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Solving equations involving exponents and logarithms

Solving equations involving exponents and logarithms. Let’s review some terms. When we write log 5 125 5 is called the base 125 is called the argument. Logarithmic form of 5 2 = 25 is log 5 25 = 2. For all the laws a , M and N > 0 a ≠ 1 r is any real. Remember ln and log.

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Solving equations involving exponents and logarithms

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  1. Solving equations involving exponents and logarithms

  2. Let’s review some terms.When we write log5 1255 is called the base125 is called the argument

  3. Logarithmic form of 52= 25 islog525 = 2

  4. For all the lawsa, M and N > 0 a≠ 1 r is any real

  5. Remember ln and log • ln is a short cut for loge • log means log10

  6. Log laws

  7. Find the pattern your equation resembles If your variable is in an exponent or in the argument of a logarithm

  8. Find the pattern Fit your equation to match the pattern Switch to the equivalent form Solve the result Check (be sure you remember the domain of a log) If your variable is in an exponent or in the argument of a logarithm

  9. log(2x) = 3

  10. log(2x) = 3 It fits

  11. log(2x) = 3 103=2x Switch Did you remember that log(2x) means log10(2x)?

  12. log(2x) = 3 103=2x 500 = x Divide by 2

  13. ln(x+3) = ln(-7x)

  14. ln(x+3) = ln(-7x) It fits

  15. ln(x+3) = ln(-7x) Switch

  16. ln(x+3) = ln(-7x) x + 3 = -7x Switch

  17. ln(x+3) = ln(-7x) x + 3 = -7x x = - ⅜ Solve the result (and check)

  18. ln(x) + ln(3) = ln(12)

  19. ln(x) + ln(3) = ln(12) x + 3 = 12

  20. ln(x) + ln(3) = ln(12) x + 3 = 12 Oh NO!!! That’s wrong!

  21. ln(x) + ln(3) = ln(12) ln(3x) = ln (12) You need to use log laws

  22. ln(x) + ln(3) = ln(12) ln(3x) = ln (12) 3x = 12 Switch

  23. ln(x) + ln(3) = ln(12) ln(3x) = ln (12) 3x = 12 x = 4 Solve the result

  24. log3(x+2) + 4 = 9

  25. log3(x+2) + 4 = 9 It will fit

  26. log3(x+2) + 4 = 9 log3(x+2) = 5 Subtract 4 to make it fit

  27. log3(x+2) + 4 = 9 log3(x+2) = 5 Switch

  28. log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 Switch

  29. log3(x+2) + 4 = 9 log3(x+2) = 5 35 = x + 2 x = 241 Solve the result

  30. 5(10x) = 19.45

  31. 5(10x) = 19.45 10x = 3.91 Divide by 5 to fit

  32. 5(10x) = 19.45 10x = 3.91 Switch

  33. 5(10x) = 19.45 10x = 3.91 log(3.91) = x Switch

  34. 5(10x) = 19.45 10x = 3.91 log(3.91) = x ≈ 0.592 Exact log(3.91) Approx 0.592

  35. 2 log3(x) = 8

  36. 2 log3(x) = 8 It will fit

  37. 2 log3(x) = 8 log3(x) = 4 Divide by 2 to fit

  38. 2 log3(x) = 8 log3(x) = 4 Switch

  39. 2 log3(x) = 8 log3(x) = 4 34=x Switch

  40. 2 log3(x) = 8 log3(x) = 4 34=x x = 81 Then Simplify

  41. log2(x-1) + log2(x-1)= 3

  42. log2(x-1) + log2(x-1)= 3 Need to use a log law

  43. log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3

  44. log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 Switch

  45. log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) Switch

  46. log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 and finish

  47. log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 But -3 does not check!

  48. log2(x-1) + log2(x+1)= 3 log2{(x-1)(x+1)} = 3 23=(x-1)(x+1) = x2 -1 x = +3 or -3 Exclude -3 (it would cause you to have a negative argument)

  49. There’s more than one way to do this

  50. Can you find why each step is valid?

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