1. Copyright © Cengage Learning. All rights reserved. 9 Nonlinear Functions and Models

2. Copyright © Cengage Learning. All rights reserved. 9.3 Logarithmic Functions and Models

3. 3 Today, computers and calculators have done away with that use of logarithms, but many other uses remain. In particular, the logarithm is used to model real-world phenomena in numerous fields, including physics, finance, and economics. From the equation 2 3 = 8 we can see that the power to which we need to raise 2 in order to get 8 is 3.

4. 4 Logarithmic Functions and Models We abbreviate the phrase “the power to which we need to raise 2 in order to get 8” as “log 2 8.” Thus, another way of writing the equation 2 3 = 8 is log 2 8 = 3. This is read “the base 2 logarithm of 8 is 3” or “the log, base 2, of 8 is 3.” The power to which we need to raise 2 in order to get 8 is 3.

5. 5 Logarithmic Functions and Models Here is the general definition. Base b Logarithm The base b logarithm of x, log b x, is the power to which we need to raise b in order to get x. Symbolically, log b x = y means b y = x. Logarithmic form Exponential form

6. 6 Logarithmic Functions and Models Quick Examples The following table lists some exponential equations and their equivalent logarithmic forms:

7. 7 Logarithmic Functions and Models Common Logarithm, Natural Logarithm The following are standard abbreviations. Base 10: log 10 x = log x Base e: log e x = ln x Quick Examples Logarithmic Form Exponential Form 1. log 10,000 = 4 10 4 = 10,000 2. ln e = 1 e 1 = e Common Logarithmlog (x) Natural Logarithmln (x)

8. 8 Logarithmic Functions and Models Change-of-Base Formula Quick Examples Change-of-base formula log(9) / log(11) ln(9) / ln(11)

9. 9 Example 1 – Solving Equations with Unknowns in the Exponent Solve the following equations a. 5 –x = 125 b. 3 2x – 1 = 6 c. 100(1.005) 3x = 200 Solution: a. Write the given equation 5 –x = 125 in logarithmic form: –x = log 5 125 This gives x = –log 5 125 = –3.

10. 10 Example 1 – Solution b. In logarithmic form, 3 2x – 1 = 6 becomes 2x – 1 = log 3 6 2x = 1 + log 3 6 giving cont’d

11. 11 Example 1 – Solution c. We cannot write the given equation, 100(1.005) 3x = 200, directly in exponential form. We must first divide both sides by 100: cont’d

12. 12 Logarithmic Functions and Models Logarithmic Function A logarithmic function has the form f (x) = log b x + C or, alternatively, f (x) = A ln x + C. Quick Examples 1. f (x) = log x 2. g(x) = ln x – 5 (b and C are constants with b > 0, b  1) (A, C constants with A  0)

13. 13 Logarithmic Functions and Models Logarithm Identities The following identities hold for all positive bases a  1 and b  1, all positive numbers x and y, and every real number r. These identities follow from the laws of exponents. IdentityQuick Examples

14. 14 Logarithmic Functions and Models Relationship with Exponential Functions The following two identities demonstrate that the operations of taking the base b logarithm and raising b to a power are inverse to each other. IdentityQuick Examples 1. log b (b x ) = x log 2 (2 7 ) = 7 In words: The power to which you raise b in order to get bx is x (!)

15. 15 Logarithmic Functions and Models IdentityQuick Examples 2. b log b x = x5 log 5 8 = 8 In words: Raising b to the power to which it must be raised to get x, yields x. (!)

16. 16 Applications

17. 17 Example 3 – Investments: How Long? Global bonds sold by Mexico are yielding an average of 2.51% per year. At that interest rate, how long will it take a $1,000 investment to be worth $1,200 if the interest is compounded monthly? Solution: Substituting A = 1,200, P = 1,000, r = 0.0251, and m = 12 in the compound interest equation gives

18. 18 Example 3 – Solution ≈ 1,000(1.004333) 12t and we must solve for t. We first divide both sides by 1,000, getting an equation in exponential form: 1.2 = 1.002092 12t. In logarithmic form, this becomes 12t = log 1.002092 (1.2). cont’d

19. 19 Example 3 – Solution We can now solve for t: ≈ 7.3 years Thus, it will take approximately 7.3 years for a $1,000 investment to be worth $1,200. cont’d log(1.2)/(log(1.002092)*12)

20. 20 Applications Exponential Decay Model and Half-Life An exponential decay function has the form Q(t) = Q 0 e −kt. Q 0 represents the value of Q at time t = 0, and k is the decay constant. The decay constant k and half-life t h for Q are related by t h k = ln 2. Q 0, k both positive

21. 21 Applications Quick Example If t h = 10 years, then 10k = ln 2, so k = ≈ 0.06931 and the decay model is Q(t) = Q 0 e −0.06931t. Exponential Growth Model and Doubling Time An exponential growth function has the form Q(t) = Q 0 e kt. Q 0 represents the value of Q at time t = 0, and k is the growth constant. Q 0, k both positive

22. 22 Applications The growth constant k and doubling time t d for Q are related by t d k = ln 2. Quick Example P(t) = 1,000e 0.05t $1,000 invested at 5% annually with interest compounded continuously

23. 23 Logarithmic Regression

24. 24 Logarithmic Regression If we start with a set of data that suggests a logarithmic curve we can, by repeating the methods, use technology to find the logarithmic regression curve y = log b x + C approximating the data.

25. 25 Example 5 – Research & Development The following table shows the total spent on research and development by universities and colleges in the U.S., in billions of dollars, for the period 1998–2008 (t is the number of years since 1990). Find the best-fit logarithmic model of the form S(t) = A ln t + C and use the model to project total spending on research by universities and colleges in 2012, assuming the trend continues.

26. 26 Example 5 – Solution We use technology to get the following regression model: S(t) = 19.3 ln t – 12.8. Because 2012 is represented by t = 22, we have S(22) = 19.3 ln(22) – 12.8 ≈ 47. So, research and development spending by universities and colleges is projected to be around $47 billion in 2012. Coefficients rounded Why did we round the result to two significant digits?