1. Algebra 2 Write 2 ln 12 – ln 9 as a single natural logarithm. 2 ln 12 – ln 9 = ln 12 2 – ln 9Power Property = lnQuotient Property 12 2 9 = ln 16Simplify. Lesson 8-6 Natural Logarithms Additional Examples

2. Algebra 2 Find the velocity of a spacecraft whose booster rocket has a mass ratio 22, an exhaust velocity of 2.3 km/s, and a firing time of 50 s. Can the spacecraft achieve a stable orbit 300 km above Earth? Let R = 22, c = 2.3, and t = 50.Find v. v = –0.0098t + c ln RUse the formula. = –0.0098(50) + 2.3 ln 22Substitute. –0.49 + 2.3(3.091)Use a calculator. 6.62Simplify. The velocity is 6.6 km/s is less than the 7.7 km/s needed for a stable orbit. Therefore, the spacecraft cannot achieve a stable orbit at 300 km above Earth. Lesson 8-6 Natural Logarithms Additional Examples

3. Algebra 2 Solve ln (2x – 4) 3 = 6. ln (2x – 4) 3 = 6 3 ln (2x – 4) = 6Power Property ln (2x – 4) = 2Divide each side by 3. 2x – 4 = e 2 Rewrite in exponential form. x = Solve for x. e 2 + 4 2 x 5.69Use a calculator. Check: ln (2 5.69 – 4) 3 6 ln 401.95 6 5.996 6 Lesson 8-6 Natural Logarithms Additional Examples

4. Algebra 2 Use natural logarithms to solve 4e 3x + 1.2 = 14. 4e 3x + 1.2 = 14 4e 3x = 12.8Subtract 1.2 from each side. e 3x = 3.2Divide each side by 4. ln e 3x = ln 3.2Take the natural logarithm of each side. 3x = ln 3.2Simplify. x = Solve for x. ln 3.2 3 x 0.388 Lesson 8-6 Natural Logarithms Additional Examples

5. Algebra 2 An initial investment of $200 is now valued at $254.25. The interest rate is 6%, compounded continuously. How long has the money been invested? A = Pe rt Continuously compounded interest formula. 254.25 = 200e 0.06t Substitute 254.25 for A, 200 for P, and 0.06 for r. 1.27125 = e 0.06t Divide each side by 200. ln 1.27125 = ln e 0.06t Take the natural logarithm of each side. ln 1.27125 = 0.06tSimplify. The money has been invested for 4 years. = tSolve for t. ln 1.27125 0.06 4 tUse a calculator. Lesson 8-6 Natural Logarithms Additional Examples