1. 2.6.1 MATHPOWER TM 12, WESTERN EDITION 2.6 Chapter 2 Exponents and Logarithms

2. The Product and Quotient Laws Product Law:log b (mn) = log b m + log b n Quotient Law: Express as a sum and difference of logarithms: = log 3 A + log 3 B - log 3 C Evaluate: log 2 10 + log 2 12.8 = log 2 (10 x 12.8) = log 2 (128) = log 2 (2 7 ) = 7 2.6.2

3. Solve: x = log 5 50 - log 5 10 Given that log 7 9 = 1.129, find the value of log 7 63: log 7 63 = log 7 (9 x 7) = log 7 9 + log 7 7 = 1.129 + 1 = 2.129 Evaluate: x = log 4 5a + log 4 8a 3 - log 4 10a 4 x = log 4 4 x = 1 2.6.3 Simplifying Logarithms x = log 5 5 = 1

4. Power Law: log b m n = n log b m Express as a single log: = log 5 216 2.6.4 The Power Law

5. Evaluate: = 2(1) + 4(1) Given that log 6 2 = 0.387 and log 6 5 = 0.898 solve = 0.418 2.6.5 Applying the Power Laws

6. 2.6.6 Applying the Power Laws Evaluate: = 4.135 If log 2 8 = x, express each in terms of x: a) log 2 512 = log 2 8 3 = 3log 2 8 = 3x b) log 2 2 log 2 8 = x log 2 2 3 = x 3log 2 2 = x

7. The pH of a substance is defined by the equation pH = -log[H + ], where H + is the concentration of hydrogen ions in moles per litre (mol/L). Determine the pH if the hydrogen ion concentration is 0.000 012 5 mol/L. pH = -log[H + ] = -log[1.25 x 10 -5 ] = -(log1.25 + -5log10) = -(log1.25 - 5) = -(0.0969 - 5) = 4.9 Therefore, the pH is 4.9. 2.6.7 Logarithm Applications Note: -log 0.000 012 5 = 4.9

8. Suggested Questions: Pages 106 and 107 1-18, 25, 31, 33-42, 47, 53 2.6.8