7.5 Apply Properties of logarithms Algebra II

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

Ex. 4: log37 = log 7 ≈ log 3 1.771 ln 7 ≈ ln 3 1.771 (base 10) Use the change of base to evaluate: log37 = (base 10) log 7 ≈ log 3 1.771 (base e) ln 7 ≈ ln 3 1.771

Ex. 1Use log53≈.683 and log57≈1.209 log53/7 = log521 = log53 – log57 ≈ A.)Approximate: log53/7 = log53 – log57 ≈ .683 – 1.209 = -.526 B.)Approximate: log521 = log5(3·7)= log53 + log57≈ .683 + 1.209 = 1.892

Use log53≈.683 and log57≈1.209 C.) Approximate: log549 = log572 = 2 log57 ≈ 2(1.209)= 2.418

Ex. 2a) Expanding Logarithms You can use the properties to expand logarithms. log2 = log27x3 - log2y = log27 + log2x3 – log2y = log27 + 3·log2x – log2y

log 5mn = log 5 + log m + log n log58x3 = log58 + 3·log5x Ex. 2 2b.) Expand: log 5mn = log 5 + log m + log n 2c.) Expand: log58x3 = log58 + 3·log5x

Ex. 3a.) Condensing Logarithms log 6 + 2 log2 – log 3 = log 6 + log 22 – log 3 = log (6·22) – log 3 = log = log 8

Ex. 3 continued log57 + 3·log5t = log57t3 3log2x – (log24 + log2y)= 3b.) Condense: log57 + 3·log5t = log57t3 3c.) Condense: 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)

Assignment