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

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

Vocabulary exponential equation logarithmic equation

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.

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

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.

Example 1A Continued Check 98 – x = 27x – 3 98 – 5 275 – 3 93 272 729 729  The solution is x = 5.

Example 1B: Solving Exponential Equations Solve and check. 4x – 1 = 5 log 4x – 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 Divide both sides by log 4. x = 1 + ≈ 2.161 log5 log4 Check Use a calculator. The solution is x ≈ 2.161.

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 32x = 33 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.

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

21 is not a power of 7, so take the log of both sides. 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 (–x)log 7 = log 21 Apply the Power Property of Logarithms. –x = log21 log7 Divide both sides by log 7. x = – ≈ –1.565 log21 log7

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

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

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

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. Solve 2n > 106 Write 1,000,000 in scientific annotation. log 2n > log 106 Take the log of both sides.

Use the Power of Logarithms. Example 2 Continued nlog 2 > log 106 Use the Power of Logarithms. nlog 2 > 6 log 106 is 6. 6 log 2 n > Divide both sides by log 2. 6 0.301 n > 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.

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

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–1 cents. Solve 3n – 1 > 1 x 108 Write 100,000,000 in scientific annotation. log 3n – 1 > log 108 Take the log of both sides.

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. 8 log 3 n – 1 > Divide both sides by log 3. 8 log3 n > + 1 Evaluate by using a calculator. n > ≈ 17.8 Round up to the next whole number. Beginning on day 18, you would receive more than a million dollars.

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.

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.

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

Example 3A: Solving Logarithmic Equations Solve. log6(2x – 1) = –1 6 log6 (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. 7 12 x = Simplify.

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

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

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

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

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

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

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

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 and 8192x as Y2. In the table, find the x-values where Y1 is greater than Y2. In the graph, find the x-value at the point of intersection. The solution set is {x | x > 16}.

Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2. Example 4B log(x + 70) = 2log( ) x 3 Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2. x 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.

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

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

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

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

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

Chapter 7 Section 5 Page 526 # 2-20, 21-36 Homework! Chapter 7 Section 5 Page 526 # 2-20, 21-36