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# LESSON 3–4

Download Presentation ## LESSON 3–4

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1. LESSON 3–4 Exponential and Logarithmic Equations

2. A. Solve 4x + 2 = 16x – 3. 4x + 2= 16x – 3Original equation 4x + 2= (42)x – 342 = 16 4x + 2=42x – 6Power of a Power x + 2 = 2x – 6 One-to-One Property 2 = x – 6 Subtract x from each side. 8 = x Add 6 to each side. Answer:8 Key Concept 1

3. One-to-One Property. B. Solve . Original equation Power of a Power Solve Exponential Equations Using One-to-One Property Example 1

4. A. 1 B. C. 2 D. –2 Solve 25x + 2 = 54x. Example 1

5. Solve Logarithmic Equations Using One-to-One Property A. Solve 2 ln x = 18. Round to the nearest hundredth. Method 1 Use exponentiation. 2 ln x = 18 Original equation ln x = 9 Divide each side by 2. eln x= e9 Exponentiate each side. x = e9 Inverse Property x ≈ 8103.08 Use a calculator. Method 2 Write in exponential form. 2 ln x = 18 Original equation ln x = 9 Divide each side by 2. x = e9 Write in exponential form. x ≈ 8103.08 Use a calculator. Answer:8103.08 Example 2

6. = x = 10–3. Answer: Solve Logarithmic Equations Using One-to-One Property B.Solve 7 – 3 log 10x = 13. 7 – 3 log 10x = 13 Original equation –3 log 10x = 6 Subtract 7 from each side. log 10x = –2 Divide each side by –3. 10–2=10x Write in exponential form. 10–3= x Divide each side by 10. Example 2

7. Solve 2 log2x3 = 18. A. 81 B. 27 C. 9 D. 8 Example 2

8. log25 = Quotient Property 5 = One-to-One Property A. Solve log2 5 = log2 10 – log2 (x – 4). log25 = log210 – log2(x – 4) Original equation 5x – 20 = 10 Multiply each side by x – 4.5 5x = 30 Add 20 to each side. x = 6 Divide each side by 5. Answer:6 Key Concept 2

9. Solve Exponential Equations Using One-to-One Property B. Solve log5 (x2 + x) = log5 20. log5(x2 + x) = log520 Original equation x2 + x = 20 One-to-One Property x2 + x – 20 = 0 Subtract 20 from each side. (x – 4)(x + 5) = 0 Factor x2 + x – 20 into linear factors. x = –5 or 4 Solve for x. Check this solution. Answer:–5, 4 Example 3

10. Solve log315 = log3x + log3(x – 2). A. 5 B. –3 C. –3, 5 D. no solution Example 3

11. x = or about 1.77 Divide each side by log 3 and use a calculator. x = or about 0.54 Solve for x and use a calculator. Solve Exponential Equations A. Solve 3x= 7. Round to the nearest hundredth. 3x= 7 Original equation log 3x= log 7 Take the common logarithm of each side. x log 3 = log 7 Power Property Answer:1.77 B. Solve e2x+ 1 = 8. Round to the nearest hundredth. e2x + 1= 8 Original equation ln e2x + 1= ln 8 Take the natural logarithm of each side. 2x + 1 = ln 8 Inverse Property Answer:0.54 Example 4

12. Solve 4x = 9. Round to the nearest hundredth. A. 0.63 B. 1.58 C. 2.25 D. 0.44 Example 4

13. Solve in Logarithmic Terms Solve 36x– 3 = 24– 4x. Round to the nearest hundredth. 36x – 3= 24 – 4xOriginal equation ln 36x – 3= ln 24 – 4xTake the natural logarithm of each side. (6x – 3) ln 3 = (4 – 4x) ln 2 Power Property 6x ln 3 – 3 ln3 = 4 ln 2 – 4x ln 2 Distributive Property 6x ln 3 + 4x ln 2 = 4 ln 2 + 3 ln3 Isolate the variable on the left side of the equation. x(6 ln 3 + 4 ln 2) = 4 ln 2 + 3 ln3 Distributive Property (could just divide here… hard to type in) Example 5

14. x = Divide each side by ln 11,664. Solve in Logarithmic Terms x(ln 36 + ln 24) = ln 24 + ln 33Power Property x ln [36(24)] = ln [24(33)] Product Property x ln 11,664 = ln 432 36(24) = 11,664 and 24(33) = 432 x ≈ 0.65 Use a calculator. Answer:0.65 Example 5

15. Solve 4x+ 2 = 32– x. Round to the nearest hundredth. A. 1.29 B. 1.08 C. 0.68 D. –0.23 Example 5

16. Solve Exponential Equations in Quadratic Form Solve e2x – ex – 2 = 0. e2x – ex – 2 = 0 Original equation u2 – u – 2 = 0 Write in quadratic form by letting u = ex. (u – 2)(u + 1) = 0 Factor. u = 2 or u = –1 Zero Product Property ex= 2 ex = –1 Replace u with ex. ln ex= ln 2 ln ex = ln (–1) Take the natural logarithm of each side. x = ln 2 x = ln (–1) Inverse Property or about 0.69 The only solution is x = ln 2 because ln (–1) is extraneous. Answer:0.69 Example 6

17. Solve e2x + ex – 12 = 0. A. ln 3 B. ln 3, ln 4 C. ln 4 D. ln 3, ln (–4) Example 6

18. Solve Logarithmic Equations Solve log x + log (x – 3) = log 28. log x + log (x – 3) = log 28 Original equation log x(x – 3) = log 28 Product Property log (x2 – 3x)= log 28 Simplify. x2 – 3x = 28 One-to-One Property x2 – 3x – 28 = 0 Subtract 28 from each side. (x – 7)(x + 4) = 0 Factor. x = 7 or x = – 4 Zero Product Property The only solution is x = 7 because –4 is an extraneous solution. Answer:7 Example 7

19. = 10 = 1 = log 101 = log 10 Check for Extraneous Solutions Solve log (3x – 4) = 1 + log (2x + 3). log (3x – 4) = 1 + log (2x + 3) log (3x – 4) – log (2x + 3) = 1 3x – 4 = 10(2x + 3) 3x – 4 = 20x + 30 –17x = 34 x = –2 But wait… must check for extraneous solutions Example 8

20. Check for Extraneous Solutions Check log (3x – 4) = 1 + log (2x + 3) log (3(–2) – 4) = 1 + log (2(–2) + 3) log (–10) = 1 = log (–1) Since neither log (–10) nor log (–1) is defined, x = –2 is an extraneous solution. Answer:no solution Example 8

21. Model Exponential Growth A. CELL PHONES This table shows the number of cell phones a new store sold in March and August of the same year. If the number of phones sold per month is increasing at an exponential rate, identify the continuous rate of growth. Then write the exponential equation to model this situation. Example 9

22. = e5k Divide each side by 88. Model Exponential Growth Let N(t) represent the number of cell phones sold at the end of t months and assume continuous growth. Then the initial number N0 is 88 cell phones sold and the number of cell phones sold N after a time of 5 months, the number of months from March to August, is 177. Use this information to find the continuous growth rate k. N(t) = N0ektExponential Growth Formula 177 = 88e5kN(5) = 177, N0 = 88, and t = 5 Example 9

23. = ln e5k Take the natural logarithm of each side. = 5k Inverse Property = k Divide each side by 5 Model Exponential Growth 0.1398 ≈ k Use a calculator. Answer:13.98%; N(t) = 88e0.1398t Example 9

24. = 0.1398t = t = e0.1398t = ln e0.1398t Model Exponential Growth B. CELL PHONES Use your model to predict the number of months it will take for the store to sell 500 phones in one month. N(t) = 88e0.1398t N(t) = 88e0.1398t 500 = 88e0.1398t 12.4 ≈ t According to this model, the store will sell 500 phones in a month in about 12.4 months. Answer:12.4 months Example 9