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Solving Logarithmic Equations — Transcript

by Mathispower4u · 1,157 words · 137 segments · language en · Watch on YouTube
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  1. 0:01- WELCOME TO A LESSON ON SOLVING LOGARITHMIC EQUATIONS.
  2. 0:05THERE ARE REALLY TWO TYPES OF LOG EQUATIONS
  3. 0:06WE'RE GOING TO TAKE A LOOK AT IN THIS VIDEO.
  4. 0:08THE FIRST TYPE IS WHEN YOU HAVE AN EQUATION
  5. 0:10THAT HAS ONLY TWO LOGS OF THE SAME BASE.
  6. 0:14AND IF THAT'S THE CASE,
  7. 0:15IF WE SET THE LOGS EQUAL TO EACH OTHER,
  8. 0:18AS WE SEE HERE,
  9. 0:19THEN IT FOLLOWS THAT THE ARGUMENT OF THE FIRST LOG
  10. 0:22MUST EQUAL THE ARGUMENT OF THE SECOND LO.
  11. 0:24SO WE'LL SET THOSE EQUAL TO EACH OTHER AND THEN SOLVE.
  12. 0:28NOW, WE DO HAVE TO CHECK OUR ANSWERS THOUGH,
  13. 0:30BECAUSE THE ARGUMENT OF THE LOG
  14. 0:32HAS TO BE GREATER THAN ZERO.
  15. 0:34SO IF WE OBTAIN A VALUE THAT MAKES THIS LESS THAN
  16. 0:37OR EQUAL TO ZERO,
  17. 0:38WE WOULD TO EXCLUDE IT.
  18. 0:39LET'S GO AHEAD AND TRY A COUPLE OF THESE FIRST.
  19. 0:42SO NOTICE WE HAVE TWO LOGS OF BASE 5 EQUAL TO EACH OTHER,
  20. 0:46THEREFORE, IT FOLLOWS THAT 12 MUST EQUAL 2X - 3.
  21. 0:51SO NOW WE CAN SOLVE THIS FOR X.
  22. 0:53WE'LL ADD THREE TO BOTH SIDES.
  23. 0:56THAT WOULD GIVE US 15 = 2X
  24. 0:59AND DIVIDING BOTH SIDES BY 2, WE HAVE X = 15 HALVES,
  25. 1:04OR WE COULD SAY 7.5.
  26. 1:07NOW, LET'S GO AHEAD AND CHECK THIS,
  27. 1:09AND THIS ONE IS PRETTY STRAIGHT FORWARD.
  28. 1:112 x 7.5, THAT WOULD BE 15 - 3, THAT'D BE 12.
  29. 1:15SO WE HAVE LOG 12 ON THE RIGHT AND LOG 12 ON THE LEFT,
  30. 1:19SO THAT SOLUTION CHECKS.
  31. 1:22THIS NEXT ONE IS VERY SIMILAR,
  32. 1:23EXCEPT WE HAVE TO SET THE TWO LOGS
  33. 1:25EQUAL TO EACH OTHER FIRST.
  34. 1:26SO IF WE ADD LOG BASE 7 OF 10 TO BOTH SIDES OF THE EQUATION,
  35. 1:30WE WOULD HAVE LOG BASE 7 OF X SQUARED - 3X
  36. 1:33= LOG BASE 7 OF 10.
  37. 1:36AND IF THE LOGS ARE EQUAL TO EACH OTHER,
  38. 1:37THAN X SQUARED - 3X MUST EQUAL 10.
  39. 1:42WELL, NOW WE HAVE A QUADRATIC EQUATION.
  40. 1:43WE'LL SET IT EQUAL TO ZERO AND SEE IF IT FACTORS,
  41. 1:47SUBTRACTING 10 ON BOTH SIDES.
  42. 1:49THERE ARE FACTORS OF -10 THAT ADD TO -3.
  43. 1:53THAT WOULD BE X - 5 AND X + 2.
  44. 1:58SO FROM THIS WE HAVE A SOLUTION OF X = 5 AND X = -2.
  45. 2:06NOW, WE DO NEED TO CHECK THESE SOLUTIONS.
  46. 2:08LET'S GO AHEAD AND CHECK THE EQUATION IN THIS FORM.
  47. 2:11LET'S JUST MAKE SURE THIS IS EQUAL TO +10
  48. 2:13FOR BOTH OF THESE X VALUES.
  49. 2:15WHEN X = 5, WE'D HAVE 5 SQUARED - 3 x 5.
  50. 2:20WE'D HAVE 25 - 15 WHICH = 10, AND THAT CHECKS.
  51. 2:24AND NOT LET'S TRY X = -2.
  52. 2:27-2 SQUARED IS 4 - 3 x -2, THAT WOULD BECOME 6.
  53. 2:334 + 6 DOES = 10, SO BOTH SOLUTIONS CHECK.
  54. 2:37NOW, THERE ARE GUIDELINES FOR SOLVING
  55. 2:39THE MORE GENERAL TYPE OF LOG EQUATION
  56. 2:41AND HERE ARE THE STEPS.
  57. 2:43IF THERE IS MORE THAN ONE LOG,
  58. 2:44COMBINE THEM USING THE PROPERTIES OF LOGS.
  59. 2:47THEN WE WANT TO ISOLATE THE SINGLE LOG,
  60. 2:49AND THEN WE'RE GOING TO WRITE THE EQUATION
  61. 2:50IS EXPONENTIAL FORM, SOLVE, AND CHECK.
  62. 2:53THESE ARE THE PROPERTIES THAT WE DO HAVE TO BE AWARE OF.
  63. 2:55THE PRODUCT, QUOTIENT, AND POWER PROPERTY,
  64. 2:58AND SOMETIMES THE CASE OF BASE FORMULA.
  65. 3:00LET'S TAKE A LOOK AT SOME OF THESE PROBLEMS.
  66. 3:03IN ORDER TO ISOLATE THE LOG,
  67. 3:04WE'LL FIRST DIVIDE BOTH SIDES OF THE EQUATION BY 2.
  68. 3:08THAT'S GOING TO GIVE US LOG BASE 3 OF 3X = 3.
  69. 3:14SO NOW THAT WE HAVE A SINGLE LOG,
  70. 3:16WE'RE GOING TO REWRITE THIS IN EXPONENTIAL FORM.
  71. 3:19SO 3 TO THE POWER OF 3 MUST EQUAL 3X,
  72. 3:22AND NOW WE'RE GOING TO SOLVE THIS EQUATION
  73. 3:23AND THEN CHECK OUR SOLUTION.
  74. 3:25WELL, 3 TO THE 3RD WOULD BE 27.
  75. 3:29SO NOW WE DIVIDE BOTH SIDES BY 3, AND WE HAVE X = 9.
  76. 3:35LET'S GO AHEAD AND CHECK X = 9 IN THIS FORM OF THE EQUATION.
  77. 3:40SO IF X = 9, WE'D HAVE THE LOG BASE 3 OF 27 = 3,
  78. 3:44AND 3 TO THE 3RD DOES = 27, SO IT CHECKS.
  79. 3:50LET'S TRY ANOTHER.
  80. 3:53SO WE WANT TO COMBINE THESE TWO LOGS.
  81. 3:56SINCE THE TWO LOGS ARE BEING ADDED TOGETHER,
  82. 3:59WE CAN MULTIPLY THE NUMBER PART OF THE LOG.
  83. 4:02SO WE CAN COMBINE THESE INTO LOG BASE 5
  84. 4:05OF X - 2 x X + 2 = 1.
  85. 4:12LET'S GO AHEAD AND MULTIPLY THIS OUT.
  86. 4:15THIS IS GOING TO BE X SQUARED + 2x - 2X, THAT'S 0
  87. 4:20AND THEN -4 = 1.
  88. 4:24WE HAVE A SINGLE LOG,
  89. 4:25SO NOW WE'LL REWRITE THIS IN EXPONENTIAL FORM.
  90. 4:285 TO THE 1ST POWER MUST EQUAL X SQUARED - 4.
  91. 4:33LET'S GO AHEAD AND ADD 4 TO BOTH SIDES.
  92. 4:35WE'LL HAVE X SQUARED = 9.
  93. 4:39NOW WE CAN SQUARE ROOT BOTH SIDES OF THE EQUATION.
  94. 4:43REMEMBER, WE'RE GOING TO HAVE TWO SOLUTIONS FROM THIS.
  95. 4:46X = +/-3.
  96. 4:51LET'S GO AHEAD AND CHECK THOSE SOLUTIONS.
  97. 4:54WHEN X = 3, WE'RE GOING TO HAVE LOG BASE 5 OF 1,
  98. 4:58THAT'S 0,
  99. 5:00PLUS LOG BASE 5 OF 3 + 2, THAT'D BE 5.
  100. 5:04LOG BASE 5 OF 5 = 1, THAT'S TRUE, SO THAT SOLUTION CHECKS.
  101. 5:08X = -3 IS GOING TO BE A PROBLEM
  102. 5:10BECAUSE WHEN WE SUB IN -3 HERE FOR X,
  103. 5:13WE'RE GOING TO HAVE LOG BASE 5 OF -5,
  104. 5:16WHICH IS NOT IN THE DOMAIN OF THIS LOGARITHM,
  105. 5:17THEREFORE, IT'S NOT A SOLUTION.
  106. 5:19WE ONLY HAVE ONE SOLUTION TO THIS EQUATION.
  107. 5:22LET'S GO AHEAD AND TRY ONE MORE.
  108. 5:24NOW, WE WANT TO COMBINE THESE TWO LOGS,
  109. 5:26BUT IT'S A DIFFERENCE.
  110. 5:27REMEMBER, IF IT'S A DIFFERENCE,
  111. 5:29THEN THIS IS GOING TO END UP BEING THE LOG OF A QUOTIENT.
  112. 5:32WE'RE GONNA HAVE LOG BASE 16 OF X DIVIDED BY X - 1 = 1/4.
  113. 5:41WE HAVE A SINGLE LOG NOW,
  114. 5:43SO WE'LL REWRITE THIS IN EXPONENTIAL FORM.
  115. 5:4516 TO THE POWER OF 1/4 MUST EQUAL THIS QUOTIENT.
  116. 5:5216 TO THE 1/4 POWER IS EQUAL TO 2.
  117. 5:55WE CAN REWRITE 16 AS 2 TO THE 4TH
  118. 6:00WHICH WOULD GIVE US JUST 2 TO THE 1ST.
  119. 6:01THIS IS 2 MUST EQUAL X DIVIDED BY X - 1.
  120. 6:08WHAT I WOULD PROBABLY DO HERE IS PUT THIS OVER 1.
  121. 6:11WE HAVE A PROPORTION SO WE CAN CROSS MULTIPLY.
  122. 6:152 x X - 1 MUST EQUAL 1 x X WHICH IS X.
  123. 6:23SO NOW WE CAN SOLVE THIS EQUATION.
  124. 6:252X - 2 = X.
  125. 6:31LET'S GO AHEAD AND SUBTRACT 2X ON BOTH SIDES.
  126. 6:35-2 = -X.
  127. 6:38NOW WE CAN DIVIDE BY -1 IF WE WANT.
  128. 6:41X = 2.
  129. 6:46NOTICE WHEN X = 2
  130. 6:48WE HAVE LOG BASE 16 OF 2 - LOG BASE 16 OF 1.
  131. 6:53THIS WOULD BE 0.
  132. 6:5416 TO THE 1/4 = 2 SO THIS DOES CHECK.
  133. 6:59THE MAIN REASON FOR CHECKING THESE
  134. 7:00IS THE ARGUMENT OF THE LOG HAS TO BE GREATER THAN ZERO,
  135. 7:02AND IF IT'S NOT, IT WOULD NOT BE A SOLUTION.
  136. 7:06OKAY. I HOPE YOU FOUND THIS VIDEO HELPFUL.
  137. 7:08HAVE A NICE DAY.

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