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Chapter 5: Exponential and Logarithmic Functions 5.6: Solving Exponential Logarithmic Equations Day 1 Essential Question: Give examples of equations that can be solved by using the properties of exponents and logarithms.

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5.6: Solving Exponential and Logarithmic Equations Powers of the Same Base Solve the equation 8 x = 2 x+1 8 x = 2 x+1 (2 3 ) x = 2 x+1 2 3x = 2 x+1 Set the exponents equal to each other 3x = x+1 2x = 1 x = 1 / 2

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5.6: Solving Exponential and Logarithmic Equations Powers of the Different Bases Solve the equation 5 x = 2 5 x = 2 log 5 2 = x log 2 / log 5 = x x = 0.4307

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5.6: Solving Exponential and Logarithmic Equations Powers of the Different Bases Solve the equation 2 4x-1 = 3 1-x Take one base and make it into a log problem log 2 3 1-x = 4x-1 (1 – x)log 2 3 = 4x-1 (1 – x)( log 3 / log 2 ) = 4x – 1 (1 – x)(1.5850) = 4x – 1 Calculate log 3 / log 2 1.5850 – 1.5850x = 4x – 1 Distribute on left 2.5850 – 1.5850x = 4x Add 1 to both sides 2.5850 = 5.5850x Add 1.5850x to both sides x = 0.4628 Divide by 5.5850

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5.6: Solving Exponential and Logarithmic Equations Using Substitution Solve the equation e x – e -x = 4 e x – e -x = 4 Multiply all terms by e x to remove the negative exponent e 2x – 1 = 4e x Set everything equal to 0, substitute u = e x e 2x – 4e x – 1 = 0 u 2 – 4u – 1 = 0This is now a… Quadratic Equation

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5.6: Solving Exponential and Logarithmic Equations Using Substitution Set u back to e x, and solve

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5.6: Solving Exponential and Logarithmic Equations Assignment Page 386 Problems 1-31, odd problems Show work

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Chapter 5: Exponential and Logarithmic Functions 5.6: Solving Exponential Logarithmic Equations Day 2 Essential Question: Give examples of equations that can be solved by using the properties of exponents and logarithms.

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5.6: Solving Exponential and Logarithmic Equations Applications of Exponential Equations Radiocarbon Dating The half-life of carbon-14 is 5730 years, so the amount of carbon-14 remaining at time t is given by Many of these problems will deal with percentage of carbon-14 remaining, so P = 1 (i.e. 100%), and the amount remaining will be the percentage left.

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5.6: Solving Exponential and Logarithmic Equations Applications: Carbon Dating The skeleton of a mastodon has lost 58% of its original carbon-14. When did the mastodon die? If 58% has been lost, then 42% remains

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5.6: Solving Exponential and Logarithmic Equations Applications: Compound Interest If $3000 is to be invested at 8% per year, compounded quarterly, in how many years will the investment be wroth $10,680?

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5.6: Solving Exponential and Logarithmic Equations Assignment Page 386 Problems 53-67, odd problems Show work

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Chapter 5: Exponential and Logarithmic Functions 5.6: Solving Exponential Logarithmic Equations Day 3 Essential Question: Give examples of equations that can be solved by using the properties of exponents and logarithms.

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5.6: Solving Exponential and Logarithmic Equations Applications: Population Growth A culture started at 1000 bacteria. 7 hours later, there are 5000 bacteria. Find the function and when there are 1 billion bacteria. Function is based off A = Pe rt. Need to find r.

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5.6: Solving Exponential and Logarithmic Equations Applications: Population Growth To find A=1,000,000, need to find t

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5.6: Solving Exponential and Logarithmic Equations Logarithmic Equations Solve the equation ln(x – 3) + ln(2x + 1) = 2(ln x) ln[(x – 3)(2x + 1)] = ln x 2 ln(2x 2 – 5x – 3) = ln x 2 Natural logs cancel each other out 2x 2 – 5x – 3 = x 2 x 2 – 5x – 3 = 0 Use quadratic equation

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5.6: Solving Exponential and Logarithmic Equations Logarithmic Equations Solve the equation ln(x – 3) + ln(2x + 1) = 2(ln x) Because = -0.5414, its undefined for ln(x – 3), so theres only one solution

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5.6: Solving Exponential and Logarithmic Equations Equations with logarithmic & constant terms Solve ln(x – 3) = 5 – ln(x – 3) ln(x – 3) + ln(x – 3) = 5 2 ln(x – 3) = 5 ln (x – 3) = 2.5 e 2.5 = x – 3 e 2.5 + 3 = x x = 15.1825

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5.6: Solving Exponential and Logarithmic Equations Equations with logarithmic & constant terms Solve log(x – 16) = 2 – log(x – 1) log(x – 16) + log(x – 1) = 2 log [(x – 16)(x – 1)] = 2 log (x 2 – 17x + 16) = 2 10 2 = x 2 – 17x + 16 0 = x 2 – 17x – 84 0 = (x – 21)(x + 4) x = 21 or x = -4 x = -4 would give log(-4 – 16) = log -20, which is undefined There is only one solution, x = 21

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5.6: Solving Exponential and Logarithmic Equations Assignment Page 386 Problems 35-51 & 69-75, odd problems Show work

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