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Learn to simplify and evaluate expressions, solve equations, and understand the properties of logarithms. Practice examples and solve real-world problems.
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Properties of Logarithmic Functions Objectives: Simplify and evaluate expressions involving logarithms Solve equations involving logarithms
Properties of Logarithms For m > 0, n > 0, b > 0, and b 1: Product Property logb (mn) = logb m + logb n
Example 1 given: log5 12 1.5440 log5 10 1.4307 log5 120 = log5 (12)(10) = log5 12 + log5 10 1.5440 + 1.4307 2.9747
logb = logb m – logb n m n Properties of Logarithms For m > 0, n > 0, b > 0, and b 1: Quotient Property
12 = log5 10 Example 2 given: log5 12 1.5440 log5 10 1.4307 log5 1.2 = log5 12 – log5 10 1.5440 – 1.4307 0.1133
Properties of Logarithms For m > 0, n > 0, b > 0, and any real number p: Power Property logb mp = p logb m
Example 3 given: log5 12 1.5440 log5 10 1.4307 log5 1254 5x = 125 = 4 log5 125 53 = 125 =4 3 x = 3 = 12
Practice Write each expression as a single logarithm. 1) log2 14 – log2 7 2) log3 x + log3 4 – log3 2 3) 7 log3 y – 4 log3 x
4 minutes Warm-Up Write each expression as a single logarithm. Then simplify, if possible. 1) log6 6 + log6 30 – log6 5 2) log6 5x + 3(log6 x – log6 y)
Properties of Logarithms For b > 0 and b 1: Exponential-Logarithmic Inverse Property logb bx = x and b logbx = x for x > 0
Example 1 Evaluate each expression. a) b)
Practice Evaluate each expression. 1) 7log711 – log3 81 2) log8 85 + 3log38
Properties of Logarithms For b > 0 and b 1: One-to-One Property of Logarithms If logb x = logb y, then x = y
Example 2 Solve log2(2x2 + 8x – 11) = log2(2x + 9) for x. log2(2x2 + 8x – 11) = log2(2x + 9) 2x2 + 8x – 11 = 2x + 9 2x2 + 6x – 20 = 0 2(x2 + 3x – 10) = 0 2(x – 2)(x + 5) = 0 x = -5,2 Check: log2(2x2 + 8x – 11) = log2(2x + 9) log2 (–1) = log2 (-1) undefined log2 13 = log2 13 true
Practice Solve for x. 1) log5 (3x2 – 1) = log5 2x 2) logb (x2 – 2) + 2 logb 6 = logb 6x
Solving Equations and Modeling Objectives: Solve logarithmic and exponential equations by using algebra and graphs Model and solve real-world problems involving logarithmic and exponential relationships
) ( m logb = logb m – logb n n Summary of Exponential-Logarithmic Definitions and Properties Definition of logarithm y = logb x only if by = x Product Property logb mn = logb m + logb n Quotient Property Power Property logb mp = p logb m
logb a logb c logc a = Summary of Exponential-Logarithmic Definitions and Properties Exp-Log Inverse b logb x = x for x > 0 logb bx = x for all x 1-to-1 for Exponents bx = by; x = y logb x = logb y; x = y 1-to-1 for Logarithms Change-of-Base
x = log 4 + 2 log 3 log 3 – log 4 Example 1 Solve for x. 3x – 2 = 4x + 1 log 3x – 2 = log 4x + 1 (x – 2) log 3 = (x + 1) log 4 x log 3 – 2 log 3 = x log 4 + log 4 x log 3 – x log 4 = log 4 + 2 log 3 x (log 3 – log 4) = log 4 + 2 log 3 x –12.46
Example 2 Solve for x. log x + log (x + 3) = 1 log [x(x + 3)] = 1 101 = x(x + 3) 101 = x2 + 3x x2 + 3x – 10 = 0 (x + 5)(x – 2) = 0 x = 2,-5
Example 2 Solve for x. log x + log (x + 3) = 1 Check: x = 2,-5 Let x = -5 Let x = 2 log x + log (x + 3) = 1 log x + log (x + 3) = 1 log -5 + log (-5 + 3) = 1 log 2 + log (2 + 3) = 1 log -5 + log -2 = 1 log 2 + log 5 = 1 1 = 1 undefined x = 2
ln 7 + 5 x = 2 Example 3 Solve for x. 8e2x-5 = 56 e2x-5 = 7 ln e2x-5 = ln 7 2x - 5 = ln 7 x 3.47
Example 4 Suppose that the magnitude, M, of an earthquake measures 7.5 on the Richter scale. Use the formula below to find the amount of energy, E, released by this earthquake.