Copyright © Cengage Learning. All rights reserved. Exponential and Logarithmic Functions

Copyright © Cengage Learning. All rights reserved. 4.4 Laws Of Logarithms

3 Objectives ► Laws of Logarithms ► Expanding and Combining Logarithmic Expressions ► Change of Base Formula

4 Laws of Logarithms

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

6 Example 1 – Using the Laws of Logarithms to Evaluate Expressions Evaluate each expression. (a) log log 4 32 (b) log 2 80 – log 2 5 (c) log 8 Solution: (a) log log 4 32 = log 4 (2  32) = log 4 64 = 3 Law 1 Because 64 = 4 3

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

8 Expanding and Combining Logarithmic Expressions

9 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, is illustrated in the next example.

10 Example 2 – Expanding Logarithmic Expressions Use the Laws of Logarithms to expand each expression. (a) log 2 (6x) (b) log 5 (x 3 y 6 ) (c) ln Solution: (a) log 2 (6x) = log log 2 x (b) log 5 (x 3 y 6 ) = log 5 x 3 + log 5 y 6 = 3 log 5 x + 6 log 5 y Law 1 Law 3

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

12 Expanding and Combining Logarithmic Expressions The Laws of Logarithms also allow us to reverse the process of expanding that was done in Example 2. 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.

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

14 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? Hermann Ebbinghaus (1850–1909) studied this phenomenon and formulated the law described in the next example.

15 Example 5 – The Law of Forgetting If a task is learned at a performance level P 0, then after a time interval t the performance level P satisfies log P = log P 0 – 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.)

16 Example 5 – Solution (a) We first combine the right-hand side. log P = log P 0 – c log(t + 1) log P = log P 0 – log(t + 1) c log P = P = Given equation Law 3 Law 2 Because log is one-to-one

17 Example 5 – Solution (b) Here P 0 = 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. cont’d

18 Change of Base Formula

19 Change of Base Formula For some purposes we find it useful to change from logarithms in one base to logarithms in another base. Suppose we are given log a x and want to find log b x. Let y = log b x

20 Change of Base Formula We write this in exponential form and take the logarithm, with base a, of each side. b y = x log a (b y ) = log a x ylog a b = log a x Exponential form Take log a of each side Law 3 Divide by log a b

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

22 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) log 8 5 (b) log 9 20 Solution: (a) We use the Change of Base Formula with b = 8 and a = 10: log 8 5 = 

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