Warm Up Lesson Presentation Lesson Quiz. Holt McDougal Algebra 2

Transcription

1 4-5 Warm Up Lesson Presentation Lesson Quiz Algebra 2

2 Warm Up Solve. 1. log 16 x = log x = log10,000 = x 4

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

4 Vocabulary exponential equation logarithmic equation

5 Logarithm Properties

6 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.

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

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

9 Example 1A Continued Check 9 8 x = 27 x The solution is x = 5.

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

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

12 Check It Out! Example 1a Continued Check 3 2x = (1.5) The solution is x = 1.5.

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

14 Check It Out! Example 1b Continued Check Use a calculator. The solution is x

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

16 Check It Out! Example 1c Continued Check Use a calculator. The solution is x

17 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 2 0 bacteria. At hour one, there are two bacteria, or 2 1 bacteria, and so on. So, at hour n there will be 2 n bacteria. Solve 2 n > 10 6 log 2 n > log 10 6 Write 1,000,000 in scientific annotation. Take the log of both sides.

18 Example 2 Continued nlog 2 > log 10 6 Use the Power of Logarithms. nlog 2 > 6 log 10 6 is 6. 6 n > log 2 Divide both sides by log 2. n > Evaluate by using a calculator. n > Round up to the next whole number. It will take about 20 hours for the number of bacteria to exceed 1,000,000.

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

20 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 3 0 cents. On day 2, you would receive 3 cents or 3 1 cents, and so on. So, on day n you would receive 3 n 1 cents. Solve 3 n 1 > 1 x 10 8 log 3 n 1 > log 10 8 Write 100,000,000 in scientific annotation. Take the log of both sides.

21 Check It Out! Example 2 Continued (n 1) log 3 > log 10 8 Use the Power of Logarithms. (n 1)log 3 > 8 log 10 8 is 8. 8 n 1 > log 3 Divide both sides by log 3. 8 n > log3 + 1 n > 17.8 Evaluate by using a calculator. Round up to the next whole number. Beginning on day 18, you would receive more than a million dollars.

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

23 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.

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

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

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

27 Example 3C: Solving Logarithmic Equations Solve. log 5 x 4 = 8 4log 5 x = 8 Power Property of Logarithms. log 5 x = 2 Divide both sides by 4 to isolate log 5 x. x = 5 2 Definition of a logarithm. x = 25

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

29 x 2 + x 12 = 0 (x 3)(x + 4) = 0 x 3 = 0 or x + 4 = 0 x = 3 or x = 4 log 12 x + log 12 (x +1) = 1 log log 12 (3 + 1) 1 log log Example 3 Continued log Multiply and collect terms. Factor. Set each of the factors equal to zero. Solve. Check Check both solutions in the original equation. log 12 x + log 12 (x +1) = 1 log 12 ( 4) + log 12 ( 4 +1) 1 log 12 ( 4) is undefined. The solution is x = 3. x

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

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

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

33 Example 4A: Using Tables and Graphs to Solve Equations and Inequalities Use a table and graph to solve 2 x + 1 > 8192x. 4-5 Use a graphing calculator. Enter 2^(x + 1) as Y1 and 8192x as Y2. In the table, find the x-values where Y1 is greater than Y2. The solution set is {x x > 16}. In the graph, find the x-value at the point of intersection.

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

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

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

37 Check It Out! Example 4c Use a table and graph to solve log x 2 = 6. Use a graphing calculator. Enter log(x 2 ) as Y1 and 6 as Y2. In the table, find the x-values where Y1 is equal to Y2. The solution is x = In the graph, find the x-value at the point of intersection.

38 Lesson Quiz: Part I Solve x 1 = 8 x+1 x = x 1 = 20 x log 7 (5x + 3) = 3 x = log(3x + 1) log 4 = 2 x = log 4 (x 1) + log 4 (3x 1) = 2 x = 3

39 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 2 3x = 3 3x 1. x 0.903