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Chapter 5: Exponential and Logarithmic Functions 5.5.A: Logarithmic Functions to Other Bases. Essential Question: What must you do to solve a logarithmic function in a base other than 10, with a calculator?. 5.5.A: Logarithmic Functions to Other Bases.
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Chapter 5: Exponential and Logarithmic Functions5.5.A: Logarithmic Functions to Other Bases Essential Question: What must you do to solve a logarithmic function in a base other than 10, with a calculator?
5.5.A: Logarithmic Functions to Other Bases • Other logarithmic bases work the same as regular logarithms • log381 = 4 34 = 81 • log464 = 3 43 = 64 • log1255 = 1/3 1251/3 = 5 • log8(1/4) = -2/3 8-2/3 = 1/4
5.5.A: Logarithmic Functions to Other Bases • Solving Logarithmic Equations • log216 • Can be rewritten as 2x = 16. • Because 24 = 16, x = 4 • log5(-25) • Rewritten as 5x = -25, which isn’t possible. • Undefined • log5x = 3 • Can be rewritten as 53 = x, so x = 125
5.5.A: Logarithmic Functions to Other Bases • Basic Properties of Other Bases • Same as with regular logs • logbv is defined only when v > 0 • logb1 = 0 • logbbk = k for every real number k • blogbv = v for every v > 0 • Solving Logarithmic Equations • log3(x – 1) = 4 • Rewritten as 34 = x – 1 • 81 = x – 1 • 82 = x
5.5.A: Logarithmic Functions to Other Bases • Laws of Logarithms to Other Bases • Same as with regular logs • Product Law: logb(vw) = logbv + logbw • Quotient Law: logb(v/w) = logbv – logbw • Power Law: logb(vk) = k logbv
5.5.A: Logarithmic Functions to Other Bases • Applications of Laws to Other Bases • Given: log72 = 0.3562 log73 = 0.5646 log75 = 0.8271 • Find log710, log72.5, & log748 • log710 = log7(2 • 5) = log72 + log75 = 0.3562 + 0.8271 = 1.1833 • log72.5 = log7(5 / 2) = log75 – log72 = 0.8271 – 0.3562 = 0.4709
5.5.A: Logarithmic Functions to Other Bases • Applications of Laws to Other Bases • Given: log72 = 0.3562 log73 = 0.5646 log75 = 0.8271 • log748 = log7(3 • 16) = log7(3 • 24) = log73 + log724 = log73 + 4 log72 = 0.5646 + 4(0.3562) = 1.9894
5.5.A: Logarithmic Functions to Other Bases • Change-of-Base Formula • and • Proof: • v = blogbv • ln v = ln (blogbv) *take ln of both sides = logbv ln b *power rule • *divide both sides by ln b • Proof using log works the same way
5.5.A: Logarithmic Functions to Other Bases • Change-of-Base Formula (Application) • Find log89
5.5.A: Logarithmic Functions to Other Bases • Transforming Logarithmic Functions • Involving other bases works no differently from regular logarithmic transformations • Describe the transformation from g(x) = log2x to h(x) = log2(x + 1) – 3 • +1: close to the x, therefore horizontal • Shifts one unit to the left (horizontal → opposite) • - 3: away from the x, therefore vertical • Shifts three units down
5.5: Properties and Laws of Logarithms • Assignment • Page 377 • Problems 41-71, odd problems • Show work