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

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Warm Up Solve. 1. log 16 x = 2. log x 1.331 = 3 3. log10,000 = x 64 1.1 4 3 2

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve exponential and logarithmic equations and equalities. Solve problems involving exponential and logarithmic equations. Objectives

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities exponential equation logarithmic equation Vocabulary

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities When you use a rounded number in a check, the result will not be exact, but it should be reasonable. Helpful Hint

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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 3. 3 16 – 2x = 3 3x – 9 To raise a power to a power, multiply exponents. Example 1A: Solving Exponential Equations 16 – 2x = 3x – 9 Bases are the same, so the exponents must be equal. x = 5 Solve for x.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Example 1A Continued Check 9 8 – x = 27 x – 3 9 8 – 5 27 5 – 3 9 3 27 2 729 The solution is x = 5.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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. Example 1B: Solving Exponential Equations Divide both sides by log 4. Check Use a calculator. The solution is x ≈ 2.161. x = 1 + ≈ 2.161 log5 log4 x –1 = log5 log4

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve and check. 3 2x = 27 (3) 2x = (3) 3 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. Check It Out! Example 1a 2x = 3 Bases are the same, so the exponents must be equal. x = 1.5 Solve for x.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Check 3 2x = 27 3 2(1.5) 27 3 3 27 27 The solution is x = 1.5. Check It Out! Example 1a Continued

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve and check. 7 –x = 21 log 7 –x = log 21 21 is not a power of 7, so take the log of both sides. (–x)log 7 = log 21 Apply the Power Property of Logarithms. Check It Out! Example 1b Divide both sides by log 7. x = – ≈ –1.565 log21 log7 – x = log21 log7

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Check Use a calculator. The solution is x ≈ –1.565. Check It Out! Example 1b Continued

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve and check. 2 3x = 15 log2 3x = 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. Check It Out! Example 1c Divide both sides by log 2, then divide both sides by 3. x ≈ 1.302 3x = log15 log2

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Check Use a calculator. The solution is x ≈ 1.302. Check It Out! Example 1c Continued

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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? Example 2: Biology Application Solve 2 n > 10 6 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. Write 1,000,000 in scientific annotation. Take the log of both sides. log 2 n > log 10 6

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Example 2 Continued Use the Power of Logarithms. log 10 6 is 6. nlog 2 > 6 nlog 2 > log 10 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.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Example 2 Continued Check In 20 hours, there will be 2 20 bacteria. 2 20 = 1,048,576 bacteria.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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. Solve 3 n – 1 > 1 x 10 8 $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. Write 100,000,000 in scientific annotation. Take the log of both sides. log 3 n – 1 > log 10 8 Check It Out! Example 2

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Use the Power of Logarithms. log 10 8 is 8. (n – 1)log 3 > 8 (n – 1) log 3 > log 10 8 8 log 3 n – 1 > Divide both sides by log 3. 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 Continued 8 log3 n > + 1

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Check On day 18, you would receive 3 18 – 1 or over a million dollars. 3 17 = 129,140,163 cents or 1.29 million dollars. Check It Out! Example 2

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Review the properties of logarithms from Lesson 7-4. Remember!

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve. Example 3A: Solving Logarithmic Equations Use 6 as the base for both sides. log 6 (2x – 1) = –1 6 log 6 (2x –1) = 6 –1 2x – 1 = 1 6 7 12 x =x = Use inverse properties to remove 6 to the log base 6. Simplify.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve. Example 3B: Solving Logarithmic Equations Write as a quotient. log 4 100 – log 4 (x + 1) = 1 x = 24 Use 4 as the base for both sides. Use inverse properties on the left side. 100 x + 1 log 4 ( ) = 1 4 log 4 = 4 1 100 x + 1 ( ) = 4 100 x + 1

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve. Example 3C: Solving Logarithmic Equations Power Property of Logarithms. log 5 x 4 = 8 x = 25 Definition of a logarithm. 4log 5 x = 8 log 5 x = 2 x = 5 2 Divide both sides by 4 to isolate log 5 x.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve. Example 3D: Solving Logarithmic Equations Product Property of Logarithms. log 12 x + log 12 (x + 1) = 1 Exponential form. Use the inverse properties. log 12 x(x + 1) = 1 log 12 x(x +1) 12 = 12 1 x(x + 1) = 12

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Example 3 Continued Multiply and collect terms. Factor. Solve. x 2 + x – 12 = 0 log 12 x + log 12 (x +1) = 1 (x – 3)(x + 4) = 0 x – 3 = 0 or x + 4 = 0 Set each of the factors equal to zero. x = 3 or x = –4 log 12 x + log 12 (x +1) = 1 log 12 3 + log 12 (3 + 1) 1 log 12 3 + log 12 4 1 log 12 12 1 The solution is x = 3. 1 log 12 ( –4) + log 12 (–4 +1) 1 log 12 ( –4) is undefined. x Check Check both solutions in the original equation.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve. 3 = log 8 + 3log x Check It Out! Example 3a 3 = log 8 + 3log x 3 = log 8 + log x 3 3 = log (8x 3 ) 10 3 = 10 log (8x 3 ) 1000 = 8x 3 125 = x 3 5 = x Use 10 as the base for both sides. Use inverse properties on the right side. Product Property of Logarithms. Power Property of Logarithms.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Solve. 2log x – log 4 = 0 Check It Out! Example 3b Write as a quotient. x = 2 Use 10 as the base for both sides. Use inverse properties on the left side. 2log ( ) = 0 x 4 2(10 log ) = 10 0 x 4 2( ) = 1 x 4

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Watch out for calculated solutions that are not solutions of the original equation. Caution

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Use a table and graph to solve 2 x + 1 > 8192x. Example 4A: Using Tables and Graphs to Solve Exponential and Logarithmic Equations and Inequalities 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}.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities log(x + 70) = 2log ( ) In the table, find the x-values where Y1 equals Y2. In the graph, find the x-value at the point of intersection. x 3 Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2. x 3 The solution is x = 30. Example 4B

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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. 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. The solution is x = 2.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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. 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. The solution is x < 2.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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. 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. The solution is x = 1000.

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities Lesson Quiz: Part I Solve. 1. 4 3x–1 = 8 x+1 2. 3 2x–1 = 20 3. log 7 (5x + 3) = 3 4. log(3x + 1) – log 4 = 2 5. log 4 (x – 1) + log 4 (3x – 1) = 2 x ≈ 1.86 x = 68 x = 133 x = 3 x = 5 3

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Holt Algebra 2 7-5 Exponential and Logarithmic Equations and Inequalities 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? 7. Use a table and graph to solve the equation 2 3x = 3 3x–1. 70 min x ≈ 0.903

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