Laws of Logarithms Since logarithms are exponents, the Laws of Exponents give rise to the Laws of Logarithms.

Example 1 – Using the Laws of Logarithms to Evaluate Expressions Evaluate each expression. (a) log4 2 + log4 32 (b) log2 80 – log2 5 (c) log 8 Solution: (a) log4 2 + log4 32 = log4(2  32) = log4 64 = 3 Law 1 Because 64 = 43

Example 1 – Solution (b) log2 80 – log2 5 = log2 = log216 = 4 cont’d (b) log2 80 – log2 5 = log2 = log216 = 4 (c) log 8 = log 8–1/3 = log  –0.301 Law 2 Because 16 = 24 Law 3 Property of negative exponents Calculator

Expanding and Combining Logarithmic Expressions The Laws of Logarithms allow us to write the logarithm of a product or a quotient as the sum or difference of logarithms. This process, called expanding a logarithmic expression

Example 2 – Expanding Logarithmic Expressions Use the Laws of Logarithms to expand each expression. (a) log2(6x) (b) log5(x3y6) (c) ln Solution: (a) log2(6x) = log2 6 + log2 x (b) log5(x3y6) = log5 x3 + log5 y6 = 3 log5 x + 6 log5 y Law 1 Law 1 Law 3

Example 2 – Solution (c) = ln(ab) – = ln a + ln b – ln c1/3 cont’d (c) = ln(ab) – = ln a + ln b – ln c1/3 = ln a + ln b – ln c Law 2 Law 1 Law 3

Expanding and Combining Logarithmic Expressions The Laws of Logarithms also allow us to reverse the process of expanding That is, we can write sums and differences of logarithms as a single logarithm. This process, called combining logarithmic expressions, is illustrated in the next example.

Example 3 – Combining Logarithmic Expressions Combine 3 log x + log (x + 1) into a single logarithm. Solution: 3log x + log(x + 1) = log x3 + log(x + 1)1/2 = log(x3(x + 1)1/2) Law 3 Law 1

Expanding and Combining Logarithmic Expressions Logarithmic functions are used to model a variety of situations involving human behavior. One such behavior is how quickly we forget things we have learned. For example, if you learn algebra at a certain performance level (say, 90% on a test) and then don’t use algebra for a while, how much will you retain after a week, a month, or a year?

Example 5 – The Law of Forgetting If a task is learned at a performance level P0, then after a time interval t the performance level P satisfies log P = log P0 – c log(t + 1) where c is a constant that depends on the type of task and t is measured in months. (a) Solve for P. (b) If your score on a history test is 90, what score would you expect to get on a similar test after two months? After a year? (Assume that c = 0.2.)

Example 5 – Solution (a) We first combine the right-hand side. log P = log P0 – c log(t + 1) log P = log P0 – log(t + 1)c log P = P = Given equation Law 3 Law 2 Because log is one-to-one

Example 5 – Solution cont’d (b) Here P0 = 90, c = 0.2, and t is measured in months. In two months: t = 2 and In one year: t = 12 and Your expected scores after two months and one year are 72 and 54, respectively.

Change of Base Formula to change from logarithms in one base to logarithms in another base. y = logb x We write this in exponential form and take the logarithm, with base a, of each side. by = x Exponential form loga (by) = loga x Take loga of each side yloga b = loga x Law 3 Divide by logab

Change of Base Formula This proves the following formula. In particular, if we put x = a, then loga a = 1, and this formula becomes

Example 6 – Evaluating Logarithms with the Change of Base Formula Use the Change of Base Formula and common or natural logarithms to evaluate each logarithm, correct to five decimal places. (a) log85 (b) log920 Solution: (a) We use the Change of Base Formula with b = 8 and a = 10: log8 5 =  0.77398

Example 6 – Solution cont’d (b) We use the Change of Base Formula with b = 9 and a = e: log9 20 =  1.36342