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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. 5.6: Solving Exponential and Logarithmic Equations.
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Chapter 5: Exponential and Logarithmic Functions5.6: Solving Exponential Logarithmic EquationsDay 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 = 0.4307
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 • 1.5850 – 1.5850x = 4x – 1 Distribute on left • 2.5850 – 1.5850x = 4xAdd 1 to both sides • 2.5850 = 5.5850x Add 1.5850x to both sides • x = 0.4628 Divide by 5.5850
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 Functions5.6: Solving Exponential Logarithmic EquationsDay 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 Functions5.6: Solving Exponential Logarithmic EquationsDay 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 • Logarithmic Equations • Solve the equationln(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 • Logarithmic Equations • Solve the equationln(x – 3) + ln(2x + 1) = 2(ln x) • Because = -0.5414, 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 • e2.5 + 3 = x • x = 15.1825
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 35-51 & 69-75, odd problems • Show work