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4.5 5.5  notes 1
 

4.5 5.5 notes 1

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    4.5 5.5  notes 1 4.5 5.5 notes 1 Presentation Transcript

    • CHAPTER 4/5: Exponential and Logarithmic Functions4.2/5.2 Exponential Functions and Graphs4.3/5.3 Logarithmic Functions and Graphs4.4/5.4 Properties of Logarithmic Functions4.5/5.5 Solving Exponential and Logarithmic Equations4.6/5.6 Applications and Models: Growth and Decay; and Compound Interest
    • 4.5/5.5 Solving Exponential and Logarithmic Equations• Solve exponential equations.• Solve logarithmic equations.
    • Solving Exponential EquationsEquations with variables in the exponents, such as 3x = 20 and 25x = 64,are called exponential equations.Use the following property to solve exponentialequations.Base-Exponent Property For any a > 0, a ≠ 1, ax = ay ↔ x = y. Slide 4.5/5.5 - 4
    • ExampleSolve 2 3x−7 = 32.Solution:Write each side as a power of the same number (base). 2 3x−7 = 2 5Since the bases are the same number, 2, we can use thebase-exponent property and set the exponents equal: 3 x −7 3x − 7 = 5 Check x = 4: 2 = 32. 3x = 12 2 3(4 )−7 ? 32 x=4 212−7 25 The solution is 4. 32 32 TRUE Slide 4.5/5.5 - 5
    • Another PropertyProperty of Logarithmic Equality For any M > 0, N > 0, a > 0, and a ≠ 1, loga M = loga N ↔ M = N. Slide 4.5/5.5 - 6
    • ExampleSolve: 3x = 20.Solution: 3x = 20 This is an exact answer. We log 3 = log 20 x cannot simplify further, but we x log 3 = log 20 can approximate using a calculator. log 20 x= ≈ 2.7268 log 3 We can check by finding 32.7268 ≈ 20. Slide 4.5/5.5 - 7
    • ExampleSolve: e0.08t = 2500.Solution: e0.08 t = 2500 ln e0.08 t = ln 2500 0.08t = ln(2500) ln(2500) t= 0.08 t ≈ 97.8 The solution is about 97.8. Slide 4.5/5.5 - 8
    • Solving Logarithmic EquationsEquations containing variables in logarithmicexpressions, such as log2 x = 4 and log x + log (x + 3) = 1,are called logarithmic equations.To solve logarithmic equations algebraically, we first tryto obtain a single logarithmic expression on one side andthen write an equivalent exponential equation. Slide 4.5/5.5 - 9
    • ExampleSolve: log3 x = −2. 1 Check: x = 9Solution: log 3 x = −2 log 3 x = −2 1 log 3 ? −2 3−2 = x 9 log 3 3−2 1 2 =x −2 − 2 TRUE 3 1 =x 1 9 The solution is . 9 Slide 4.5/5.5 - 10
    • ExampleSolve: log x + log (x + 3) = 1.Solution: log x + log (x + 3) = 1 log ⎡ x (x + 3)⎤ = 1 ⎣ ⎦ x (x + 3) = 101 x + 3x = 10 2 x 2 + 3x − 10 = 0 (x − 2 )(x + 5 ) = 0 x − 2 = 0 or x + 5 = 0 x = 2 or x = −5 Slide 4.5/5.5 - 11
    • Example (continued) Check x = 2: Check x = –5: log x + log (x + 3) = 1 log x + log (x + 3) = 1log 2 + log (2 + 3) ? 1 log (−5 ) + log (−5 + 3) ? 1 log 2 + log 5 FALSE log (2 ⋅ 5 ) log10 1 1 TRUE The number –5 is not a solution because negative numbers do not have real number logarithms. The solution is 2. Slide 4.5/5.5 - 12
    • Example Solve: ln (4x + 6 ) − ln (x + 3) = ln x. Solution:ln (4x + 6 ) − ln (x + 3) = ln x 0 = x2 + x − 6 4x + 6 0 = (x + 3)(x − 2 ) ln = ln x x+5 x + 3 = 0 or x − 2 = 0 4x + 6 =x x = −3 or x = 2 x+5 4x + 6(x + 5 )⋅ = x (x + 5 ) Only the value 2 checks x+5 and it is the only solution. 4x + 6 = x 2 + 5x Slide 4.5/5.5 - 13
    • Example - Using the Graphing CalculatorSolve: e0.5x – 7.3 = 2.08x + 6.2.Solve:Graphy1 = e0.5x – 7.3 andy2 = 2.08x + 6.2and use the Intersectmethod.The approximatesolutions are –6.471 and6.610. Slide 4.5/5.5 - 14