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8.4A Logarithmic and Exponential Equations

This text provides strategies and restrictions for solving logarithmic and exponential equations, using logarithmic and exponential forms, graphically, and verifying solutions algebraically.

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8.4A Logarithmic and Exponential Equations

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  1. 8.4A Logarithmic and Exponential Equations Write in logarithmic form Write in logarithmic form Write in exponential form Write in exponential form Math 30-1

  2. Strategy One: Exponents When solving a logarithmic equation, try rewriting the equation in exponential form. Rewrite the problem in exponential form. Math 30-1

  3. Strategy Two: Equating Logarithms 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. Restrictions Math 30-1

  4. Strategy Three: Graphically. Restrictions Since the bases are both ‘3’ we set the arguments equal. extraneous No solution Math 30-1

  5. Solving Log Equations 1.log272 = log2x + log212 log272 - log212 = log2x Restrictions x = 6 Math 30-1

  6. 2. Restrictions extraneous Math 30-1

  7. Solving Log Equations Restrictions 3.log7(2x + 2) - log7(x - 1) = log7(x + 1) 2x + 2 = (x- 1)(x + 1) 2x + 2 = x2 - 1 0 = x2 - 2x - 3 0 = (x - 3)(x + 1) x - 3 = 0 or x + 1 = 0 x = 3 x = -1 Therefore, x = 3 Verify Algebraically: extraneous log7(2x + 2) - log7(x - 1) = log7(x + 1) log7(2x + 2) - log7(x - 1) = log7(x + 1) log7(2(-1) + 2) - log7(-1 - 1) = log7(-1 + 1) log7(2(3) + 2) - log7(3 - 1) = log7(3 + 1) log70 - log7(-2) = log7(0) Negative logarithms and logs of 0 are undefined. Math 30-1 log74 = log74

  8. Solving Log Equations Restrictions log7(x + 1) + log7(x - 5) = 1 log7[(x + 1)(x - 5)] = log77 (x + 1)(x - 5) = 7 x2 - 4x - 5 = 7 x2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 x - 6 = 0 or x + 2 = 0 x = 6 x = -2 4. extraneous x= 6 Math 30-1

  9. Solving Exponential Equations Unlike Bases 6. Solve for x: 5.2x = 8 log 2x = log 8 xlog2 = log 8 xlog2 = log12 x = 3 x = 3.58 23.58 = 12 Math 30-1

  10. Solving Log Equations 7. 8. Solve log5(x - 6) = 1 - log5(x - 2) log5(x - 6) + log5(x - 2) = 1 log5(x - 6)(x - 2) = 1 log5(x - 6)(x - 2) = log551 (x - 6)(x - 2) = 5 x2 - 8x + 12 = 5 x2 - 8x + 7 = 0 (x - 7)(x - 1) = 0 x = 7 or x = 1 xlog7 = 2log40 x = 3.79 Since x > 6, the value of x = 1 is extraneous therefore, the solution is x = 7. Math 30-1

  11. Solving Log Equations 9.3x = 2x + 1 log(3x) = log(2x + 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 10.2(18)x = 6x + 1 log[2(18)x] = log(6x + 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 = 1.71 x =1 Math 30-1

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

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