lesson 11 homework 5.6 answer key

• Daily Agenda
• Staying Informed

What is the "Go Math!" curriculum?

Online textbook and glossary.

• Operations and Algebraic Thinking
• Number and Operations in Base Ten
• Number and Operations—Fractions
• ​Number and Operations in Base Ten
• ​​ Number and Operations—Fractions
• ​ Measurement and Data
• Geometry Operations
• Algebraic Thinking
• ​​ Measurement and Data
• ​​ Measurement
• Data Geometry

• 888-309-8227
• 732-384-0146

New User Registration

Forgot Password

Textbook Resources

Textbook resources.

• Call us toll-free
• FAQs – Frequently Asked Questions
• Contact Lumos Learning – Proven Study Programs by Expert Teachers

Follow us: Lumos Learning -->

• 2024 © Lumos Learning
• Privacy Policy - Terms of Service - Disclaimers

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc... Read More

PARCC® is a registered trademark of PARCC, Inc. Lumos Learning, is not owned by or affiliated in any fashion with PARCC, Inc., the Partnership for the Assessment of Readiness for College and Careers, nor any state of the Union. Neither PARCC, Inc., nor The Partnership for the Assessment of Readiness for College and Careers, nor any member state has endorsed this product. No portion of any fees or charges paid for any products or services Lumos Learning offers will be paid or inure to the benefit of PARCC, Inc., or any state of the Union

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not aff... Read More

SBAC is a copyright of The Regents of the University of California – Smarter Balanced Assessment Consortium, which is not affiliated to Lumos Learning. The Regents of the University of California – Smarter Balanced Assessment Consortium, was not involved in the production of, and does not endorse these products or this site.

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC, was not... Read More

ACT® Aspire™ is a registered trademark of ACT Aspire LLC., which is not affiliated to Lumos Learning. ACT Aspire LLC,was not involved in the production of, and does not endorse these products or this site.

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the... Read More

Florida Department of Education is not affiliated to Lumos Learning. Florida department of education, was not involved in the production of, and does not endorse these products or this site.

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the... Read More

Indiana Department of Education is not affiliated to Lumos Learning. Indiana department of education, was not involved in the production of, and does not endorse these products or this site.

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved... Read More

Mississippi Department of Education is not affiliated to Lumos Learning. Mississippi department of education, was not involved in the production of, and does not endorse these products or this site.

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the prod... Read More

Ohio Department of Education is not affiliated to Lumos Learning. Ohio department of education, was not involved in the production of, and does not endorse these products or this site.

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved... Read More

Tennessee Department of Education is not affiliated to Lumos Learning. Tennessee department of education, was not involved in the production of, and does not endorse these products or this site.

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved... Read More

Georgia Department of Education is not affiliated to Lumos Learning. Georgia department of education, was not involved in the production of, and does not endorse these products or this site.

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved... Read More

Missouri Department of Education is not affiliated to Lumos Learning. Missouri department of education, was not involved in the production of, and does not endorse these products or this site.

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved... Read More

Louisiana Department of Education is not affiliated to Lumos Learning. Louisiana department of education, was not involved in the production of, and does not endorse these products or this site.

• Texas Go Math
• Big Ideas Math
• enVision Math
• EngageNY Math
• McGraw Hill My Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Texas Go Math Grade 5 Lesson 5.6 Answer Key Add and Subtract Mixed Numbers

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 5.6 Answer Key Add and Subtract Mixed Numbers.

Unlock the Problem

Denise mixed 1\(\frac{4}{5}\) ounces of blue paint with 2\(\frac{1}{10}\) ounces of yellow paint. How many ounces of paint did Denise mix?

• What operation should you use to solve the problem?
• Do the fractions have the same denominator?

Add. 1\(\frac{4}{5}\) + 2\(\frac{1}{10}\) To find the sum of mixed numbers with unequal denominators, you can use a common denominator. STEP 1: Estimate the sum.

STEP 2: Find a common denominator. Use the common denominator to write equivalent fractions with equal denominators.

Math Talk Mathematical Processes

Did you use the least common denominator? Explain. Answer: yes Explanation: The sum of mixed numbers with unequal denominators  can use a common denominator.

Question 1. Explain how you know whether your answer is reasonable. Answer: Yes my answer is reasonable Because the sum of two mixed numbers is solved and it is proved

Go Math Grade 5 Lesson 5.6 Answer Key Question 2. What other common denominator could you have used? Answer: 50 multiply 5 x 10 = 50

Subtract. 4\(\frac{5}{6}\) – 2\(\frac{3}{4}\)

You can also use a common denominator to find the difference between mixed numbers with unequal denominators. STEP 1: Estimate the difference.

Question 3. Explain how you know whether your answer is reasonable. Answer: Used the common denominator to write equivalent fractions with equal denominators. so, my answer is reasonable.

Share and Show

Find the sum. Write your answer in simplest form.

Question 2. 2\(\frac{3}{4}\) + 3\(\frac{3}{10}\) Answer: 2\(\frac{3}{4}\) + 3\(\frac{3}{10}\)  = \(\frac{11}{4}\) + \(\frac{33}{10}\) = \(\frac{55}{20}\) + \(\frac{66}{20}\) = \(\frac{121}{20}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take the least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Lesson 5.6 Answer Key Go Math Grade 5 Question 3. 5\(\frac{3}{4}\) + 1\(\frac{1}{3}\) Answer: 5\(\frac{3}{4}\) + 1\(\frac{1}{3}\) = \(\frac{23}{4}\) + \(\frac{4}{3}\) = \(\frac{69}{12}\)+ \(\frac{16}{12}\) = \(\frac{85}{12}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 4. 3\(\frac{4}{5}\) + 2\(\frac{3}{10}\) Answer: 3\(\frac{4}{5}\) + 2\(\frac{3}{10}\) = \(\frac{19}{5}\) + \(\frac{23}{10}\) = \(\frac{38}{10}\) + \(\frac{23}{10}\) = \(\frac{61}{10}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Problem Solving

Practice: Copy and Solve Find the sum or difference. Write your answer in simplest form.

Question 5. 1\(\frac{5}{12}\) + 4\(\frac{1}{6}\) Answer: 1\(\frac{5}{12}\) + 4\(\frac{1}{6}\) = \(\frac{17}{12}\) + \(\frac{21}{6}\) = \(\frac{17}{12}\) + \(\frac{42}{12}\) = \(\frac{59}{12}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take the least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Go Math 5th Grade Lesson 5.6 Add and Subtract Mixed Numbers Question 6. 8\(\frac{1}{2}\) + 6\(\frac{3}{5}\) Answer: 8\(\frac{1}{2}\) + 6\(\frac{3}{5}\) = \(\frac{17}{2}\) + \(\frac{33}{5}\) = \(\frac{85}{10}\) + \(\frac{66}{10}\) = \(\frac{151}{10}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take the least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 7. 2\(\frac{1}{6}\) + 4\(\frac{5}{9}\) Answer: 2\(\frac{1}{6}\) + 4\(\frac{5}{9}\) = \(\frac{13}{6}\) + \(\frac{36}{9}\) = \(\frac{39}{18}\) + \(\frac{72}{18}\) = \(\frac{41}{18}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 8. 3\(\frac{5}{8}\) + \(\frac{5}{12}\) Answer: 3\(\frac{5}{8}\) + \(\frac{5}{12}\) = \(\frac{29}{8}\) + \(\frac{5}{12}\) = \(\frac{87}{24}\) + \(\frac{10}{24}\) = \(\frac{97}{24}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Lesson 5.6 Add and Subtract Mixed Numbers Answer Key Question 9. 3\(\frac{2}{3}\) – 1\(\frac{1}{6}\) Answer: 3\(\frac{2}{3}\) – 1\(\frac{1}{6}\) = \(\frac{11}{3}\) – \(\frac{7}{6}\) = \(\frac{22}{6}\) – \(\frac{7}{6}\) = \(\frac{15}{6}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take the least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 10. 5\(\frac{6}{7}\) – 1\(\frac{2}{3}\) Answer: 5\(\frac{6}{7}\) – 1\(\frac{2}{3}\) = \(\frac{41}{7}\) – \(\frac{5}{3}\) = \(\frac{123}{21}\) – \(\frac{35}{21}\) = \(\frac{88}{21}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 11. 2\(\frac{7}{8}\) – \(\frac{1}{2}\) Answer: 2\(\frac{7}{8}\) – \(\frac{1}{2}\) = \(\frac{23}{8}\) – \(\frac{1}{2}\) = \(\frac{23}{8}\) – \(\frac{4}{8}\) =  \(\frac{19}{8}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 12. 4\(\frac{7}{12}\) – 1\(\frac{2}{9}\) Answer: 4\(\frac{7}{12}\) – 1\(\frac{2}{9}\) = \(\frac{55}{12}\) – \(\frac{7}{5}\) = \(\frac{275}{60}\) – \(\frac{72}{60}\) = \(\frac{203}{60}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Go Math 5th Grade Lesson 5.6 Answers Question 13. Communicate Why do you need to write equivalent fractions with common denominators to add 4\(\frac{5}{6}\) and \(\frac{11}{8}\)? Explain. Answer: 4\(\frac{5}{6}\) + \(\frac{11}{8}\) = \(\frac{25}{24}\) + \(\frac{11}{24}\) = \(\frac{36}{24}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take the least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Problem-Solving

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 16. Dr. Whether-or-Not collects two hailstones during a storm in California. One hailstone weighs 2\(\frac{3}{8}\) pounds, and the other hailstone weighs 1\(\frac{3}{10}\) pounds. How much heavier is the larger hailstone than the smaller hailstone? (A) \(\frac{3}{40}\) pounds (B) 1\(\frac{27}{40}\) pounds (C) 1\(\frac{3}{40}\) pounds (D) 3\(\frac{27}{40}\) pounds Answer: D Explanation: Dr. Whether-or-Not collects two hailstones during a storm in California. One hailstone weighs 2\(\frac{3}{8}\) pounds, and the other hailstone weighs 1\(\frac{3}{10}\) pounds. 2\(\frac{3}{8}\) + 1\(\frac{3}{10}\) pounds \(\frac{19}{8}\) + \(\frac{13}{10}\) \(\frac{294}{80}\) 3\(\frac{27}{40}\) pounds

Go Math Lesson 5.6 5th Grade Answer Key Question 17. Apply Jason is making a fruit salad. He mixes in 3\(\frac{1}{4}\) cups of orange melon and 2\(\frac{2}{3}\) cups of green melon. How many cups of melon does Jason put in the fruit salad? (A) 5\(\frac{1}{4}\) cups (B) 5\(\frac{1}{3}\) cups (C) 5\(\frac{7}{12}\) cups (D) 5\(\frac{11}{12}\) cups Answer: A Explanation: Apply Jason is making a fruit salad. He mixes in 3\(\frac{1}{4}\) cups of orange melon and 2\(\frac{2}{3}\) cups of green melon. 3\(\frac{1}{4}\) + 2\(\frac{2}{3}\) \(\frac{13}{4}\) + \(\frac{8}{3}\) \(\frac{39}{12}\) + \(\frac{24}{12}\) \(\frac{63}{12}\) 5\(\frac{1}{4}\)

Question 18. Multi-Step Dakota makes a salad dressing by combining 6\(\frac{1}{3}\) fluid ounces of oil and 2\(\frac{3}{8}\) fluid ounces of vinegar in a jar. She then pours 2\(\frac{1}{4}\) fluid ounces of the dressing onto her salad. How much dressing remains in the jar? (A) 6\(\frac{1}{8}\) fluid ounces (B) 6\(\frac{3}{8}\) fluid ounces (C) 6\(\frac{11}{24}\) fluid ounces (D) 6\(\frac{17}{24}\) fluid ounces Answer: C Dakota makes a salad dressing by combining 6\(\frac{1}{3}\) fluid ounces of oil and 2\(\frac{3}{8}\) fluid ounces of vinegar in a jar. She then pours 2\(\frac{1}{4}\) fluid ounces of the dressing onto her salad. 6\(\frac{1}{3}\) + 2\(\frac{3}{8}\) – 2\(\frac{1}{4}\) \(\frac{155}{24}\) 6\(\frac{11}{24}\) fluid ounces

Texas Test Prep

Question 19. Yolanda walked 3\(\frac{6}{10}\) miles. Then she walked 4\(\frac{1}{2}\) more miles. How many miles did Yolanda walk? (A) 7\(\frac{1}{10}\) miles (B) 8\(\frac{7}{10}\) miles (C) 8\(\frac{1}{10}\) miles (D) 7\(\frac{7}{10}\) miles Answer: C Explanation: Yolanda walked 3\(\frac{6}{10}\) miles. Then she walked 4\(\frac{1}{2}\) more miles 3\(\frac{6}{10}\) + 4\(\frac{1}{2}\) \(\frac{72}{20}\) + \(\frac{90}{20}\) \(\frac{162}{20}\)

Texas Go Math Grade 5 Lesson 5.6 Homework and Practice Answer Key

Find the sum or difference. Write your answer in simplest form.

Question 1. 1\(\frac{1}{4}\) + 2\(\frac{2}{3}\) _____________ Answer: 1\(\frac{1}{4}\) + 2\(\frac{2}{3}\) = \(\frac{5}{4}\) + \(\frac{8}{3}\) = \(\frac{15}{12}\) + \(\frac{32}{12}\) = \(\frac{47}{12}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 2. 3\(\frac{3}{4}\) + 4\(\frac{5}{12}\) _____________ Answer: 3\(\frac{3}{4}\) + 4\(\frac{5}{12}\) = \(\frac{36}{12}\) + \(\frac{53}{12}\) = \(\frac{89}{12}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 3. 1\(\frac{1}{3}\) + 2\(\frac{1}{6}\) _____________ Answer: 1\(\frac{1}{3}\) + 2\(\frac{1}{6}\) = \(\frac{8}{6}\) + \(\frac{13}{6}\) = \(\frac{21}{6}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Go Math Lesson 5.6 5th Grade Add and Subtract Mixed Numbers Question 4. 4\(\frac{1}{2}\) + 3\(\frac{4}{5}\) _____________ Answer: 4\(\frac{1}{2}\) + 3\(\frac{4}{5}\) = \(\frac{45}{10}\) + \(\frac{38}{10}\) = \(\frac{83}{10}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 5. 5\(\frac{5}{6}\) + 4\(\frac{2}{9}\) ____________ Answer: 5\(\frac{5}{6}\) + 4\(\frac{2}{9}\) = \(\frac{35}{6}\) + \(\frac{38}{9}\) = \(\frac{105}{18}\) + \(\frac{76}{18}\) = \(\frac{181}{18}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 6. 7\(\frac{1}{4}\) + 3\(\frac{2}{5}\) ___________ Answer: 7\(\frac{1}{4}\) + 3\(\frac{2}{5}\) = \(\frac{29}{4}\) + \(\frac{17}{5}\) = \(\frac{145}{20}\) + \(\frac{68}{20}\) = \(\frac{213}{20}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 7. 3\(\frac{2}{7}\) + 8\(\frac{1}{3}\) _____________ Answer: 3\(\frac{2}{7}\) + 8\(\frac{1}{3}\) = \(\frac{23}{7}\) + \(\frac{25}{3}\) = \(\frac{69}{21}\) + \(\frac{175}{21}\) = \(\frac{244}{21}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 8. 4\(\frac{3}{7}\) + 3\(\frac{1}{2}\) ____________ Answer: 4\(\frac{3}{7}\) + 3\(\frac{1}{2}\)  = \(\frac{31}{7}\) + \(\frac{7}{2}\) = \(\frac{62}{14}\) + \(\frac{49}{14}\) = \(\frac{114}{14}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 9. 2\(\frac{4}{5}\) – 1\(\frac{1}{2}\) ____________ Answer: 2\(\frac{4}{5}\) – 1\(\frac{1}{2}\) = \(\frac{14}{5}\) – \(\frac{3}{2}\) = \(\frac{28}{10}\) – \(\frac{15}{10}\) = \(\frac{13}{10}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 10. 5\(\frac{3}{8}\) – 1\(\frac{1}{4}\) ____________ Answer: 5\(\frac{3}{8}\) – 1\(\frac{1}{4}\) = \(\frac{43}{8}\) – \(\frac{5}{4}\) = \(\frac{43}{8}\) – \(\frac{10}{8}\) = \(\frac{33}{8}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 11. 4\(\frac{1}{3}\) – 3\(\frac{1}{6}\) _____________ Answer: 4\(\frac{1}{3}\) – 3\(\frac{1}{6}\) = \(\frac{13}{3}\) – \(\frac{19}{6}\) = \(\frac{26}{6}\) – \(\frac{19}{6}\) = \(\frac{7}{6}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 12. 6\(\frac{5}{6}\) – 5\(\frac{7}{9}\) _____________ Answer: 6\(\frac{5}{6}\) – 5\(\frac{7}{9}\) = \(\frac{41}{6}\) – \(\frac{53}{9}\) = \(\frac{123}{18}\) – \(\frac{106}{18}\) = \(\frac{17}{18}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Go Math Grade 5 Practice and Homework Lesson 5.6 Answer Key Question 13. 4\(\frac{1}{3}\) – 2\(\frac{1}{4}\) ____________ Answer: 4\(\frac{1}{3}\) – 2\(\frac{1}{4}\) = \(\frac{13}{3}\) – \(\frac{9}{4}\) = \(\frac{52}{12}\) – \(\frac{27}{12}\) = \(\frac{25}{12}\) Explanation: Step I: We add the whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 14. 3\(\frac{1}{4}\) – 1\(\frac{1}{6}\) _____________ Answer: 3\(\frac{1}{4}\) – 1\(\frac{1}{6}\) = \(\frac{13}{4}\) – \(\frac{7}{6}\)  = \(\frac{39}{12}\) – \(\frac{14}{12}\) = \(\frac{25}{12}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 15. 6\(\frac{3}{4}\) – 2\(\frac{5}{16}\) _____________ Answer: 6\(\frac{3}{4}\) – 2\(\frac{5}{16}\) = \(\frac{23}{4}\) – \(\frac{37}{16}\)= \(\frac{92}{16}\) – \(\frac{37}{16}\) = \(\frac{55}{16}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 16. 7\(\frac{3}{5}\) – 2\(\frac{1}{4}\) _____________ Answer: 7\(\frac{3}{5}\) – 2\(\frac{1}{4}\) = \(\frac{38}{5}\) – \(\frac{9}{4}\) = \(\frac{152}{20}\) – \(\frac{45}{20}\) = \(\frac{107}{20}\) Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 17. Use two mixed numbers to write an equation with a sum of 4\(\frac{1}{4}\). Answer: Explanation: Step I: We add the  whole numbers, separately. We change the mixed fractions into improper fractions. Step II: To add fractions, we take least common denominators and change the fractions into like fractions. Step III: We find the sum of the whole numbers and the fractions in the simplest form.

Question 18. Lucas says his twin baby brothers have a total weight of 15\(\frac{1}{8}\) pounds. Jackson weighs pounds, and Jeremy weighs 8\(\frac{7}{8}\) pounds. Explain how you can use estimation to tell if the total weight is reasonable. Answer:

Question 19. The gas tank in Rebecca’s old car held 14\(\frac{1}{5}\) gallons. The gas tank in Rebecca’s new car holds 18\(\frac{1}{2}\) gallons. How many more gallons will the tank in Rebecca’s new car hold than her old car? Answer: 4\(\frac{3}{10}\) Explanation: 18\(\frac{1}{2}\) – 14\(\frac{1}{5}\) = 4\(\frac{3}{10}\)

Lesson Check

Question 20. If Ioana attaches her chain to the end of Gabrielle’s chain, what will be the length of the combined chain? (A) 13\(\frac{3}{4}\) feet (B) 13\(\frac{1}{4}\) feet (C) 12\(\frac{1}{4}\) feet (D) 12\(\frac{1}{2}\) feet Answer: B Explanation: the length of the combined chain is 13\(\frac{1}{4}\) feet 7\(\frac{1}{2}\) feet + 5\(\frac{3}{4}\) feet= \(\frac{30+23}{4}\) feet \(\frac{53}{4}\)

Question 21. How much longer is Oksana’s chain than Gabrielle’s chain? (A) 15\(\frac{7}{12}\) feet (B) 14\(\frac{1}{12}\) feet (C) 4\(\frac{1}{4}\) feet (D) 4\(\frac{1}{12}\) feet Answer: D 4\(\frac{1}{12}\) feet is longer than Oksana’s chain than Gabrielle’s chain Explanation: 9\(\frac{5}{6}\) feet + 5\(\frac{3}{4}\) feet= \(\frac{118-69}{12}\) feet \(\frac{49}{12}\) 4\(\frac{1}{12}\)

Question 22. Mia hiked 2\(\frac{1}{2}\) miles farther than Jacob. Which could be the two distances each person hiked? (A) Mia: 2\(\frac{1}{2}\) miles; Jacob: 1\(\frac{1}{4}\) miles (B) Mia: 2\(\frac{1}{2}\) miles; Jacob: 7\(\frac{1}{2}\) miles (C) Mia: 3\(\frac{2}{5}\) miles; Jacob: 5\(\frac{9}{10}\) miles (D) Mia: 5\(\frac{9}{10}\) miles; Jacob: 3\(\frac{2}{5}\) miles Answer: A Explanation: 2\(\frac{1}{2}\) than jacob if Mia: 2\(\frac{1}{2}\) miles; Jacob: 1\(\frac{1}{4}\) miles

Question 23. Multi-Step Mr. Carter owned a ranch with 7\(\frac{1}{4}\) acres. Last year, he bought 3\(\frac{1}{5}\) acres of land from his neighbor. Then he sold 2\(\frac{1}{4}\) acres. How many acres does Mr. Carter own now? (A) 10\(\frac{9}{20}\) acres (B) 8\(\frac{1}{5}\) acres (C) 12\(\frac{7}{10}\) acres (D) 6\(\frac{3}{10}\) acres Answer: B Mr. Carter owned a ranch with 7\(\frac{1}{4}\) acres. Last year, he bought 3\(\frac{1}{5}\) acres of land from his neighbor Then he sold 2\(\frac{1}{4}\) acres. 7\(\frac{1}{4}\) + 3\(\frac{1}{5}\)  – 2\(\frac{1}{4}\) \(\frac{164}{20}\) = 8\(\frac{1}{5}\) acres

Question 24. Multi-Step This week, Maddie worked 2\(\frac{1}{2}\) hours on Monday, 2\(\frac{2}{3}\) hours on Tuesday, and 3\(\frac{1}{4}\) hours on Wednesday. How many more hours will Maddie need to work this week to make her goal of 10\(\frac{1}{2}\) hours a week? (A) 2\(\frac{1}{12}\) hours (B) 8\(\frac{5}{12}\) hours (C) 18\(\frac{11}{12}\) hours (D) 5\(\frac{1}{3}\) hours Answer: A Explanation: This week, Maddie worked 2\(\frac{1}{2}\) hours on Monday, 2\(\frac{2}{3}\) hours on Tuesday, and 3\(\frac{1}{4}\) hours on Wednesday. 2\(\frac{1}{2}\)+2\(\frac{2}{3}\) + 3\(\frac{1}{4}\) -10\(\frac{1}{2}\) \(\frac{25}{12}\) =2\(\frac{1}{12}\) hours

Share this:

Leave a comment cancel reply.

You must be logged in to post a comment.

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Eureka Math Grade 5 Module 6 Lesson 19 Answer Key

Engage ny eureka math 5th grade module 6 lesson 19 answer key, eureka math grade 5 module 6 lesson 19 sprint answer key.

Question 1. \(\frac{2}{4}\) = Answer: \(\frac{2}{4}\) =\(\frac{1}{2}\)

Question 2. \(\frac{2}{6}\) = Answer: \(\frac{2}{6}\) = \(\frac{1}{3}\)

Question 3. \(\frac{2}{8}\) = Answer: \(\frac{2}{8}\) = \(\frac{1}{4}\)

Question 4. \(\frac{5}{10}\) = Answer: \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 5. \(\frac{5}{15}\) = Answer: \(\frac{5}{15}\) = \(\frac{1}{3}\)

Question 6. \(\frac{5}{20}\) = Answer: \(\frac{5}{20}\) = \(\frac{1}{4}\)

Question 7. \(\frac{4}{8}\) = Answer: \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 8. \(\frac{4}{12}\) = Answer: \(\frac{4}{12}\) = \(\frac{1}{3}\)

Question 9. \(\frac{4}{16}\) = Answer: \(\frac{4}{16}\) = \(\frac{1}{4}\)

Question 10. \(\frac{3}{6}\) = Answer: \(\frac{3}{6}\) = \(\frac{1}{2}\) 

Question 11. \(\frac{3}{9}\) = Answer: \(\frac{3}{9}\) = \(\frac{1}{3}\)

Question 12. \(\frac{3}{12}\) = Answer: \(\frac{3}{12}\) = \(\frac{1}{4}\)

Question 13. \(\frac{4}{6}\) = Answer: \(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 14. \(\frac{6}{12}\) = Answer: \(\frac{6}{12}\) = \(\frac{1}{2}\)

Question 15. \(\frac{6}{18}\) = Answer: \(\frac{6}{18}\) = \(\frac{1}{3}\)

Question 16. \(\frac{6}{30}\) = Answer: \(\frac{6}{30}\) = \(\frac{1}{5}\)

Question 17. \(\frac{6}{9}\) = Answer: \(\frac{6}{9}\) = \(\frac{2}{3}\)

Question 18. \(\frac{7}{14}\) = Answer: \(\frac{7}{14}\) = \(\frac{1}{2}\)

Question 19. \(\frac{7}{21}\) = Answer: latex]\frac{7}{21}[/latex] = latex]\frac{1}{3}[/latex]

Question 20. \(\frac{7}{42}\) = Answer: \(\frac{7}{42}\) = \(\frac{1}{6}\)

Question 21. \(\frac{8}{12}\) = Answer: \(\frac{8}{12}\) = \(\frac{2}{3}\)

Question 22. \(\frac{9}{18}\) = Answer: \(\frac{9}{18}\) = \(\frac{1}{2}\)

Question 23. \(\frac{9}{27}\) = Answer: \(\frac{9}{27}\) = \(\frac{1}{3}\)

Question 24. \(\frac{9}{63}\) = Answer: \(\frac{9}{63}\) = \(\frac{1}{7}\)

Question 25. \(\frac{8}{12}\) = Answer: \(\frac{8}{12}\) = \(\frac{2}{3}\)

Question 26. \(\frac{8}{16}\) = Answer: \(\frac{8}{16}\) = \(\frac{1}{2}\)

Question 27. \(\frac{8}{24}\) = Answer: \(\frac{8}{24}\) = \(\frac{1}{3}\)

Question 28. \(\frac{8}{64}\) = Answer: \(\frac{8}{64}\) = \(\frac{1}{8}\)

Question 29. \(\frac{12}{18}\) = Answer: \(\frac{12}{18}\) = \(\frac{2}{3}\)

Question 30. \(\frac{12}{16}\) = Answer: \(\frac{12}{16}\) = \(\frac{3}{4}\)

Question 31. \(\frac{9}{12}\) = Answer: \(\frac{9}{12}\) = \(\frac{3}{4}\)

Question 32. \(\frac{6}{8}\) = Answer: \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 33. \(\frac{10}{12}\) = Answer: \(\frac{10}{12}\) = \(\frac{5}{6}\)

Question 34. \(\frac{15}{18}\) = Answer: \(\frac{15}{18}\) = \(\frac{5}{6}\)

Question 35. \(\frac{8}{10}\) = Answer: \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 36. \(\frac{16}{20}\) = Answer \(\frac{16}{20}\) = \(\frac{4}{5}\)

Question 37. \(\frac{12}{15}\) = Answer: \(\frac{12}{15}\) = \(\frac{4}{5}\)

Question 38. \(\frac{18}{27}\) = Answer: \(\frac{18}{27}\) = \(\frac{6}{9}\)

Question 39. \(\frac{27}{36}\) = Answer: \(\frac{27}{36}\) = \(\frac{3}{4}\)

Question 40. \(\frac{32}{40}\) = Answer: \(\frac{32}{40}\) = \(\frac{4}{5}\)

Question 41. \(\frac{45}{54}\) = Answer: \(\frac{45}{54}\) = \(\frac{5}{6}\)

Question 42. \(\frac{24}{36}\) = Answer: \(\frac{24}{36}\) = \(\frac{6}{9}\)

Question 43. \(\frac{60}{72}\) = Answer: \(\frac{60}{72}\) = \(\frac{5}{6}\)

Question 44. \(\frac{48}{60}\) = Answer: \(\frac{48}{60}\) = \(\frac{4}{5}\)

Question 1. \(\frac{5}{10}\) = Answer: \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 2. \(\frac{5}{15}\) = Answer: \(\frac{5}{15}\) = \(\frac{1}{3}\)

Question 3. \(\frac{5}{20}\) = Answer: \(\frac{5}{20}\) = \(\frac{1}{4}\)

Question 4. \(\frac{2}{4}\) = Answer: \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 5. \(\frac{2}{6}\) = Answer: \(\frac{2}{6}\) = \(\frac{1}{3}\)

Question 6. \(\frac{2}{8}\) = Answer: \(\frac{2}{8}\) = \(\frac{1}{4}\)

Question 7. \(\frac{3}{6}\) = Answer: \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 8. \(\frac{3}{9}\) = Answer: \(\frac{3}{9}\) = \(\frac{1}{3}\)

Question 9. \(\frac{3}{12}\) = Answer: \(\frac{3}{12}\) = \(\frac{1}{4}\)

Question 10. \(\frac{4}{8}\) = Answer: \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 11. \(\frac{4}{12}\) = Answer: \(\frac{4}{12}\) = \(\frac{1}{3}\)

Question 12. \(\frac{4}{16}\) = Answer: \(\frac{4}{16}\) = \(\frac{1}{4}\)

Question 14. \(\frac{7}{14}\) = Answer: \(\frac{7}{14}\) = \(\frac{1}{2}\)

Question 15. \(\frac{7}{21}\) = Answer: \(\frac{7}{21}\) = \(\frac{1}{3}\)

Question 16. \(\frac{7}{35}\) = Answer: \(\frac{7}{35}\) = \(\frac{1}{5}\)

Question 18. \(\frac{6}{12}\) = Answer: \(\frac{6}{12}\) = \(\frac{1}{2}\)

Question 19. \(\frac{6}{18}\) = Answer: \(\frac{6}{18}\) = \(\frac{1}{3}\)

Question 20. \(\frac{6}{36}\) = Answer: \(\frac{6}{36}\) = \(\frac{1}{6}\)

Question 22. \(\frac{8}{16}\) = Answer: \(\frac{8}{16}\) = \(\frac{1}{2}\)

Question 23. \(\frac{8}{24}\) = Answer: \(\frac{8}{24}\) = \(\frac{1}{3}\)

Question 24. \(\frac{8}{56}\) = Answer: \(\frac{8}{56}\) = \(\frac{1}{7}\)

Question 26. \(\frac{9}{18}\) = Answer: \(\frac{9}{18}\) = \(\frac{1}{2}\)

Question 27. \(\frac{9}{27}\) = Answer: \(\frac{9}{27}\) = \(\frac{1}{3}\) =

Question 28. \(\frac{9}{72}\) = Answer: \(\frac{9}{72}\) = \(\frac{1}{8}\)

Question 29. \(\frac{12}{18}\) = Answer: \(\frac{12}{18}\) = \(\frac{4}{6}\)

Question 30. \(\frac{6}{8}\) = Answer: \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 32. \(\frac{12}{16}\) = Answer: \(\frac{12}{16}\) = \(\frac{3}{4}\)

Question 33. \(\frac{8}{10}\) = Answer: \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 34. \(\frac{16}{20}\) = Answer: \(\frac{16}{20}\) = \(\frac{4}{5}\)

Question 35. \(\frac{12}{15}\) = Answer: \(\frac{12}{15}\) = \(\frac{4}{5}\)

Question 36. \(\frac{10}{12}\) = Answer: \(\frac{10}{12}\) = \(\frac{5}{6}\)

Question 37. \(\frac{15}{18}\) = Answer: \(\frac{15}{18}\) = \(\frac{5}{6}\)

Question 38. \(\frac{16}{24}\) = Answer: \(\frac{16}{24}\) = \(\frac{2}{3}\)

Question 39. \(\frac{24}{32}\) = Answer: \(\frac{24}{32}\) = \(\frac{3}{4}\)

Question 40. \(\frac{36}{45}\) = Answer: \(\frac{36}{45}\) = \(\frac{4}{5}\)

Question 41. \(\frac{40}{48}\) = Answer: \(\frac{40}{48}\) = \(\frac{5}{6}\)

Question 42. \(\frac{24}{36}\) = Answer: \(\frac{24}{36}\) = \(\frac{2}{3}\)

Question 43. \(\frac{48}{60}\) = Answer: \(\frac{48}{60}\) = \(\frac{4}{5}\)

Question 44. \(\frac{60}{72}\) = Answer: \(\frac{60}{72}\) = \(\frac{5}{6}\)

Eureka Math Grade 5 Module 6 Lesson 19 Problem Set Answer Key

Eureka Math Grade 5 Module 6 Lesson 19 Exit Ticket Answer Key

Eureka Math Grade 5 Module 6 Lesson 19 Homework Answer Key

Leave a Comment Cancel Reply

You must be logged in to post a comment.

6.1 Exponential Functions

g ( x ) = 0.875 x g ( x ) = 0.875 x and j ( x ) = 1095.6 − 2 x j ( x ) = 1095.6 − 2 x represent exponential functions.

5.5556 5.5556

About 1.548 1.548 billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.

( 0 , 129 ) ( 0 , 129 ) and ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t

f ( x ) = 2 ( 1.5 ) x f ( x ) = 2 ( 1.5 ) x

f ( x ) = 2 ( 2 ) x . f ( x ) = 2 ( 2 ) x . Answers may vary due to round-off error. The answer should be very close to 1.4142 ( 1.4142 ) x . 1.4142 ( 1.4142 ) x .

y ≈ 12 ⋅ 1.85 x y ≈ 12 ⋅ 1.85 x

about $3,644,675.88

e − 0.5 ≈ 0.60653 e − 0.5 ≈ 0.60653

$3,659,823.44

3.77E-26 (This is calculator notation for the number written as 3.77 × 10 − 26 3.77 × 10 − 26 in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

6.2 Graphs of Exponential Functions

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 0 , ∞ ) ; ( 0 , ∞ ) ; the horizontal asymptote is y = 0. y = 0.

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 3 , ∞ ) ; ( 3 , ∞ ) ; the horizontal asymptote is y = 3. y = 3.

x ≈ − 1.608 x ≈ − 1.608

f ( x ) = − 1 3 e x − 2 ; f ( x ) = − 1 3 e x − 2 ; the domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( − ∞ , −2 ) ; ( − ∞ , −2 ) ; the horizontal asymptote is y = −2. y = −2.

6.3 Logarithmic Functions

• ⓐ log 10 ( 1 , 000 , 000 ) = 6 log 10 ( 1 , 000 , 000 ) = 6 is equivalent to 10 6 = 1 , 000 , 000 10 6 = 1 , 000 , 000
• ⓑ log 5 ( 25 ) = 2 log 5 ( 25 ) = 2 is equivalent to 5 2 = 25 5 2 = 25
• ⓐ 3 2 = 9 3 2 = 9 is equivalent to log 3 ( 9 ) = 2 log 3 ( 9 ) = 2
• ⓑ 5 3 = 125 5 3 = 125 is equivalent to log 5 ( 125 ) = 3 log 5 ( 125 ) = 3
• ⓒ 2 − 1 = 1 2 2 − 1 = 1 2 is equivalent to log 2 ( 1 2 ) = − 1 log 2 ( 1 2 ) = − 1

log 121 ( 11 ) = 1 2 log 121 ( 11 ) = 1 2 (recalling that 121 = ( 121 ) 1 2 = 11 121 = ( 121 ) 1 2 = 11 )

log 2 ( 1 32 ) = − 5 log 2 ( 1 32 ) = − 5

log ( 1 , 000 , 000 ) = 6 log ( 1 , 000 , 000 ) = 6

log ( 123 ) ≈ 2.0899 log ( 123 ) ≈ 2.0899

The difference in magnitudes was about 3.929. 3.929.

It is not possible to take the logarithm of a negative number in the set of real numbers.

6.4 Graphs of Logarithmic Functions

( 2 , ∞ ) ( 2 , ∞ )

( 5 , ∞ ) ( 5 , ∞ )

The domain is ( 0 , ∞ ) , ( 0 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

The domain is ( − 4 , ∞ ) , ( − 4 , ∞ ) , the range ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the asymptote x = – 4. x = – 4.

The domain is ( 2 , ∞ ) , ( 2 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 2. x = 2.

The domain is ( − ∞ , 0 ) , ( − ∞ , 0 ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

x ≈ 3.049 x ≈ 3.049

x = 1 x = 1

f ( x ) = 2 ln ( x + 3 ) − 1 f ( x ) = 2 ln ( x + 3 ) − 1

6.5 Logarithmic Properties

log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k

log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 ) log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 )

2 ln x 2 ln x

− 2 ln ( x ) − 2 ln ( x )

log 3 16 log 3 16

2 log x + 3 log y − 4 log z 2 log x + 3 log y − 4 log z

2 3 ln x 2 3 ln x

1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 ) 1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 )

log ( 3 ⋅ 5 4 ⋅ 6 ) ; log ( 3 ⋅ 5 4 ⋅ 6 ) ; can also be written log ( 5 8 ) log ( 5 8 ) by reducing the fraction to lowest terms.

log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) ) log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) )

log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; this answer could also be written log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 . log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 .

The pH increases by about 0.301.

ln 8 ln 0.5 ln 8 ln 0.5

ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861 ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861

6.6 Exponential and Logarithmic Equations

x = − 2 x = − 2

x = − 1 x = − 1

x = 1 2 x = 1 2

The equation has no solution.

x = ln 3 ln ( 2 3 ) x = ln 3 ln ( 2 3 )

t = 2 ln ( 11 3 ) t = 2 ln ( 11 3 ) or ln ( 11 3 ) 2 ln ( 11 3 ) 2

t = ln ( 1 2 ) = − 1 2 ln ( 2 ) t = ln ( 1 2 ) = − 1 2 ln ( 2 )

x = ln 2 x = ln 2

x = e 4 x = e 4

x = e 5 − 1 x = e 5 − 1

x ≈ 9.97 x ≈ 9.97

x = 1 x = 1 or x = − 1 x = − 1

t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years . t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years .

6.7 Exponential and Logarithmic Models

f ( t ) = A 0 e − 0.0000000087 t f ( t ) = A 0 e − 0.0000000087 t

less than 230 years, 229.3157 to be exact

f ( t ) = A 0 e ln 2 3 t f ( t ) = A 0 e ln 2 3 t

6.026 hours

895 cases on day 15

Exponential. y = 2 e 0.5 x . y = 2 e 0.5 x .

y = 3 e ( ln 0.5 ) x y = 3 e ( ln 0.5 ) x

6.8 Fitting Exponential Models to Data

• ⓐ The exponential regression model that fits these data is y = 522.88585984 ( 1.19645256 ) x . y = 522.88585984 ( 1.19645256 ) x .
• ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.
• ⓐ The logarithmic regression model that fits these data is y = 141.91242949 + 10.45366573 ln ( x ) y = 141.91242949 + 10.45366573 ln ( x )
• ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.
• ⓐ The logistic regression model that fits these data is y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x . y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x .
• ⓑ If the population continues to grow at this rate, there will be about 25,634   25,634   seals in 2020.
• ⓒ To the nearest whole number, the carrying capacity is 25,657.

6.1 Section Exercises

Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal .

exponential; the population decreases by a proportional rate. .

not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

The forest represented by the function B ( t ) = 82 ( 1.029 ) t . B ( t ) = 82 ( 1.029 ) t .

After t = 20 t = 20 years, forest A will have 43 43 more trees than forest B.

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

exponential growth; The growth factor, 1.06 , 1.06 , is greater than 1. 1.

exponential decay; The decay factor, 0.97 , 0.97 , is between 0 0 and 1. 1.

f ( x ) = 2000 ( 0.1 ) x f ( x ) = 2000 ( 0.1 ) x

f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x

$ 10 , 250 $ 10 , 250

$ 13 , 268.58 $ 13 , 268.58

P = A ( t ) ⋅ ( 1 + r n ) − n t P = A ( t ) ⋅ ( 1 + r n ) − n t

$ 4,572.56 $ 4,572.56

continuous growth; the growth rate is greater than 0. 0.

continuous decay; the growth rate is less than 0. 0.

$ 669.42 $ 669.42

f ( − 1 ) = − 4 f ( − 1 ) = − 4

f ( − 1 ) ≈ − 0.2707 f ( − 1 ) ≈ − 0.2707

f ( 3 ) ≈ 483.8146 f ( 3 ) ≈ 483.8146

y = 3 ⋅ 5 x y = 3 ⋅ 5 x

y ≈ 18 ⋅ 1.025 x y ≈ 18 ⋅ 1.025 x

y ≈ 0.2 ⋅ 1.95 x y ≈ 0.2 ⋅ 1.95 x

APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; I ( n ) = ( 1 + r n ) n − 1 I ( n ) = ( 1 + r n ) n − 1

Let f f be the exponential decay function f ( x ) = a ⋅ ( 1 b ) x f ( x ) = a ⋅ ( 1 b ) x such that b > 1. b > 1. Then for some number n > 0 , n > 0 , f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x . f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x .

47 , 622 47 , 622 fox

1.39 % ; 1.39 % ; $ 155 , 368.09 $ 155 , 368.09

$ 35 , 838.76 $ 35 , 838.76

$ 82 , 247.78 ; $ 82 , 247.78 ; $ 449.75 $ 449.75

6.2 Section Exercises

An asymptote is a line that the graph of a function approaches, as x x either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

g ( x ) = 4 ( 3 ) − x ; g ( x ) = 4 ( 3 ) − x ; y -intercept: ( 0 , 4 ) ; ( 0 , 4 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

g ( x ) = − 10 x + 7 ; g ( x ) = − 10 x + 7 ; y -intercept: ( 0 , 6 ) ; ( 0 , 6 ) ; Domain: all real numbers; Range: all real numbers less than 7. 7.

g ( x ) = 2 ( 1 4 ) x ; g ( x ) = 2 ( 1 4 ) x ; y -intercept: ( 0 , 2 ) ; ( 0 , 2 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

y -intercept: ( 0 , − 2 ) ( 0 , − 2 )

Horizontal asymptote: h ( x ) = 3 ; h ( x ) = 3 ; Domain: all real numbers; Range: all real numbers strictly greater than 3. 3.

As x → ∞ x → ∞ , f ( x ) → − ∞ f ( x ) → − ∞ ; As x → − ∞ x → − ∞ , f ( x ) → − 1 f ( x ) → − 1

As x → ∞ x → ∞ , f ( x ) → 2 f ( x ) → 2 ; As x → − ∞ x → − ∞ , f ( x ) → ∞ f ( x ) → ∞

f ( x ) = 4 x − 3 f ( x ) = 4 x − 3

f ( x ) = 4 x − 5 f ( x ) = 4 x − 5

f ( x ) = 4 − x f ( x ) = 4 − x

y = − 2 x + 3 y = − 2 x + 3

y = − 2 ( 3 ) x + 7 y = − 2 ( 3 ) x + 7

g ( 6 ) = 800 + 1 3 ≈ 800.3333 g ( 6 ) = 800 + 1 3 ≈ 800.3333

h ( − 7 ) = − 58 h ( − 7 ) = − 58

x ≈ − 2.953 x ≈ − 2.953

x ≈ − 0.222 x ≈ − 0.222

The graph of G ( x ) = ( 1 b ) x G ( x ) = ( 1 b ) x is the refelction about the y -axis of the graph of F ( x ) = b x ; F ( x ) = b x ; For any real number b > 0 b > 0 and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b ) x ( 1 b ) x is the the reflection about the y -axis, F ( − x ) . F ( − x ) .

The graphs of g ( x ) g ( x ) and h ( x ) h ( x ) are the same and are a horizontal shift to the right of the graph of f ( x ) ; f ( x ) ; For any real number n , real number b > 0 , b > 0 , and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b n ) b x ( 1 b n ) b x is the horizontal shift f ( x − n ) . f ( x − n ) .

6.3 Section Exercises

A logarithm is an exponent. Specifically, it is the exponent to which a base b b is raised to produce a given value. In the expressions given, the base b b has the same value. The exponent, y , y , in the expression b y b y can also be written as the logarithm, log b x , log b x , and the value of x x is the result of raising b b to the power of y . y .

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation b y = x , b y = x , and then properties of exponents can be applied to solve for x . x .

The natural logarithm is a special case of the logarithm with base b b in that the natural log always has base e . e . Rather than notating the natural logarithm as log e ( x ) , log e ( x ) , the notation used is ln ( x ) . ln ( x ) .

a c = b a c = b

x y = 64 x y = 64

15 b = a 15 b = a

13 a = 142 13 a = 142

e n = w e n = w

log c ( k ) = d log c ( k ) = d

log 19 y = x log 19 y = x

log n ( 103 ) = 4 log n ( 103 ) = 4

log y ( 39 100 ) = x log y ( 39 100 ) = x

ln ( h ) = k ln ( h ) = k

x = 2 − 3 = 1 8 x = 2 − 3 = 1 8

x = 3 3 = 27 x = 3 3 = 27

x = 9 1 2 = 3 x = 9 1 2 = 3

x = 6 − 3 = 1 216 x = 6 − 3 = 1 216

x = e 2 x = e 2

14.125 14.125

2 . 7 0 8 2 . 7 0 8

0.151 0.151

No, the function has no defined value for x = 0. x = 0. To verify, suppose x = 0 x = 0 is in the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) . Then there is some number n n such that n = log ( 0 ) . n = log ( 0 ) . Rewriting as an exponential equation gives: 10 n = 0 , 10 n = 0 , which is impossible since no such real number n n exists. Therefore, x = 0 x = 0 is not the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) .

Yes. Suppose there exists a real number x x such that ln x = 2. ln x = 2. Rewriting as an exponential equation gives x = e 2 , x = e 2 , which is a real number. To verify, let x = e 2 . x = e 2 . Then, by definition, ln ( x ) = ln ( e 2 ) = 2. ln ( x ) = ln ( e 2 ) = 2.

No; ln ( 1 ) = 0 , ln ( 1 ) = 0 , so ln ( e 1.725 ) ln ( 1 ) ln ( e 1.725 ) ln ( 1 ) is undefined.

6.4 Section Exercises

Since the functions are inverses, their graphs are mirror images about the line y = x . y = x . So for every point ( a , b ) ( a , b ) on the graph of a logarithmic function, there is a corresponding point ( b , a ) ( b , a ) on the graph of its inverse exponential function.

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

Domain: ( − ∞ , 1 2 ) ; ( − ∞ , 1 2 ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( − 17 4 , ∞ ) ; ( − 17 4 , ∞ ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( 5 , ∞ ) ; ( 5 , ∞ ) ; Vertical asymptote: x = 5 x = 5

Domain: ( − 1 3 , ∞ ) ; ( − 1 3 , ∞ ) ; Vertical asymptote: x = − 1 3 x = − 1 3

Domain: ( − 3 , ∞ ) ; ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3

Domain: ( 3 7 , ∞ ) ( 3 7 , ∞ ) ; Vertical asymptote: x = 3 7 x = 3 7 ; End behavior: as x → ( 3 7 ) + , f ( x ) → − ∞ x → ( 3 7 ) + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ x → ∞ , f ( x ) → ∞

Domain: ( − 3 , ∞ ) ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3 ; End behavior: as x → − 3 + x → − 3 + , f ( x ) → − ∞ f ( x ) → − ∞ and as x → ∞ x → ∞ , f ( x ) → ∞ f ( x ) → ∞

Domain: ( 1 , ∞ ) ; ( 1 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 1 ; x = 1 ; x -intercept: ( 5 4 , 0 ) ; ( 5 4 , 0 ) ; y -intercept: DNE

Domain: ( − ∞ , 0 ) ; ( − ∞ , 0 ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( − e 2 , 0 ) ; ( − e 2 , 0 ) ; y -intercept: DNE

Domain: ( 0 , ∞ ) ; ( 0 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( e 3 , 0 ) ; ( e 3 , 0 ) ; y -intercept: DNE

f ( x ) = log 2 ( − ( x − 1 ) ) f ( x ) = log 2 ( − ( x − 1 ) )

f ( x ) = 3 log 4 ( x + 2 ) f ( x ) = 3 log 4 ( x + 2 )

x = 2 x = 2

x ≈ 2 .303 x ≈ 2 .303

x ≈ − 0.472 x ≈ − 0.472

The graphs of f ( x ) = log 1 2 ( x ) f ( x ) = log 1 2 ( x ) and g ( x ) = − log 2 ( x ) g ( x ) = − log 2 ( x ) appear to be the same; Conjecture: for any positive base b ≠ 1 , b ≠ 1 , log b ( x ) = − log 1 b ( x ) . log b ( x ) = − log 1 b ( x ) .

Recall that the argument of a logarithmic function must be positive, so we determine where x + 2 x − 4 > 0 x + 2 x − 4 > 0 . From the graph of the function f ( x ) = x + 2 x − 4 , f ( x ) = x + 2 x − 4 , note that the graph lies above the x -axis on the interval ( − ∞ , − 2 ) ( − ∞ , − 2 ) and again to the right of the vertical asymptote, that is ( 4 , ∞ ) . ( 4 , ∞ ) . Therefore, the domain is ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) . ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) .

6.5 Section Exercises

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, log b ( x 1 n ) = 1 n log b ( x ) . log b ( x 1 n ) = 1 n log b ( x ) .

log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y ) log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y )

log b ( 13 ) − log b ( 17 ) log b ( 13 ) − log b ( 17 )

− k ln ( 4 ) − k ln ( 4 )

ln ( 7 x y ) ln ( 7 x y )

log b ( 4 ) log b ( 4 )

log b ( 7 ) log b ( 7 )

15 log ( x ) + 13 log ( y ) − 19 log ( z ) 15 log ( x ) + 13 log ( y ) − 19 log ( z )

3 2 log ( x ) − 2 log ( y ) 3 2 log ( x ) − 2 log ( y )

8 3 log ( x ) + 14 3 log ( y ) 8 3 log ( x ) + 14 3 log ( y )

ln ( 2 x 7 ) ln ( 2 x 7 )

log ( x z 3 y ) log ( x z 3 y )

log 7 ( 15 ) = ln ( 15 ) ln ( 7 ) log 7 ( 15 ) = ln ( 15 ) ln ( 7 )

log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b

log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1 log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1

2.81359 2.81359

0.93913 0.93913

− 2.23266 − 2.23266

x = 4 ; x = 4 ; By the quotient rule: log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1. log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1.

Rewriting as an exponential equation and solving for x : x :

6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4 6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4

Checking, we find that log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) is defined, so x = 4. x = 4.

Let b b and n n be positive integers greater than 1. 1. Then, by the change-of-base formula, log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) . log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) .

6.6 Section Exercises

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

x = − 1 3 x = − 1 3

n = − 1 n = − 1

b = 6 5 b = 6 5

x = 10 x = 10

No solution

p = log ( 17 8 ) − 7 p = log ( 17 8 ) − 7

k = − ln ( 38 ) 3 k = − ln ( 38 ) 3

x = ln ( 38 3 ) − 8 9 x = ln ( 38 3 ) − 8 9

x = ln 12 x = ln 12

x = ln ( 3 5 ) − 3 8 x = ln ( 3 5 ) − 3 8

no solution

x = ln ( 3 ) x = ln ( 3 )

10 − 2 = 1 100 10 − 2 = 1 100

n = 49 n = 49

k = 1 36 k = 1 36

x = 9 − e 8 x = 9 − e 8

n = 1 n = 1

x = ± 10 3 x = ± 10 3

x = 0 x = 0

x = 3 4 x = 3 4

x = 9 x = 9

x = e 2 3 ≈ 2.5 x = e 2 3 ≈ 2.5

x = − 5 x = − 5

x = e + 10 4 ≈ 3.2 x = e + 10 4 ≈ 3.2

x = 11 5 ≈ 2.2 x = 11 5 ≈ 2.2

x = 101 11 ≈ 9.2 x = 101 11 ≈ 9.2

about $ 27 , 710.24 $ 27 , 710.24

about 5 years

ln ( 17 ) 5 ≈ 0.567 ln ( 17 ) 5 ≈ 0.567

x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078 x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078

x ≈ 2.2401 x ≈ 2.2401

x ≈ − 44655 . 7143 x ≈ − 44655 . 7143

about 5.83 5.83

t = ln ( ( y A ) 1 k ) t = ln ( ( y A ) 1 k )

t = ln ( ( T − T s T 0 − T s ) − 1 k ) t = ln ( ( T − T s T 0 − T s ) − 1 k )

6.7 Section Exercises

Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 10 2 10 2 times, or 2 orders of magnitude greater, than the mass of Earth.

f ( 0 ) ≈ 16.7 ; f ( 0 ) ≈ 16.7 ; The amount initially present is about 16.7 units.

exponential; f ( x ) = 1.2 x f ( x ) = 1.2 x

logarithmic

about 1.4 1.4 years

about 7.3 7.3 years

4 4 half-lives; 8.18 8.18 minutes

M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2 M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2

Let y = b x y = b x for some non-negative real number b b such that b ≠ 1. b ≠ 1. Then,

ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b ) ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b )

A = 125 e ( − 0.3567 t ) ; A ≈ 43 A = 125 e ( − 0.3567 t ) ; A ≈ 43 mg

about 60 60 days

A ( t ) = 250 e ( − 0.00822 t ) ; A ( t ) = 250 e ( − 0.00822 t ) ; half-life: about 84 84 minutes

r ≈ − 0.0667 , r ≈ − 0.0667 , So the hourly decay rate is about 6.67 % 6.67 %

f ( t ) = 1350 e ( 0.03466 t ) ; f ( t ) = 1350 e ( 0.03466 t ) ; after 3 hours: P ( 180 ) ≈ 691 , 200 P ( 180 ) ≈ 691 , 200

f ( t ) = 256 e ( 0.068110 t ) ; f ( t ) = 256 e ( 0.068110 t ) ; doubling time: about 10 10 minutes

about 88 88 minutes

T ( t ) = 90 e ( − 0.008377 t ) + 75 , T ( t ) = 90 e ( − 0.008377 t ) + 75 , where t t is in minutes.

about 113 113 minutes

log ( x ) = 1.5 ; x ≈ 31.623 log ( x ) = 1.5 ; x ≈ 31.623

MMS magnitude: 5.82 5.82

N ( 3 ) ≈ 71 N ( 3 ) ≈ 71

6.8 Section Exercises

Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

The y -intercept on the graph of a logistic equation corresponds to the initial population for the population model.

P ( 0 ) = 22 P ( 0 ) = 22 ; 175

p ≈ 2.67 p ≈ 2.67

y -intercept: ( 0 , 15 ) ( 0 , 15 )

about 6.8 6.8 months.

About 38 wolves

About 8.7 years

f ( x ) = 776.682 ( 1.426 ) x f ( x ) = 776.682 ( 1.426 ) x

f ( x ) = 731.92 e -0.3038 x f ( x ) = 731.92 e -0.3038 x

When f ( x ) = 250 , x ≈ 3.6 f ( x ) = 250 , x ≈ 3.6

y = 5.063 + 1.934 log ( x ) y = 5.063 + 1.934 log ( x )

When f ( 10 ) ≈ 2.3 f ( 10 ) ≈ 2.3

When f ( x ) = 8 , x ≈ 0.82 f ( x ) = 8 , x ≈ 0.82

f ( x ) = 25.081 1 + 3.182 e − 0.545 x f ( x ) = 25.081 1 + 3.182 e − 0.545 x

When f ( x ) = 68 , x ≈ 4.9 f ( x ) = 68 , x ≈ 4.9

f ( x ) = 1.034341 ( 1.281204 ) x f ( x ) = 1.034341 ( 1.281204 ) x ; g ( x ) = 4.035510 g ( x ) = 4.035510 ; the regression curves are symmetrical about y = x y = x , so it appears that they are inverse functions.

f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b

Review Exercises

exponential decay; The growth factor, 0.825 , 0.825 , is between 0 0 and 1. 1.

y = 0.25 ( 3 ) x y = 0.25 ( 3 ) x

$ 42 , 888.18 $ 42 , 888.18

continuous decay; the growth rate is negative.

domain: all real numbers; range: all real numbers strictly greater than zero; y -intercept: (0, 3.5);

g ( x ) = 7 ( 6.5 ) − x ; g ( x ) = 7 ( 6.5 ) − x ; y -intercept: ( 0 , 7 ) ; ( 0 , 7 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

17 x = 4913 17 x = 4913

log a b = − 2 5 log a b = − 2 5

x = 64 1 3 = 4 x = 64 1 3 = 4

log ( 0 .000001 ) = − 6 log ( 0 .000001 ) = − 6

ln ( e − 0.8648 ) = − 0.8648 ln ( e − 0.8648 ) = − 0.8648

Domain: x > − 5 ; x > − 5 ; Vertical asymptote: x = − 5 ; x = − 5 ; End behavior: as x → − 5 + , f ( x ) → − ∞ x → − 5 + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ . x → ∞ , f ( x ) → ∞ .

log 8 ( 65 x y ) log 8 ( 65 x y )

ln ( z x y ) ln ( z x y )

log y ( 12 ) log y ( 12 )

ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2 ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2

log 7 ( v 3 w 6 u 3 ) log 7 ( v 3 w 6 u 3 )

x = log ( 125 ) log ( 5 ) + 17 12 = 5 3 x = log ( 125 ) log ( 5 ) + 17 12 = 5 3

x = − 3 x = − 3

x = ln ( 11 ) x = ln ( 11 )

a = e 4 − 3 a = e 4 − 3

x = ± 9 5 x = ± 9 5

about 5.45 5.45 years

f − 1 ( x ) = 2 4 x − 1 3 f − 1 ( x ) = 2 4 x − 1 3

f ( t ) = 300 ( 0.83 ) t ; f ( t ) = 300 ( 0.83 ) t ; f ( 24 ) ≈ 3.43     g f ( 24 ) ≈ 3.43     g

about 45 45 minutes

about 8.5 8.5 days

exponential

y = 4 ( 0.2 ) x ; y = 4 ( 0.2 ) x ; y = 4 e -1.609438 x y = 4 e -1.609438 x

about 7.2 7.2 days

logarithmic; y = 16.68718 − 9.71860 ln ( x ) y = 16.68718 − 9.71860 ln ( x )

Practice Test

About 13 dolphins.

$ 1,947 $ 1,947

y -intercept: ( 0 , 5 ) ( 0 , 5 )

8.5 a = 614.125 8.5 a = 614.125

x = ( 1 7 ) 2 = 1 49 x = ( 1 7 ) 2 = 1 49

ln ( 0.716 ) ≈ − 0.334 ln ( 0.716 ) ≈ − 0.334

Domain: x < 3 ; x < 3 ; Vertical asymptote: x = 3 ; x = 3 ; End behavior: x → 3 − , f ( x ) → − ∞ x → 3 − , f ( x ) → − ∞ and x → − ∞ , f ( x ) → ∞ x → − ∞ , f ( x ) → ∞

log t ( 12 ) log t ( 12 )

3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3 3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3

x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497 x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497

a = ln ( 4 ) + 8 10 a = ln ( 4 ) + 8 10

x = ln ( 9 ) x = ln ( 9 )

x = ± 3 3 2 x = ± 3 3 2

f ( t ) = 112 e − .019792 t ; f ( t ) = 112 e − .019792 t ; half-life: about 35 35 days

T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F

exponential; y = 15.10062 ( 1.24621 ) x y = 15.10062 ( 1.24621 ) x

logistic; y = 18.41659 1 + 7.54644 e − 0.68375 x y = 18.41659 1 + 7.54644 e − 0.68375 x

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Algebra and Trigonometry
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-6

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Home > CC2 > Chapter 5 > Lesson 5.2.6

Lesson 5.1.1, lesson 5.1.2, lesson 5.2.1, lesson 5.2.2, lesson 5.2.3, lesson 5.2.4, lesson 5.2.5, lesson 5.2.6, lesson 5.3.1, lesson 5.3.2, lesson 5.3.3, lesson 5.3.4, lesson 5.3.5.

© 2022 CPM Educational Program. All rights reserved.