Example 3, continued Divide to isolate k and then apply the Symmetric Property of Equality. k ≈ 0.088Evaluate using a calculator. The growth rate, k, is approximately 0.088. 21 4.3.2: Natural Logarithms

22. Guided Practice: Example 3, continued 2.Determine the number of bacteria after 24 hours. Use the exponential growth/decay formula, P = P 0 e kt. The value for P 0 remains the same (5,000). Let the growth rate, k, be 0.088. Let the time, t, be 24. Substitute these values into the given formula and solve for P, the final population. 22 4.3.2: Natural Logarithms

23. Guided Practice: Example 3, continued P = P 0 e kt Exponential growth/decay formula P = (5000)e (0.088)(24) Substitute 5,000 for P 0, 0.088 for k, and 24 for t. P = 5000e 2.112 Simplify the exponent. P ≈ 41,324Evaluate using a calculator. After 24 hours, there will be approximately 41,324 bacteria on the sink. 23 4.3.2: Natural Logarithms

24. Guided Practice: Example 3, continued 3.Determine the number of bacteria after 48 hours. Use the exponential growth/decay formula, P = P 0 e kt. The values for P 0 and k remain the same (5,000 and 0.088, respectively). Let the time, t, be 48. Substitute these values into the given formula and solve for P, the final population. 24 4.3.2: Natural Logarithms

25. Guided Practice: Example 3, continued P = P 0 e kt Exponential growth/decay formula P = (5000)e (0.088)(48) Substitute 5,000 for P 0, 0.088 for k, and 48 for t. P = 5000e 4.224 Simplify the exponent. P ≈ 341,531Evaluate using a calculator. After 48 hours, there will be approximately 341,531 bacteria on the sink. 25 4.3.2: Natural Logarithms ✔

26. Guided Practice: Example 3, continued 26 4.3.2: Natural Logarithms