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Exponential and Logarithmic Equations Mastery Lesson

Learn to solve exponential and logarithmic equations, practice with examples, and master key concepts in Holt McDougal Algebra 2 warm-up lesson.

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Exponential and Logarithmic Equations Mastery Lesson

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  1. Exponential and Logarithmic Equations and Inequalities 7-5 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

  2. 3 2 Warm Up Solve. 1.log16x = 2. logx1.331 = 3 3. log10,000 =x 64 1.1 4

  3. Objectives Solve exponential and logarithmic equations and equalities. Solve problems involving exponential and logarithmic equations.

  4. Vocabulary exponential equation logarithmic equation

  5. An exponential equation is an equation containing one or more expressions that have a variable as an exponent. To solve exponential equations: • Try writing them so that the bases are all the same. • Take the logarithm of both sides.

  6. Helpful Hint When you use a rounded number in a check, the result will not be exact, but it should be reasonable.

  7. Example 1A: Solving Exponential Equations Solve and check. 98 – x = 27x – 3 Rewrite each side with the same base; 9 and 27 are powers of 3. (32)8 – x = (33)x – 3 To raise a power to a power, multiply exponents. 316– 2x = 33x – 9 16 – 2x = 3x – 9 Bases are the same, so the exponents must be equal. x = 5 Solve for x.

  8. Example 1A Continued Check 98 – x = 27x – 3 98 – 5275 – 3 93 272  729 729 The solution is x = 5.

  9. log5 log5 x–1 = log4 log4 x = 1 + ≈ 2.161 Example 1B: Solving Exponential Equations Solve and check. 4x – 1= 5 5 is not a power of 4, so take the log of both sides. log 4x – 1 = log 5 (x– 1)log 4 = log 5 Apply the Power Property of Logarithms. Divide both sides by log 4. CheckUse a calculator. The solution is x ≈ 2.161.

  10. Check It Out! Example 1a Solve and check. 32x = 27 Rewrite each side with the same base; 3 and 27 are powers of 3. (3)2x = (3)3 To raise a power to a power, multiply exponents. 32x = 33 2x = 3 Bases are the same, so the exponents must be equal. x = 1.5 Solve for x.

  11. Check It Out! Example 1a Continued Check 32x = 27 32(1.5)27 33 27  27 27 The solution is x = 1.5.

  12. log21 log21 –x = log7 log7 x = – ≈ –1.565 Check It Out! Example 1b Solve and check. 7–x = 21 21 is not a power of 7, so take the log of both sides. log 7–x = log 21 Apply the Power Property of Logarithms. (–x)log 7 = log 21 Divide both sides by log 7.

  13. Check It Out! Example 1b Continued CheckUse a calculator. The solution is x ≈ –1.565.

  14. log15 3x = log2 Check It Out! Example 1c Solve and check. 23x = 15 15 is not a power of 2, so take the log of both sides. log23x = log15 Apply the Power Property of Logarithms. (3x)log 2 = log15 Divide both sides by log 2, then divide both sides by 3. x ≈ 1.302

  15. Check It Out! Example 1c Continued CheckUse a calculator. The solution is x ≈ 1.302.

  16. Example 2: Biology Application Suppose a bacteria culture doubles in size every hour. How many hours will it take for the number of bacteria to exceed 1,000,000? At hour 0, there is one bacterium, or 20 bacteria. At hour one, there are two bacteria, or 21 bacteria, and so on. So, at hour n there will be 2n bacteria. Write 1,000,000 in scientific annotation. Solve 2n > 106 log 2n > log 106 Take the log of both sides.

  17. 6 6 log 2 0.301 n > n > Example 2 Continued nlog 2 > log 106 Use the Power of Logarithms. nlog 2 > 6 log 106 is 6. Divide both sides by log 2. Evaluate by using a calculator. n > ≈ 19.94 Round up to the next whole number. It will take about 20 hours for the number of bacteria to exceed 1,000,000.

  18. Example 2 Continued Check In 20 hours, there will be 220 bacteria. 220 = 1,048,576 bacteria.

  19. Check It Out! Example 2 You receive one penny on the first day, and then triple that (3 cents) on the second day, and so on for a month. On what day would you receive a least a million dollars. $1,000,000 is 100,000,000 cents. On day 1, you would receive 1 cent or 30 cents. On day 2, you would receive 3 cents or 31 cents, and so on. So, on day n you would receive 3n–1cents. Write 100,000,000 in scientific annotation. Solve 3n – 1 > 1 x 108 log 3n –1 > log 108 Take the log of both sides.

  20. 8 8 log 3 log3 n – 1 > n > + 1 Check It Out! Example 2 Continued (n – 1)log 3 > log 108 Use the Power of Logarithms. (n – 1)log 3 > 8 log 108 is 8. Divide both sides by log 3. Evaluate by using a calculator. Round up to the next whole number. n > ≈ 17.8 Beginning on day 18, you would receive more than a million dollars.

  21. Check It Out! Example 2 Check On day 18, you would receive 318 – 1 or over a million dollars. 317 = 129,140,163 cents or 1.29 million dollars.

  22. A logarithmic equation is an equation with a logarithmic expression that contains a variable. You can solve logarithmic equations by using the properties of logarithms.

  23. Remember! Review the properties of logarithms from Lesson 7-4.

  24. 1 7 6 12 2x – 1= x= Example 3A: Solving Logarithmic Equations Solve. log6(2x – 1) = –1 6log6 (2x –1) = 6–1 Use 6 as the base for both sides. Use inverse properties to remove 6 to the log base 6. Simplify.

  25. 100 100 x + 1 x + 1 log4( ) = 1 ( ) 100 x + 1 4log4 = 41 = 4 Example 3B: Solving Logarithmic Equations Solve. log4100 – log4(x + 1) = 1 Write as a quotient. Use 4 as the base for both sides. Use inverse properties on the left side. x= 24

  26. Example 3C: Solving Logarithmic Equations Solve. log5x 4 = 8 4log5x = 8 Power Property of Logarithms. log5x = 2 Divide both sides by 4 to isolate log5x. x = 52 Definition of a logarithm. x= 25

  27. log12x(x +1) 12 = 121 Example 3D: Solving Logarithmic Equations Solve. log12x+ log12(x + 1) = 1 Product Property of Logarithms. log12x(x + 1) = 1 Exponential form. x(x + 1) = 12 Use the inverse properties.

  28. Example 3 Continued x2 + x – 12 = 0 Multiply and collect terms. (x – 3)(x+ 4) = 0 Factor. Set each of the factors equal to zero. x – 3 = 0 or x+ 4 = 0 x = 3 or x= –4 Solve. Check Check both solutions in the original equation. log12x+ log12(x +1) = 1 log12x+ log12(x +1) = 1 x log123+ log12(3 + 1) 1 log12( –4) + log12(–4 +1) 1 log123 + log124 1 log12( –4) is undefined. log1212 1 1 1  The solution is x = 3.

  29. Check It Out! Example 3a Solve. 3 = log 8 + 3log x 3 = log 8 + 3log x 3 = log 8 + log x3 Power Property of Logarithms. 3 = log (8x3) Product Property of Logarithms. 103 = 10log (8x3) Use 10 as the base for both sides. 1000 = 8x3 Use inverse properties on the right side. 125 = x3 5 = x

  30. x 4 2log() = 0 x 4 x 4 2(10log ) = 100 2( ) = 1 Check It Out! Example 3b Solve. 2log x– log 4 = 0 Write as a quotient. Use 10 as the base for both sides. Use inverse properties on the left side. x= 2

  31. Caution Watch out for calculated solutions that are not solutions of the original equation.

  32. Example 4A: Using Tables and Graphs to Solve Exponential and Logarithmic Equations and Inequalities Use a table and graph to solve 2x + 1 > 8192x. Use a graphing calculator. Enter 2^(x + 1)as Y1 and8192xas Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is greater than Y2. The solution set is {x | x > 16}.

  33. Use a graphing calculator. Enter log(x + 70) as Y1 and 2log() as Y2. x x 3 3 Example 4B log(x + 70) = 2log() In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 equals Y2. The solution is x = 30.

  34. Check It Out! Example 4a Use a table and graph to solve 2x = 4x – 1. Use a graphing calculator. Enter 2xas Y1 and4(x – 1)as Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is equal to Y2. The solution is x = 2.

  35. Check It Out! Example 4b Use a table and graph to solve 2x > 4x – 1. Use a graphing calculator. Enter 2xas Y1 and4(x – 1)as Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is greater than Y2. The solution is x < 2.

  36. Check It Out! Example 4c Use a table and graph to solve log x2 = 6. Use a graphing calculator. Enter log(x2)as Y1 and6 as Y2. In the graph, find the x-value at the point of intersection. In the table, find the x-values where Y1 is equal to Y2. The solution is x = 1000.

  37. 5 x = 3 Lesson Quiz: Part I Solve. 1. 43x–1 = 8x+1 x ≈ 1.86 2. 32x–1 =20 x = 68 3. log7(5x + 3) = 3 4. log(3x + 1) – log 4 = 2 x = 133 5. log4(x – 1) + log4(3x – 1) = 2 x = 3

  38. Lesson Quiz: Part II 6. A single cell divides every 5 minutes. How long will it take for one cell to become more than 10,000 cells? 70 min 7. Use a table and graph to solve the equation 23x = 33x–1. x ≈ 0.903

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