1. Write in logarithmic form Write in exponential form Write in exponential form Math 30-11

2. When solving a logarithmic equation, try rewriting the equation in exponential form. Strategy One: Exponents Rewrite the problem in exponential form. Math 30-12

3. Property of Equality for Logarithmic Equations. For equations with logarithmic expressions on both sides the equal sign, if the bases match, then the arguments must be equal. Strategy Two: Equating Logarithms Restrictions Math 30-13

4. Strategy Three: Graphically. Since the bases are both ‘3’ we set the arguments equal. Restrictions extraneous No solution Math 30-14

5. Solving Log Equations 1. log 2 72 = log 2 x + log 2 12 log 2 72 - log 2 12 = log 2 x x = 6 Restrictions Math 30-15

6. 2. Restrictions extraneous Math 30-16

7. 3. log 7 (2x + 2) - log 7 (x - 1) = log 7 (x + 1) 2x + 2 = (x- 1)(x + 1) 2x + 2 = x 2 - 1 0 = x 2 - 2x - 3 0 = (x - 3)(x + 1) x - 3 = 0 or x + 1 = 0 x = 3 x = -1 log 7 (2x + 2) - log 7 (x - 1) = log 7 (x + 1) log 7 (2(3) + 2) - log 7 (3 - 1) = log 7 (3 + 1) log 7 4 = log 7 4 log 7 (2x + 2) - log 7 (x - 1) = log 7 (x + 1) log 7 (2(-1) + 2) - log 7 (-1 - 1) = log 7 (-1 + 1) log 7 0 - log 7 (-2) = log 7 (0) Negative logarithms and logs of 0 are undefined. Therefore, x = 3 Solving Log Equations Verify Algebraically: Restrictions extraneous Math 30-17

8. 4. log 7 (x + 1) + log 7 (x - 5) = 1 log 7 [(x + 1)(x - 5)] = log 7 7 (x + 1)(x - 5) = 7 x 2 - 4x - 5 = 7 x 2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 x - 6 = 0 or x + 2 = 0 x = 6 x = -2 x = 6 Solving Log Equations Restrictions extraneous Math 30-18

9. Solving Exponential Equations Unlike Bases 5. 2 x = 8 log 2 x = log 8 xlog2 = log 8 x = 3 6. Solve for x: xlog2 = log12 x = 3.58 2 3.58 = 12 Math 30-19

10. Solving Log Equations 8. Solve log 5 (x - 6) = 1 - log 5 (x - 2) log 5 (x - 6) + log 5 (x - 2) = 1 log 5 (x - 6)(x - 2) = 1 log 5 (x - 6)(x - 2) = log 5 5 1 (x - 6)(x - 2) = 5 x 2 - 8x + 12 = 5 x 2 - 8x + 7 = 0 (x - 7)(x - 1) = 0 x = 7 or x = 1 Since x > 6, the value of x = 1 is extraneous therefore, the solution is x = 7. 7. xlog7 = 2log40 x = 3.79 Math 30-110

11. 9. 3 x = 2 x + 1 log(3 x ) = log(2 x + 1 ) x log 3 = (x + 1)log 2 x log 3 = x log 2 + 1 log 2 x log 3 - x log 2 = log 2 x(log 3 - log 2) = log 2 x = 1.71 Solving Log Equations 10. 2(18) x = 6 x + 1 log[2(18) x ] = log(6 x + 1 ) log 2 + x log 18 = (x + 1)log 6 log 2 + x log 18 = x log 6 + 1 log 6 x log 18 - x log 6 = log 6 - log 2 x(log 18 - log 6) = log 6 - log 2 x =1 Math 30-111

12. Page 412 1, 2, 3, 4b,c, 5a,c, 6, 7c,d, 8, 18 Math 30-112