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ALGEBRA II LESSON 10-3 Pg. 541-545 PROPERTIES OF LOGARITHMS Today’s Key Concepts: We should know what a logarithmic function is, now we will work with the multiplication, division, and exponential factors dealing with logarithmic functions. Objectives: (1) Simplify and evaluate expressions using the properties of logarithms (2) Solve logarithmic equations using the properties of logarithms Recall the EXPONENT lessons we had when multiplying, dividing, and power-to-a-power? A. (x2) (x5) = x2+5 = x7 We added the exponents when multiplying. B. (x14) = . (x14-3) = x11 We subtracted exponents when dividing. (x3) C. (x5)3 = x 5·3 = x15 We multiplied exponents when power-to-a-power is shown Logarithms are basically the same process! The Three Laws of Logarithms 1. Product Property of Logarithms logbxy = logbx + logby "The logarithm of a product (multiplication) is equal to the sum of the logarithms of each factor." 2. Quotient Property of Logarithms logb x = logbx − logby y “The logarithm of a quotient (division) is equal to the logarithm of the numerator minus the logarithm of the denominator." 3. Power Property of Logarithms logb x n = n logbx "The logarithm of a power of x is equal to the exponent of that power (n) times the logarithm of x." Demonstrate Examples of these Rules-----> EXAMPLE 1 (Law #1): (Product Property of Logarithms) Given that the log5 2 ≈ 0.4307 NOTE: log5 2 = log10 2 Find log5 250 (Calculator Use) log10 5 Solution: log5 250 = log5 (125)(2) = log5 125 + log5 2 = log5 53 + log5 2 = 3 + 0.4307 = 3.4307 125 = 53 ≈ 0.4307 EXAMPLE 2 (Law #2) (Quotient Property of Logarithms) Given log6 8 ≈ 1.1606 and log632 ≈ 1.9343 Find the value of log6 4 Solution: Since 32/8 = 4 Then log632 log6 8 log632 - log6 8 ≈ 1.9343 - 1.1606 log6 4 ≈ 0.7737 EXAMPLE 3 (Law #3): (Power Property of Logarithms) Given log5 6 ≈ 1.1133, Approximate the value of log5 216 Solution: Since 216 = 63 log5 216 = log5 63 = 3 log5 6 ≈ 3 (1.1133) ≈ 3.3399 ________________________________________________________________________________ PROBLEMS: Solve each of these: a) 4 log2 x – log2 5 = log2 125 x4 = 125 x4 = (125)(5) 5 x4 = 625 x = 5 b) log8 x + log8 (x – 12) = 2 (Since “2” is not log8 2, we must follow the rule and state it as 82 x(x – 12) = 82 Product Property of Logarithms x2 –12x = 64 x2 –12x – 64 = 0 (x –16)(x + 4) =0 x = 16 x ≠ - 4 (log8 –4 is undefined) SAMPLE PROBLEMS: #18 log5 30 = log5 (5)(2)(3) = log5 5 + log5 2 + log5 3 = 1 + 0.4307 + .6826 = 2.1133 #20 log5 10 = log5 10 - log5 9 = log5 (5)(2) - log5 (3)(3) 9 = (log5 5 + log5 2) – (log5 3 + log5 3) = (1 + .4307) – (0.6826 + 0.6826) = 0.0655 #26 2 log10 6 – 1/3 log10 27 = log10 x 62 271/3 = x 36 = x x = 12 3