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Solving Exponential and Logarithmic Equations Section 3.4 JMerrill, 2005 Revised, 2008

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Same Base Solve: 4x-2 = 64x 4 4x-2 = (43)x x-2 = 43x x x–2 = 3x - -2 = 2x 1 = x If b M = b N, then M = N 64 = 4 3 If the bases are already =, just solve the exponents

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You Do Solve 27 x+3 = 9 x-1 Solve 27 x+3 = 9 x-1

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Review – Change Logs to Exponents log3x = 2 logx16 = 2 log 1000 = x 3 2 = x, x = 9 x 2 = 16, x = 4 10 x = 1000, x = 3

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Using Properties to Solve Logarithmic Equations If the exponent is a variable, then take the natural log of both sides of the equation and use the appropriate property. Then solve for the variable.

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Example: Solving 2 2x = 7 problem ln2x = ln7 take ln both sides xln2 = ln7power rule x x = divide to solve for x = 2.807

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Example: Solving e ex = 72problem l lnex = ln 72take ln both sides x x lne = ln 72power rule = 4.277solution: because ln e = ?

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You Do: Solving 2 2ex + 8 = 20problem ex = 12subtract 8 e ex = 6divide by 2 l ln ex = ln 6take ln both sides x x lne = 1.792power rule x = 1.792(remember: lne = 1)

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Example Solve 5x-2 = 42x+3 ln5x-2 = ln42x+3 (x-2)ln5 = (2x+3)ln4 The book wants you to distribute… Instead, divide by ln4 (x-2)1.1609 = 2x+3 1.1609x-2.3219 = 2x+3 x6.3424

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Solving by Rewriting as an Exponential Solve log4(x+3) = 2 42 = x+3 16 = x+3 13 = x

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You Do Solve 3ln(2x) = 12 ln(2x) = 4 Realize that our base is e, so e4 = 2x x 27.299 You always need to check your answers because sometimes they dont work!

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Using Properties to Solve Logarithmic Equations 1.Condense both sides first (if necessary). 2.If the bases are the same on both sides, you can cancel the logs on both sides. 3.Solve the simple equation

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Example: Solve for x log36 = log33 + log3xproblem log36 = log33xcondense 6 6 = 3xdrop logs 2 2 = xsolution

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You Do: Solve for x log 16 = x log 2problem log 16 = log 2xcondense 1 16 = 2xdrop logs x x = 4solution

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You Do: Solve for x l log4x = log44problem = = log44condense 4drop logs cube each side X X = 64solution

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Example 7xlog 2 5 = 3xlog 2 5 + ½ log 2 25 7xlog 2 5 = 3xlog 2 5 + ½ log 2 25 log 2 5 7x = log 2 5 3x + log 2 25 ½ log 2 5 7x = log 2 5 3x + log 2 25 ½ log 2 5 7x = log 2 5 3x + log 2 5 1 log 2 5 7x = log 2 5 3x + log 2 5 1 7x = 3x + 1 7x = 3x + 1 4x = 1 4x = 1

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You Do Solve:log 7 7 + log 7 2 = log 7 x + log 7 (5x – 3) Solve:log 7 7 + log 7 2 = log 7 x + log 7 (5x – 3)

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You Do Answer Solve:log 7 7 + log 7 2 = log 7 x + log 7 (5x – 3) Solve:log 7 7 + log 7 2 = log 7 x + log 7 (5x – 3) log 7 14 = log 7 x(5x – 3) log 7 14 = log 7 x(5x – 3) 14 = 5x 2 -3x 14 = 5x 2 -3x 0 = 5x 2 – 3x – 14 0 = 5x 2 – 3x – 14 0 = (5x + 7)(x – 2) 0 = (5x + 7)(x – 2) Do both answers work? NO!!

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Final Example How long will it take for $25,000 to grow to $500,000 at 9% annual interest compounded monthly? How long will it take for $25,000 to grow to $500,000 at 9% annual interest compounded monthly?

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3.4 Exponential & Logarithmic Equations JMerrill, 2010.

3.4 Exponential & Logarithmic Equations JMerrill, 2010.

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