# Chapter 5: Exponential and Logarithmic Functions 5

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Chapter 5: Exponential and Logarithmic Functions 5
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.

5.6: Solving Exponential and Logarithmic Equations
Powers of the Same Base Solve the equation 8x = 2x+1 8x = 2x+1 (23)x = 2x+1 23x = 2x+1 Set the exponents equal to each other 3x = x+1 2x = 1 x = 1/2

5.6: Solving Exponential and Logarithmic Equations
Powers of the Different Bases Solve the equation 5x = 2 5x = 2 log52 = x log 2/log 5 = x x =

5.6: Solving Exponential and Logarithmic Equations
Powers of the Different Bases Solve the equation 24x-1 = 31-x Take one base and make it into a log problem log231-x = 4x-1 (1 – x)log23 = 4x-1 (1 – x)(log 3/log 2) = 4x – 1 (1 – x)(1.5850) = 4x – 1 Calculate log 3/log 2 – x = 4x – 1 Distribute on left – x = 4x Add 1 to both sides = x Add x to both sides x = Divide by

5.6: Solving Exponential and Logarithmic Equations
Using Substitution Solve the equation ex – e-x = 4 ex – e-x = 4 Multiply all terms by ex to remove the negative exponent e2x – 1 = 4ex Set everything equal to 0, substitute u = ex e2x – 4ex – 1 = 0 u2 – 4u – 1 = 0 This is now a… Quadratic Equation

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

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

Chapter 5: Exponential and Logarithmic Functions 5
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.

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.

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

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?

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

Chapter 5: Exponential and Logarithmic Functions 5
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.

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 = Pert. Need to find r.

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

5.6: Solving Exponential and Logarithmic Equations
Solve the equation ln(x – 3) + ln(2x + 1) = 2(ln x) ln[(x – 3)(2x + 1)] = ln x2 ln(2x2 – 5x – 3) = ln x2 Natural logs cancel each other out 2x2 – 5x – 3 = x2 x2 – 5x – 3 = 0 Use quadratic equation

5.6: Solving Exponential and Logarithmic Equations
Solve the equation ln(x – 3) + ln(2x + 1) = 2(ln x) Because = , it’s undefined for ln(x – 3), so there’s only one solution

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 e2.5 = x – 3 e = x x =

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 (x2 – 17x + 16) = 2 102 = x2 – 17x + 16 0 = x2 – 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

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