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Properties of logarithms. Math 3 Keeper 29. Properties of Logarithms. Let b, u, and v be positive numbers such that b ≠1. Product property: log b uv = log b u + log b v Quotient property: log b u/v = log b u – log b v Power property: log b u n = n log b u.
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Properties of logarithms Math 3 Keeper 29
Properties of Logarithms • Let b, u, and v be positive numbers such that b≠1. • Product property: • logbuv = logbu + logbv • Quotient property: • logbu/v = logbu– logbv • Power property: • logbun = n logbu
EXAMPLE 1: Use log53≈.683 and log57≈1.209 to approximate the following: a) log53/7 = log53 – log57 ≈ 0.683 – 1.209 = -0.526 b) log521 = log5(3·7)= log53 + log57≈ 0.683 + 1.209 = 1.892
EXAMPLE 1 (continued): Use log53≈.683 and log57≈1.209 to approximate the following: c) log549 = log572 = 2 log57 ≈ 2(1.209)= 2.418
YOUR TURN!Use log95≈0.732 and log911≈1.091 to approximate the following: d) log95/11 e) log955 f) log925 d) -0.359 e) 1.823 f) 1.464
EXAMPLE 2: Expand the given logarithm *You can use the properties to expand logarithms. a) log2 = log27x3 - log2y = log27 + log2x3 – log2y = log27 + 3·log2x – log2y
Your turn! Expand the logarithm. b) log 5mn= log 5 + logm + logn c) log58x3 = log58 + 3·log5x
EXAMPLE 3: Condense the logarithms a) log 6 + 2 log2 – log 3 = log 6 + log 22 – log 3 = log (6·22) – log 3 = log = log 8
YOUR TURN AGAIN! Condense the logarithm. b) log57 + 3·log5t = log57t3 c) 3log2x – (log24 + log2y)= log2
Change of base formula: • u, b, and c are positive numbers with b≠1 and c≠1. Then: • logcu = • logcu = (base 10) • logcu = (base e)
EXAMPLE 4: Evaluate using the change-of-base formula using 1) common log & 2) natural logarithm. a1) log37 = log 7 ≈ log 3 0.8451 ≈ 0.4771 1.771 a2) ln 7≈ ln 3 1.946 ≈ 1.099 1.771
EXAMPLE 4 (continued): Evaluate using the change-of-base formula using 1) common log & 2) natural logarithm. b1) log48 = log 8 ≈ log 4 0.903 ≈ 0.602 1.500 a2) ln 8≈ ln 4 2.079 ≈ 1.386 1.500