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Five-Minute Check (over Lesson 7–6) CCSS Then/Now New Vocabulary Key Concept: Natural Base Functions Example 1: Write Equivalent Expressions Example 2: Write Equivalent Expressions Example 3: Simplify Expressions with e and the Natural Log Example 4: Solve Base e Equations Example 5: Solve Natural Log Equations and Inequalities Key Concept: Continuously Compounded Interest Example 6: Real-World Example: Solve Base e Inequalities Lesson Menu

Use a calculator to evaluate log 3.4 to the nearest ten-thousandth. B. 1.5314 C. 1.2238 D. 29.9641 5-Minute Check 1

Use a calculator to evaluate log 3.4 to the nearest ten-thousandth. B. 1.5314 C. 1.2238 D. 29.9641 5-Minute Check 1

Solve 2x – 4 = 14. Round to the nearest ten-thousandth. B. 7.0164 C. 7.8074 D. 9.2381 5-Minute Check 2

Solve 2x – 4 = 14. Round to the nearest ten-thousandth. B. 7.0164 C. 7.8074 D. 9.2381 5-Minute Check 2

Solve 42p – 1 > 11p. Round to the nearest ten-thousandth. A. {p | p = 4} B. {p | p > 3.6998} C. {p | p < 3.4679} D. {p | p > 2.5713} 5-Minute Check 3

Solve 42p – 1 > 11p. Round to the nearest ten-thousandth. A. {p | p = 4} B. {p | p > 3.6998} C. {p | p < 3.4679} D. {p | p > 2.5713} 5-Minute Check 3

Express log4 (2. 2)3 in terms of common logarithms Express log4 (2.2)3 in terms of common logarithms. Then approximate its value to four decimal places. A. –3.4829 B. 1.5 C. 1.6845 D. 1.7063 5-Minute Check 4

Express log4 (2. 2)3 in terms of common logarithms Express log4 (2.2)3 in terms of common logarithms. Then approximate its value to four decimal places. A. –3.4829 B. 1.5 C. 1.6845 D. 1.7063 5-Minute Check 4

Solve for x: 92x = 45. A. B. C. D. x = 2 log 5 5-Minute Check 5

Solve for x: 92x = 45. A. B. C. D. x = 2 log 5 5-Minute Check 5

Mathematical Practices 7 Look for and make use of structure. Content Standards A.SSE.2 Use the structure of an expression to identify ways to rewrite it. Mathematical Practices 7 Look for and make use of structure. CCSS

You worked with common logarithms. Evaluate expressions involving the natural base and natural logarithm. Solve exponential equations and inequalities using natural logarithms. Then/Now

natural base exponential function natural logarithm Vocabulary

Concept

A. Write an equivalent logarithmic equation for ex = 23. Write Equivalent Expressions A. Write an equivalent logarithmic equation for ex = 23. ex = 23 → loge 23 = x ln 23 = x Answer: Example 1

A. Write an equivalent logarithmic equation for ex = 23. Write Equivalent Expressions A. Write an equivalent logarithmic equation for ex = 23. ex = 23 → loge 23 = x ln 23 = x Answer: ln 23 = x Example 1

B. Write an equivalent logarithmic equation for e4 = x. Write Equivalent Expressions B. Write an equivalent logarithmic equation for e4 = x. e4 = x → loge x = 4 ln x = 4 Answer: Example 1

B. Write an equivalent logarithmic equation for e4 = x. Write Equivalent Expressions B. Write an equivalent logarithmic equation for e4 = x. e4 = x → loge x = 4 ln x = 4 Answer: ln x = 4 Example 1

A. What is ex = 15 in logarithmic form? A. ln e = 15 B. ln 15 = e C. ln x = 15 D. ln 15 = x Example 1

A. What is ex = 15 in logarithmic form? A. ln e = 15 B. ln 15 = e C. ln x = 15 D. ln 15 = x Example 1

B. What is e4 = x in logarithmic form? A. ln e = 4 B. ln x = 4 C. ln x = e D. ln 4 = x Example 1

B. What is e4 = x in logarithmic form? A. ln e = 4 B. ln x = 4 C. ln x = e D. ln 4 = x Example 1

A. Write ln x ≈ 1.2528 in exponential form. Write Equivalent Expressions A. Write ln x ≈ 1.2528 in exponential form. ln x ≈ 1.2528 → loge x = 1.2528 x ≈ e1.2528 Answer: Example 2

A. Write ln x ≈ 1.2528 in exponential form. Write Equivalent Expressions A. Write ln x ≈ 1.2528 in exponential form. ln x ≈ 1.2528 → loge x = 1.2528 x ≈ e1.2528 Answer: x ≈ e1.2528 Example 2

B. Write ln 25 = x in exponential form. Write Equivalent Expressions B. Write ln 25 = x in exponential form. ln 25 = x → loge 25 = x 25 = ex Answer: Example 2

B. Write ln 25 = x in exponential form. Write Equivalent Expressions B. Write ln 25 = x in exponential form. ln 25 = x → loge 25 = x 25 = ex Answer: 25 = ex Example 2

A. Write ln x ≈ 1.5763 in exponential form. A. x ≈ 1.5763e B. x ≈ e1.5763 C. e ≈ x1.5763 D. e ≈ 1.5763x Example 2

A. Write ln x ≈ 1.5763 in exponential form. A. x ≈ 1.5763e B. x ≈ e1.5763 C. e ≈ x1.5763 D. e ≈ 1.5763x Example 2

B. Write ln 47 = x in exponential form. A. 47 = ex B. e = 47x C. x = 47e D. 47 = xe Example 2

B. Write ln 47 = x in exponential form. A. 47 = ex B. e = 47x C. x = 47e D. 47 = xe Example 2

A. Write 4 ln 3 + ln 6 as a single algorithm. Simplify Expressions with e and the Natural Log A. Write 4 ln 3 + ln 6 as a single algorithm. 4 ln 3 + ln 6 = ln 34 + ln 6 Power Property of Logarithms = ln (34 ● 6) Product Property of Logarithms = ln 486 Simplify. Answer: Example 3

A. Write 4 ln 3 + ln 6 as a single algorithm. Simplify Expressions with e and the Natural Log A. Write 4 ln 3 + ln 6 as a single algorithm. 4 ln 3 + ln 6 = ln 34 + ln 6 Power Property of Logarithms = ln (34 ● 6) Product Property of Logarithms = ln 486 Simplify. Answer: ln 486 Example 3

Check Use a calculator to verify the solution. Simplify Expressions with e and the Natural Log Check Use a calculator to verify the solution. 4 3 6 LN ENTER ) + Keystrokes: 486 6.1862  LN ENTER ) Keystrokes: Example 3

B. Write 2 ln 3 + ln 4 + ln y as a single algorithm. Simplify Expressions with e and the Natural Log B. Write 2 ln 3 + ln 4 + ln y as a single algorithm. 2 ln 3 + ln 4 + ln y = ln 32 + ln 4 + ln y Power Property of Logarithms = ln (32 ● 4 ● y) Product Property of Logarithms = ln 36y Simplify. Answer: Example 3

B. Write 2 ln 3 + ln 4 + ln y as a single algorithm. Simplify Expressions with e and the Natural Log B. Write 2 ln 3 + ln 4 + ln y as a single algorithm. 2 ln 3 + ln 4 + ln y = ln 32 + ln 4 + ln y Power Property of Logarithms = ln (32 ● 4 ● y) Product Property of Logarithms = ln 36y Simplify. Answer: ln 36y Example 3

A. Write 4 ln 2 + In 3 as a single logarithm. A. ln 6 B. ln 24 C. ln 32 D. ln 48 Example 3

A. Write 4 ln 2 + In 3 as a single logarithm. A. ln 6 B. ln 24 C. ln 32 D. ln 48 Example 3

B. Write 3 ln 3 + ln + ln x as a single logarithm. 1 3 __ 1 3 A. ln 3x B. ln 9x C. ln 18x D. ln 27x Example 3

B. Write 3 ln 3 + ln + ln x as a single logarithm. 1 3 __ 1 3 A. ln 3x B. ln 9x C. ln 18x D. ln 27x Example 3

Solve 3e–2x + 4 = 10. Round to the nearest ten-thousandth. Solve Base e Equations Solve 3e–2x + 4 = 10. Round to the nearest ten-thousandth. 3e–2x + 4 = 10 Original equation 3e–2x = 6 Subtract 4 from each side. e–2x = 2 Divide each side by 3. ln e–2x = ln 2 Property of Equality for Logarithms –2x = ln 2 Inverse Property of Exponents and Logarithms Divide each side by –2. Example 4

Solve Base e Equations x ≈ –0.3466 Use a calculator. Answer: Example 4

Answer: The solution is about –0.3466. Solve Base e Equations x ≈ –0.3466 Use a calculator. Answer: The solution is about –0.3466. Example 4

What is the solution to the equation 2e–2x + 5 = 15? B. –0.6931 C. 0.6931 D. 0.8047 Example 4

What is the solution to the equation 2e–2x + 5 = 15? B. –0.6931 C. 0.6931 D. 0.8047 Example 4

A. Solve 2 ln 5x = 6. Round to the nearest ten-thousandth. Solve Natural Log Equations and Inequalities A. Solve 2 ln 5x = 6. Round to the nearest ten-thousandth. 2 ln 5x = 6 Original equation ln 5x = 3 Divide each side by 2. eln 5x = e3 Property of Equality for Exponential Functions 5x = e3 eln x = x Divide each side by 5. x ≈ 4.0171 Use a calculator. Answer: Example 5

A. Solve 2 ln 5x = 6. Round to the nearest ten-thousandth. Solve Natural Log Equations and Inequalities A. Solve 2 ln 5x = 6. Round to the nearest ten-thousandth. 2 ln 5x = 6 Original equation ln 5x = 3 Divide each side by 2. eln 5x = e3 Property of Equality for Exponential Functions 5x = e3 eln x = x Divide each side by 5. x ≈ 4.0171 Use a calculator. Answer: about 4.0171 Example 5

ln (3x + 1)2 > 8 Original equation Solve Natural Log Equations and Inequalities B. Solve the inequality ln (3x + 1)2 > 8. Round to the nearest ten-thousandth. ln (3x + 1)2 > 8 Original equation eln (3x + 1)2 > e8 Write each side using exponents and base e. (3x + 1)2 > (e4)2 eln x = x and Power of of Power 3x + 1 > e4 Property of Inequality for Exponential Functions 3x > e4 – 1 Subtract 1 from each side. Example 5

Divide each side by 3. x > 17.8661 Use a calculator. Answer: Solve Natural Log Equations and Inequalities Divide each side by 3. x > 17.8661 Use a calculator. Answer: Example 5

Divide each side by 3. x > 17.8661 Use a calculator. Solve Natural Log Equations and Inequalities Divide each side by 3. x > 17.8661 Use a calculator. Answer: x > 17.8661 Example 5

A. Solve the equation 3 ln 6x = 12. Round to the nearest ten-thousandth. B. 8.0349 C. 9.0997 D. 11.232 Example 5

A. Solve the equation 3 ln 6x = 12. Round to the nearest ten-thousandth. B. 8.0349 C. 9.0997 D. 11.232 Example 5

B. Solve the inequality in (4x – 2) > 7 B. Solve the inequality in (4x – 2) > 7. Round to the nearest ten-thousandth. A. x > 274.66 B. x > 282.84 C. x > 286.91 D. x < 294.85 Example 5

B. Solve the inequality in (4x – 2) > 7 B. Solve the inequality in (4x – 2) > 7. Round to the nearest ten-thousandth. A. x > 274.66 B. x > 282.84 C. x > 286.91 D. x < 294.85 Example 5

Concept

A = Pert Continuously Compounded Interest formula Solve Base e Inequalities A. SAVINGS Suppose you deposit $700 into an account paying 3% annual interest, compounded continuously. What is the balance after 8 years? A = Pert Continuously Compounded Interest formula = 700e(0.03)(8) Replace P with 700, r with 0.03 and t with 8. = 700e0.24 Simplify. ≈ 889.87 Use a calculator. Answer: Example 6

A = Pert Continuously Compounded Interest formula Solve Base e Inequalities A. SAVINGS Suppose you deposit $700 into an account paying 3% annual interest, compounded continuously. What is the balance after 8 years? A = Pert Continuously Compounded Interest formula = 700e(0.03)(8) Replace P with 700, r with 0.03 and t with 8. = 700e0.24 Simplify. ≈ 889.87 Use a calculator. Answer: The balance after 8 years will be $889.87. Example 6

The balance is at least $1200. Solve Base e Inequalities B. SAVINGS Suppose you deposit $700 into an account paying 3% annual interest, compounded continuously. How long will it take for the balance in your account to reach at least $1200? The balance is at least $1200. A ≥ 1200 Write an inequality. Replace A with 700e(0.03)t. Divide each side by 700. Example 6

Property of Inequality for Logarithms Solve Base e Inequalities Property of Inequality for Logarithms Inverse Property of Exponents and Logarithms Divide each side by 0.03. t ≥ 17.97 Use a calculator. Answer: Example 6

Property of Inequality for Logarithms Solve Base e Inequalities Property of Inequality for Logarithms Inverse Property of Exponents and Logarithms Divide each side by 0.03. t ≥ 17.97 Use a calculator. Answer: It will take about 18 years for the balance to reach at least $1200. Example 6

A = Pert Continuously Compounded Interest formula Solve Base e Inequalities C. SAVINGS Suppose you deposit $700 into an account paying 3% annual interest, compounded continuously. How much would have to be deposited in order to reach a balance of $1500 after 12 years? A = Pert Continuously Compounded Interest formula 1500 = P ● e0.03 ● 12 A = 1500, r = 0.003, and t = 12 Divide each side by e0.36. Example 6

1046.51 ≈ P Use a calculator. Answer: Solve Base e Inequalities Example 6

Answer: You need to deposit $1046.51. Solve Base e Inequalities 1046.51 ≈ P Use a calculator. Answer: You need to deposit $1046.51. Example 6

A. SAVINGS Suppose you deposit $700 into an account paying 6% annual interest, compounded continuously. What is the balance after 7 years? A. $46,058.59 B. $46,680.43 C. $1065.37 D. $365.37 Example 6

A. SAVINGS Suppose you deposit $700 into an account paying 6% annual interest, compounded continuously. What is the balance after 7 years? A. $46,058.59 B. $46,680.43 C. $1065.37 D. $365.37 Example 6

B. SAVINGS Suppose you deposit $700 into an account paying 6% annual interest, compounded continuously. How long will it take for the balance in your account to reach at least $2500? A. at least 1.27 years B. at least 7.50 years C. at least 21.22 years D. at least 124.93 years Example 6

B. SAVINGS Suppose you deposit $700 into an account paying 6% annual interest, compounded continuously. How long will it take for the balance in your account to reach at least $2500? A. at least 1.27 years B. at least 7.50 years C. at least 21.22 years D. at least 124.93 years Example 6

C. SAVINGS Suppose you deposit money into an account paying 3% annual interest, compounded continuously. How much would have to be deposited in order to reach a balance of $1950 after 10 years? A. $1299.43 B. $1332.75 C. $1365.87 D. $1444.60 Example 6

C. SAVINGS Suppose you deposit money into an account paying 3% annual interest, compounded continuously. How much would have to be deposited in order to reach a balance of $1950 after 10 years? A. $1299.43 B. $1332.75 C. $1365.87 D. $1444.60 Example 6

End of the Lesson