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

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

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Page 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

Algebra 2Algebra 2

Write 2 ln 12 – ln 9 as a single natural logarithm.

2 ln 12 – ln 9 = ln 122 – ln 9 Power Property

= ln Quotient Property122

9

= ln 16 Simplify.

Lesson 8-6

Natural LogarithmsNatural Logarithms

Additional Examples

Page 2: 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

Algebra 2Algebra 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 R Use the formula.

= –0.0098(50) + 2.3 ln 22 Substitute.

–0.49 + 2.3(3.091) Use a calculator.

6.62 Simplify.

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 LogarithmsNatural Logarithms

Additional Examples

Page 3: 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

Algebra 2Algebra 2

Solve ln (2x – 4)3 = 6.

ln (2x – 4)3 = 6

3 ln (2x – 4) = 6 Power Property

ln (2x – 4) = 2 Divide each side by 3.

2x – 4 = e2 Rewrite in exponential form.

x = Solve for x.e2 + 42

x 5.69 Use a calculator.

Check: ln (2 • 5.69 – 4)3 6 ln 401.95 6 5.996 6

Lesson 8-6

Natural LogarithmsNatural Logarithms

Additional Examples

Page 4: 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

Algebra 2Algebra 2

Use natural logarithms to solve 4e3x + 1.2 = 14.

4e3x + 1.2 = 14

4e3x = 12.8 Subtract 1.2 from each side.

e3x = 3.2 Divide each side by 4.

ln e3x = ln 3.2 Take the natural logarithm of each side.

3x = ln 3.2 Simplify.

x = Solve for x.ln 3.23

x 0.388

Lesson 8-6

Natural LogarithmsNatural Logarithms

Additional Examples

Page 5: 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

Algebra 2Algebra 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 = Pert Continuously compounded interest formula.

254.25 = 200e0.06t Substitute 254.25 for A, 200 for P, and 0.06 for r.

1.27125 = e0.06t Divide each side by 200.

ln 1.27125 = ln e0.06t Take the natural logarithm of each side.

ln 1.27125 = 0.06t Simplify.

The money has been invested for 4 years.

= t Solve for t.ln 1.271250.06

4 t Use a calculator.

Lesson 8-6

Natural LogarithmsNatural Logarithms

Additional Examples