8-5 exponential and logarithmic equations

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85 Exponential and Logarithmic Equations 1 April 08, 2009 Mar 202:11 PM Objectives: Solve exponential equations. Solve logarithmic equations. 8-5 Exponential & Logarithmic Equations

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Page 1: 8-5 Exponential and Logarithmic Equations

8­5 Exponential and Logarithmic Equations

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April 08, 2009

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Objectives:• Solve exponential equations.• Solve logarithmic equations.

8-5 Exponential & Logarithmic Equations

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Evaluate each logarithm.

1. log981 log93 2. log 10 log39

3. log216 ÷ log28 4. Simplify 12523­

Check Skills You'll Need

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An equation of the form bcx=a, where the exponent includes a variable, is an exponential equation.

If m and n are positive and m = n, then log m = log n.

Therefore, you can solve an exponential equation by taking the logarithm of each side of the equation.

Solving Exponential Equations

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Example #1: Solving an Exponential Equation

Solve 73x = 20.

73x = 20log 73x = log 20 Take the common logarithm of each side.

3x log 7 = log 20 Use the power property of logarithms.

x =

x ≈ 0.5132

log 203log 7

Divide each side by 3 log 7.

Use a calculator.

Check: 73x = 20 73(0.5132) = 20 20.00382 ≈ 20

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Example #2: Solve each equation. Round to the nearest ten-thousandth. Check your answers.

a. 3x = 4 b. 62x = 21 c. 3x+4 = 101

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Solving Logarithmic Equations

To evaluate a logarithm with any base, you can use the Change of Base Formula.

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Example #3: Using the Change of Base Formula

Use the Change of Base Formula to evaluate log315.

log315 =

≈ 2.4650

log 15log 3

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Example #4: Evaluate log5400.

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An equation that includes a logarithmic expression, such as log315 = log2x is called a

logarithmic equation.

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Example #5: Solving a Logarithmic Equation

Solve log (3x + 1) = 5.

log (3x + 1) = 5

3x + 1 = 105

3x + 1 = 100,000

3x = 99,999

x = 33,333Check: log (3x + 1) = 5

log (3(33,333) + 1) = 5 log (100,000) = 5

log 105 = 5 5 = 5

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Example #6: Solve log (7 ­ 2x) = ­1. Check your answer.

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Example #7: Using Logarithmic Properties to Solve an Eqauation

Solve 2 log x ­ log 3 = 2.

2 log x ­ log 3 = 2

log = 2 Write as a single logarithm.

= 102Write in exponential form.

x2 = 3(100)

x = ±10√3 ≈ ±17.32

Log x is defined only for x>0, so the solution is 10√3 or about 17.32.

x2

3x2

3

(  )

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Example #8: Solve log 6 ­ log 3x = ­2.

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Homework: page 464(1 - 12, 23, 25 - 32 evaluate, 33 - 45)