# Lesson 19: Exponential and Logarithmic Functions

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### Lesson 19: Exponential and Logarithmic Functions

• 1. 𝐏𝐓𝐒 𝟑 Bridge to Calculus Workshop Summer 2020 Lesson 19 Exponential and Logarithmic Functions "If I have been able to see further, it was only because I stood on the shoulders of giants." – Newton -
• 2. Lehman College, Department of Mathematics Review of Rational Exponents (1 of 2) Example 0.1. Write the following as an exponential expression with rational powers: Solution. Example 0.2. Simplify 81− 3 4. Solution. 3 142 3 142 = 142 1 3 = 14 2 3 81− 3 4 = 1 81 3 4 = 1 81 1 4 3 = 1 4 81 3 = 1 33 = 1 27
• 3. Lehman College, Department of Mathematics Review of Rational Exponents (2 of 2) Example 0.3. Simplify the expression: Write your answer without using negative exponents. Assume that all variables are positive real numbers. Solution. 𝑎−4 ⋅ 𝑏 1 3 4 5 𝑎−4 ⋅ 𝑏 1 3 4 5 = 𝑎−4 4 5 ⋅ 𝑏 1 3 4 5 = 𝑎− 16 5 ⋅ 𝑏 4 15 = 𝑏 4 15 𝑎 16 5
• 4. Lehman College, Department of Mathematics Exponential Function Definition (1 of 1) The exponential function 𝑓 with base 𝑏 is defined by: where 𝑏 is a positive constant other than 1 (𝑏 > 0 and 𝑏 ≠ 1) and 𝑥 is a real number. Here are some examples of exponential functions: Each exponential function has a constant base and a variable exponent. The functions listed below are not exponential functions. For each case, state why not: 𝒇(𝒙) = 𝒃 𝒙 𝑓 𝑥 = 2 𝑥 (a) 𝑔 𝑥 = 10 𝑥 (b) ℎ 𝑥 = 3 𝑥+1 (c) 𝐹 𝑥 = 𝑥2 (d) 𝐺 𝑥 = 1 𝑥(e) 𝐻 𝑥 = −1 𝑥 (f) 𝐽 𝑥 = 𝑥 𝑥(g)
• 5. Lehman College, Department of Mathematics Graphing Exponential Functions (1 of 5) Example 1. Graph the exponential function 𝑓 𝑥 = 2 𝑥 : Solution. Set up a table of coordinates then draw graph 𝒙 𝒇 𝒙 = 𝟐 𝒙 1 𝑓 1 = 21 = 2 2 𝑓 2 = 22 = 4 3 𝑓 3 = 23 = 8 0 𝑓 0 = 20 = 1 −1 𝑓 −1 = 2−1 = 1/2 −2 𝑓 −2 = 2−2 = 1/4 −3 𝑓 −3 = 2−3 = 1/8
• 6. Lehman College, Department of Mathematics Graphing Exponential Functions (2 of 5) Example 2. Graph the function 𝑔 𝑥 = 1 2 𝑥 Solution. Set up a table of coordinates then draw graph 𝒙 𝒈 𝒙 = 𝟏/𝟐 𝒙 1 𝑔 1 = 1/2 1 = 1/2 2 𝑔 2 = 1/2 2 = 1/4 3 𝑔 3 = 1/2 3 = 1/8 0 𝑔 0 = 1/2 0 = 1 −1 𝑔 −1 = 1/2 −1 = 2 −2 𝑔 −2 = 1/2 −2 = 4 −3 𝑔 −3 = 1/2 −3 = 8
• 7. Lehman College, Department of Mathematics Graphing Exponential Functions (3 of 5) Similarly, below we show the graphs of 𝑦 = 3 𝑥 as well as 𝑦 = 1/3 𝑥 : Note that: if 𝑓 𝑥 = 3 𝑥 = 3−𝑥 𝑔 𝑥 = 1 3 𝑥 = 1 𝑥 3 𝑥 = 1 3 𝑥 = 𝑓 −𝑥
• 8. Lehman College, Department of Mathematics Graphing Exponential Functions (4 of 5) The graph of 𝑦 = 𝑎 𝑥 , when 𝑎 > 1: Domain: (−∞, ∞) Range: (0, ∞) 𝑦-intercept: (0, 1) Increasing function Continuous function 𝑎 𝑥 → 0 as 𝑥 → −∞
• 9. Lehman College, Department of Mathematics Graphing Exponential Functions (5 of 5) The graph of 𝑦 = 𝑎−𝑥 , when 𝑎 > 1: Domain: (−∞, ∞) Range: (0, ∞) 𝑦-intercept: (0, 1) Decreasing function Continuous function 𝑎−𝑥 → 0 as 𝑥 → ∞
• 10. Lehman College, Department of Mathematics Exponential Equations (1 of 4) Example 3. Solve the following exponential equation: Solution. We will use the one-to-one property: Example 4. Solve the following exponential equation: Solution. We will use the one-to-one property: 2 𝑥 = 8 2 𝑥 = 8 Write the original equation 2 𝑥 = 23 Convert 8 to a power of 2 𝑥 = 3 Equate the powers of 2 53𝑥−6 = 125 53𝑥−6 = 125 Write the original equation 53𝑥−6 = 53 Convert 125 to a power of 5 3𝑥 − 6 = Equate the powers of 53 3𝑥 = 9 It follows that: 𝑥 = 3
• 11. Lehman College, Department of Mathematics Exponential Equations (2 of 4) Example 5. Solve the following exponential equation: Solution. What common base should we use? Example 6. Solve the following exponential equation: Solution. What common base should we use? 8 𝑥+2 = 4 𝑥−3 8 𝑥+2 = 4 𝑥−3 Write the original equation 23 𝑥+2 = Convert 8 and 4 to powers of 2 3𝑥 + 6 = 2𝑥 − 6 Equate the powers of 2 22 𝑥−3 23𝑥+6 = 22𝑥−6 Simplify the exponents 𝑥 = −12 Solve for 𝑥 125 𝑥 = 25 𝑥+2 125 𝑥 = 25 𝑥+2 Write the original equation
• 12. Lehman College, Department of Mathematics Exponential Equations (3 of 4) Solution. What common base should we use? 53 𝑥 = Convert 125 and 25 to powers of 5 3𝑥 = 2𝑥 + 4 Equate the powers of 5 52 𝑥+2 53𝑥 = 52𝑥+4 Simplify the exponents 𝑥 = 4 Solve for 𝑥 125 𝑥 = 25 𝑥+2 Write the original equation
• 13. Lehman College, Department of Mathematics Exponential Equations (4 of 4) Example 7. Solve the following exponential equation: Solution. What common base should we use? It follows that 𝑥 = −3. 42 4𝑥−6 = Convert 16 to a power of 4 8𝑥 − 12 = 𝑥2 + 14𝑥 − 3 Equate the powers of 4 4 𝑥2+14𝑥−3 48𝑥−12 = 4 𝑥2+14𝑥−3 Simplify the exponents 𝑥2 + 6𝑥 + 9 = 0 Factor the quadratic 164𝑥−6 = 4 𝑥2+14𝑥−3 Write the original equation 164𝑥−6 = 4 𝑥2+14𝑥−3 𝑥 + 3 2 = 0 Simplify
• 14. Lehman College, Department of Mathematics Kovalevskaya and Lobachevsky (1 of 1) Nikolai Lobachevsky (1792-1856) - Russian Mathematician Sofya Kovalevskaya (1850-1891) - Russian Mathematician
• 15. Lehman College, Department of Mathematics Logarithmic Function Definition (1 of 2) Let 𝑏 is a positive constant other than 1. That is, 𝑏 > 0 and 𝑏 ≠ 1. Let 𝑥 and 𝑦 be real numbers, then: The logarithmic function 𝑓 with base 𝑏 is defined by: Example 8. Write each equation in exponential form: Solution. We use the equivalent definition above: Example 9. Write each equation in logarithmic form: 𝒚 = log 𝒃 𝒙 is equivalent to 𝒃 𝒚 = 𝒙 𝒇(𝒙) = log 𝒃 𝒙 3 = log7 𝑥(a) 2 = log 𝑏 25(b) log4 26 = 𝑦(c) 73 = 𝑥(a) 𝑏2 = 25(b) 4 𝑦 = 26(c) 25 = 𝑥(d) 𝑏3 = 27(e) 𝑒 𝑦 = 33(f)
• 16. Lehman College, Department of Mathematics Logarithmic Function Definition (2 of 2) Example 10. Write each equation in logarithmic form: Solution. We use the equivalent definition: 25 = 𝑥(d) 𝑏3 = 27(e) 𝑒 𝑦 = 33(f) 25 = 𝑥(d) log2 𝑥log2 25 = 5 = log2 𝑥 𝑏3 = 27(e) log 𝑏 27log 𝑏 𝑏3 = 3 = log 𝑏 27 𝑒 𝑦 = 33(f) log 𝑒 33log 𝑒 𝑒 𝑦 = 𝑦 = log 𝑒 33
• 17. Lehman College, Department of Mathematics Graphing Logarithmic Functions (1 of 1) The graph of 𝑦 = log 𝑎 𝑥, when 𝑎 > 1: Domain: (0, ∞) Range: (−∞, ∞) 𝑥-intercept: (1, 0) Increasing function Continuous function log 𝑎 𝑥 → −∞ as 𝑥 → 0+
• 18. Lehman College, Department of Mathematics Natural & Common Logarithms (1 of 2) There are two bases with special notation: Above 𝑒 ≈ 2.718 is the base of natural logarithms. Example 11. Simplify the following expressions: Solution. Rewrite in terms of powers of the bases: log 𝒙 = log 𝟏𝟎 𝒙 ln 𝒙 = log 𝒆 𝒙 log4 16 log 1000 log4 16 log 1000 = log4 42 log 103 = 2 3 (a) log3 27 ln 𝑒5(b) (a) log3 27 ln 𝑒5 =(b) log3 33 ln 𝑒5 = 3 5
• 19. Lehman College, Department of Mathematics Properties of Logarithms (1 of 13) Let 𝑏 be a base. That is, 𝑏 > 0 and 𝑏 ≠ 1. Let 𝑥 and 𝑦 be positive numbers, then: Proof. Suppose: Then, the equivalences are: It follows that: log 𝒃 𝒙𝒚 = log 𝒃 𝒙 + log 𝒃 𝒚 log 𝑏 𝑥 = 𝑝 log 𝑏 𝑦 = 𝑞and 𝑥 = 𝑏 𝑝 and 𝑦 = 𝑏 𝑞 𝑥𝑦 = 𝑏 𝑝 ⋅ 𝑏 𝑞 = 𝑏 𝑝+𝑞 log 𝑏 𝑥𝑦 = log 𝑏 𝑏 𝑝+𝑞 = 𝑝 + 𝑞 = log 𝑏 𝑥 + log 𝑏 𝑦
• 20. Lehman College, Department of Mathematics Properties of Logarithms (2 of 13) Example 12. Simplify the following expressions: Solution. Rewrite in terms of powers of the bases: log4 8 log 0.0001 log4 8 log 0.0001 = log4 4 ⋅ 2 log 10−4 = log4 4 + log4 2 −4 = 1 + log4 41/2 −4 = 1 + 1 2 −4 = − 3 8
• 21. Lehman College, Department of Mathematics Properties of Logarithms (3 of 13) Similarly, we can show that if 𝑏 is a base, and 𝑥 and 𝑦 and positive numbers, then: Finally, if 𝑏 is a base, and 𝑥 is a positive number, then: For any integer 𝑛, Proof. Let log 𝒃 𝒙 𝒚 = log 𝒃 𝒙 − log 𝒃 𝒚 log 𝒃 𝒙 𝒏 = 𝒏 ⋅ log 𝒃 𝒙 log 𝑏 𝑥 = 𝑝 then 𝑥 = 𝑏 𝑝 𝑥 𝑛 = 𝑏 𝑝 𝑛 = 𝑏 𝑛𝑝 log 𝑏 𝑥 𝑛 = log 𝑏 𝑏 𝑛𝑝 = 𝑛 ⋅ 𝑝 = 𝑛 ⋅ log 𝑏 𝑥
• 22. Lehman College, Department of Mathematics Properties of Logarithms (4 of 13) Example 13. Solve the following equations: Solution. We use the equivalent definition: 7 𝑥+1 = 1 49 (a) 2𝑥 + 1 = log 0.001(b) log2 8𝑥 = 5(c) (a) 7 𝑥+1 = 1 49 = 1 72 = 7−2 Therefore: 𝑥 + 1 = −2 2𝑥 + 1 = log 0.001(b) = log 10−3 = −3 Therefore: 2𝑥 = −4 and: 𝑥 = −2 and: 𝑥 = −3
• 23. Lehman College, Department of Mathematics Properties of Logarithms (5 of 13) Solution (cont’d). We use the equivalent definition: Solution 2. log2 8𝑥 = 5(c) Write the original equation Exponentiate Property of logarithms 2log2 8𝑥 = 25 8𝑥 = 32 𝑥 = 4 Divide both sides by 8 Write the original equation log2 8 + log2 𝑥 = 5 Property of logarithms log2 23 + log2 𝑥 = 5 3 + log2 𝑥 = 5 log2 8𝑥 = 5(c)
• 24. Lehman College, Department of Mathematics Properties of Logarithms (6 of 13) Solution 2 (cont’d). Example 14. Solve the equation 10 𝑥2−3𝑥 = 10,000. Solution. Write 10,000 as a power of the base 10. It follows that 𝑥 = 4 or 𝑥 = −1. 3 + log2 𝑥 = 5 log2 𝑥 = 2 Subtract 3 from both sides Exponentiate2log2 𝑥 = 22 𝑥 = 4 10 𝑥2−3𝑥 = 104 𝑥2 − 3𝑥 = 4 𝑥2 − 3𝑥 − 4 = 0 𝑥 − 4 𝑥 + 1 = 0
• 25. Lehman College, Department of Mathematics Properties of Logarithms (7 of 13) Example 15. Solve the equation: Solution. The roots of the quadratic are: 𝑥 = 4 and 𝑥 = −3. From the original equation, the only solution is 𝑥 = 4. log2 𝑥 − 3 + log2 𝑥 + 2 = log2 6 log2 𝑥 − 3 + log2 𝑥 + 2 = log2 6 log2 𝑥 − 3 𝑥 + 2 = log2 6 2log2 𝑥−3 𝑥+2 = 2log2 6 𝑥 − 3 𝑥 + 2 = 6 𝑥2 − 𝑥 − 6 = 6 𝑥2 − 𝑥 − 12 = 0 𝑥 − 4 𝑥 + 3 = 0
• 26. Lehman College, Department of Mathematics Properties of Logarithms (8 of 13) Example 16. Solve the equation: Solution. log7 𝑥 − 1 − log7 𝑥 + 1 = log7 𝑥 + 5 log7 𝑥 − 1 − log7 𝑥 + 1 = log7 𝑥 + 5 log7 𝑥 − 1 𝑥 + 1 = log7 𝑥 + 5 7log7 𝑥−1 𝑥+1 = 7log7 𝑥+5 𝑥 − 1 𝑥 + 1 = 𝑥 + 5 𝑥 + 1 𝑥 + 5 = 𝑥 − 1 𝑥2 + 6𝑥 + 5 = 𝑥 − 1
• 27. Lehman College, Department of Mathematics Properties of Logarithms (9 of 13) Solution (cont’d). From the previous slide: The roots of the quadratic are: 𝑥 = −2 and 𝑥 = −3. The original equation was: From the original equation, there is no solution. 𝑥2 + 6𝑥 + 5 = 𝑥 − 1 𝑥2 + 5𝑥 + 6 = 0 𝑥 + 2 (𝑥 + 3) = 0 log7 𝑥 − 1 − log7 𝑥 + 1 = log7 𝑥 + 5
• 28. Lehman College, Department of Mathematics Properties of Logarithms (10 of 13) Let 𝑎 and 𝑏 be bases. That is, 𝑎, 𝑏 > 0 and 𝑎, 𝑏 ≠ 1. Let 𝑥 be a positive number, then: Proof. Let: This implies that: It follows that: log 𝒂 𝒙 = log 𝒃 𝒙 log 𝒃 𝒂 log 𝑏 𝑥 = 𝑝 then 𝑥 = 𝑏 𝑝 and let: log 𝑎 𝑥 = 𝑞 then 𝑥 = 𝑎 𝑞 𝑥 = 𝑏 𝑝 = 𝑎 𝑞 log 𝑏 𝑥 = 𝑝 = 𝑞 ⋅ log 𝑏 𝑎 = log 𝑎 𝑥 ⋅ log 𝑏 𝑎 log 𝑏 𝑎 𝑞 = log 𝑎 𝑥 = log 𝑏 𝑥 log 𝑏 𝑎
• 29. Lehman College, Department of Mathematics Properties of Logarithms (11 of 13) Example 17. Use the change of base formula to find: Solution. The change of base formula is given by: Since most calculators only compute logarithms base 10 or base 𝑒 (denoted log or ln, respectively): log7 4 log 𝑎 𝑥 = log 𝑏 𝑥 log 𝑏 𝑎 log7 4 = ln 4 ln 7 ≈ 0.712
• 30. Lehman College, Department of Mathematics Properties of Logarithms (12 of 13) Example 18. Use the change of base formula to find 𝑥: Solution. log8 15 𝑥−2 = 5 log8 15 𝑥−2 = 5 Write the original equation (𝑥 − 2) log8 15 = 5 Property of logarithms 𝑥 − 2 = 5 log8 15 Divide by log8 15 𝑥 = 2 + 5 log8 15 𝑥 = 2 + 5 ln 15 ln 8 Add 2 to both sides Change of base formula
• 31. Lehman College, Department of Mathematics Properties of Logarithms (13 of 13) Example 19. Show that if 𝑎 and 𝑏 are bases. That is, 𝑎, 𝑏 > 0 and 𝑎, 𝑏 ≠ 1. Then: Proof. In the formula: Let 𝑥 = 𝑏, then: It follows that: log 𝒂 𝒃 = 𝟏 log 𝒃 𝒂 log 𝑎 𝑥 = log 𝑏 𝑥 log 𝑏 𝑎 log 𝑎 𝑏 = 1 log 𝑏 𝑎 log 𝑎 𝑏 = log 𝑏 𝑏 log 𝑏 𝑎
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