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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.

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Presentation on theme: "Chapter 5: Exponential and Logarithmic Functions 5.5.A: Logarithmic Functions to Other Bases Essential Question: What must you do to solve a logarithmic."— Presentation transcript:

1 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?

2 5.5.A: Logarithmic Functions to Other Bases Other logarithmic bases work the same as regular logarithms ◦ log 3 81 = 43 4 = 81 ◦ log 4 64 = 34 3 = 64 ◦ log 125 5 = 1 / 3 125 1/3 = 5 ◦ log 8 (¼) = - 2 / 3 8 -2/3 = ¼

3 5.5.A: Logarithmic Functions to Other Bases Solving Logarithmic Equations ◦ Log 2 16 = x  Can be rewritten as 2 x = 16.  Because 2 4 = 16, x = 4 ◦ log 5 (-25) = x  Rewritten as 5 x = -25, which isn’t possible.  Undefined ◦ log 5 x = 3  Can be rewritten as 5 3 = x, so x = 125

4 5.5.A: Logarithmic Functions to Other Bases Basic Properties of Other Bases ◦ Same as with regular logs  log b v is defined only when v > 0  log b 1 = 0  log b b k = k for every real number k  b log b v = v for every v > 0 ◦ Solving Logarithmic Equations  log 3 (x – 1) = 4  Rewritten as 3 4 = x – 1  81 = x – 1  82 = x

5 5.5.A: Logarithmic Functions to Other Bases Laws of Logarithms to Other Bases ◦ Same as with regular logs  Product Law:log b (vw) = log b v + log b w  Quotient Law:log b ( v / w ) = log b v – log b w  Power Law:log b (v k ) = k log b v

6 5.5.A: Logarithmic Functions to Other Bases Applications of Laws to Other Bases ◦ Given:log 7 2 = 0.3562 log 7 3 = 0.5646 log 7 5 = 0.8271  Find log 7 10, log 7 2.5, & log 7 48  log 7 10= log 7 (2 5) = log 7 2 + log 7 5 = 0.3562 + 0.8271 = 1.1833  log 7 2.5= log 7 (5 / 2) = log 7 5 – log 7 2 = 0.8271 – 0.3562 = 0.4709

7 5.5.A: Logarithmic Functions to Other Bases Applications of Laws to Other Bases ◦ Given:log 7 2 = 0.3562 log 7 3 = 0.5646 log 7 5 = 0.8271  log 7 48= log 7 (3 16) = log 7 (3 2 4 ) = log 7 3 + log 7 2 4 = log 7 3 + 4 log 7 2 = 0.5646 + 4(0.3562) = 1.9894

8 5.5.A: Logarithmic Functions to Other Bases Change-of-Base Formula ◦ and/or ◦ Proof:  v = b log b v  ln v = ln (b log b v )*take ln of both sides = log b v ln b*power rule  *divide both sides by ln b  Proof using log works the same way

9 5.5.A: Logarithmic Functions to Other Bases Change-of-Base Formula (Application) ◦ Find log 8 9 

10 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) = log 2 x to h(x) = log 2 (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

11 5.5A: Properties and Laws of Logarithms Assignment ◦ Page 377 ◦ Problems 41-71, odd problems ◦ Show work


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