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Two Logarithm Concepts

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Ahm97sic

286 posts since Apr '08
- 04 Oct `08, 5:32PM
  
  There are two logarithm concepts that are not in the usual textbooks, guides or ten years series. However, these two concepts are already known to students and teachers.
  
  Question
  
  (1) Evaluate logn 9 / logn 4
  
  (2) Given that zy = 2^[log2 (z + x)], show that z = x / [y-1]
  
  [The number or letter in red is in subscript ie the base of the log].
  
  These two questions are not dificult as these two questions are just used to illustrate the two logarithm concepts where some students might not be aware.
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic
secretliker

255 posts since Dec '06
- 04 Oct `08, 5:45PM
  
  (1)
  
  logn 9 / logn 4
  
  = (lg n / lg 9) / (lg n / lg 4)
  
  = (lg n / lg 9) * (lg 4 / lg n)
  
  = lg 4 / lg 9
  
  = lg 2 / lg 3 ?
skythewood

4,297 posts since Jul '07
- 04 Oct `08, 5:57PM
  
  try this on calculator... log 9/log 4
  
  now try this.. ln9 / ln 4
  
  one of the first things i teach about logs
Ahm97sic

286 posts since Apr '08
- 04 Oct `08, 6:21PM
  
  Hi Secretkiller and Skythewood,
  
  The concept used in question 1 is
  
  logn 9 / logn 4 = ratio of the logarithms of any same base ie
  
  logn 9 / logn 4 = log10 9 / log10 4 = loge 9 / loge 4 = log2 9 / log2 4 and so on.
  
  [The number or letter in red is in subscript ie the base of the log].
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic
  
  Edited by Ahm97sic 04 Oct `08, 6:22PM
wishboy

1,269 posts since Aug '05
- 05 Oct `08, 2:34AM
  
  zy = 2^[log2 (z + x)]
  zy = z + x
  zy - z = x
  z(y - 1) = x
  z = x/(y - 1) (shown)
Ahm97sic

286 posts since Apr '08
- 05 Oct `08, 9:56AM
  
  Hi Wishboy,
  
  Well done, your answer is perfectly correct.
  
  The concept used in the second question is
  
  a^loga y = y.
  
  So, 2^[log2 (z + x)] = z + x, 3^ (log3 4) = 4, e^ (ln x^2) = x^2, 10^ (lg 7) = 7 and so on
  
  [The number or letter in red is in subscript ie the base of the log].
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic
  
  Edited by Ahm97sic 05 Oct `08, 10:32AM
xiaobai_23

49 posts since Jan '08
- 05 Oct `08, 11:15AM
  
  for 1st qns is it tat logn 9 / logn 4=logn9-logn4???
Ahm97sic

286 posts since Apr '08
- 05 Oct `08, 11:37AM
  
  Hi Xiaobai,
  
  It is logn (9/4) = logn 9 - logn 4. This rule is found in all the textbooks, guides and
  
                                                   ten years series
  
  But, logn 9 / logn 4 is not equal to logn 9 - logn 4.
  
  logn 9 / logn 4 = log10 9 / log10 4 = loge 9 / loge 4    This rule is not found in the
  
                                                                                 usual textbooks, guides and
  
                                                                                 ten years series. However,
  
                                                                                 many students and teachers
  
                                                                                 have already known about it.
  
  [The number or letter in red is in subscript ie the base of the log].
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic
dadeadman1337

69 posts since Feb '08
- 05 Oct `08, 12:07PM
  
  So whats the ans? 1.58?
FirePig

1,189 posts since Mar '08
- 05 Oct `08, 4:20PM
  
  Originally posted by secretliker:
  
  (1)
  
  logn 9 / logn 4
  
  = (lg n / lg 9) / (lg n / lg 4)
  
  = (lg n / lg 9) * (lg 4 / lg n)
  
  = lg 4 / lg 9
  
  = lg 2 / lg 3 ?
  
  wrong la.
skythewood

4,297 posts since Jul '07
- 05 Oct `08, 4:26PM
  
  Originally posted by dadeadman1337:
  
  So whats the ans? 1.58?
  
  yah, as shown by calculator
Moderator

eagle

18,012 posts since Aug '01
- 05 Oct `08, 4:34PM
  
  thanks ahm97sic
  
  used your question in ExamWorld... under A maths logarithms, with a backlink to this thread and a courtesy thanks to you :D
Ahm97sic

286 posts since Apr '08
- 05 Oct `08, 6:27PM
  
  Hi Eagle,
  
  You are welcome to include these questions in the examworld.
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic
cantnmousegame

4 posts since Jan '08
- 08 Oct `08, 11:31AM
  
  err how do i solve these two questions?
  
  (2x)^(lg2) = (7x)^(lg7)
  
  solve (log a X)^(log b X) = x where a, b are positive real numbers except 1. leave ur answers in terms of a and b.
  
  thx
Ahm97sic

286 posts since Apr '08
- 08 Oct `08, 1:28PM
  
  Hi cantnmousegame,
  
  Question 1
  
  Solve for x in the equation
  
  (2x)^(lg2) = (7x)^(lg7)
  
  Answer
  
  (2x)^(lg2) = (7x)^(lg7)
  
  (2^lg2)(x^lg2)=(7^lg7)(x^lg7)
  
  (x^lg2)/(x^lg7)=(7^lg7)/(2^lg2)
  
  x^(lg2-lg7)=(7^lg7)/(2^lg2)
  
  Bring over the power to the other side of the equation
  
  x = [(7^lg7)/(2^lg2)]^{1/(lg2-lg7)}
  
  Question 2
  
  Solve (log a X)^(log b X) = x where a, b are positive real numbers except 1, leave the answer in terms of a and b.
  
  Answer
  
  (loga x)^(logb x) = x
  
  Let x = b^(logb x)
  
  (loga x)^(logb x) = b^(logb x)
  
  Same power, equate base
  
  loga x = b
  
         x = a^b
  
  The number in red is in subscript ie the base of the log].
  
  The second question is interesting and fun. Please post more in the forum.
  
  Dear cantnmousegame, you should post these two questions in a new thread so that others can have the fun to solve the questions.
  
  Thank your for your kind attention.
  
  Regards,
  
  ahm97sic
proasd

1 post since Oct '08
- 10 Oct `08, 12:01PM
  
  continuation of (2x)^(lg2) = (7x)^(lg7)
  
  taking lg on both side,
  
  lg2(lg2+lgx) = lg7(lg7+lgx)
  lgx(lg2-lg7)=(lg7)^2-(lg2)^2
  lgx(lg2-lg7)=(lg7-lg2)(lg7+lg2)
  lgx=-(lg7+lg2)
  x=1/14
  
  solved without calc.
LatecomerX

2,331 posts since May '07
- 10 Oct `08, 12:14PM
  
  Originally posted by Ahm97sic:
  
  There are two logarithm concepts that are not in the usual textbooks, guides or ten years series. However, these two concepts are already known to students and teachers.
  
  Question
  
  (1) Evaluate logn 9 / logn 4
  
  (2) Given that zy = 2^[log2 (z + x)], show that z = x / [y-1]
  
  [The number or letter in red is in subscript ie the base of the log].
  
  These two questions are not dificult as these two questions are just used to illustrate the two logarithm concepts where some students might not be aware.
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic
  
  I'll just solve Qn 1 here since it seems easy to me.
  
  Set n = 4.
  
  log4 9 / log4 4
  = log4 9
  = 2 log4 3
LatecomerX

2,331 posts since May '07
- 10 Oct `08, 12:19PM
  
  = log2 3 / 0.5 (log2 4)
  = log2 3 / (0.5 * 2)(log2 2)
  = log2 3
LatecomerX

2,331 posts since May '07
- 10 Oct `08, 12:28PM
  
  I just realized the answer to Qn 1 can be summarized into two steps.
  
  logn 9 / logn 4
  = 2 logn 3 / 2 logn 2 (coefficients can be cancelled out)
  = log2 3 (reversal of "change of base" method)
  
  Edited by LatecomerX 10 Oct `08, 12:29PM
Ahm97sic

286 posts since Apr '08
- 10 Oct `08, 1:30PM
  
  Hi LatecomerX,
  
  I agree with your finding ie with the ratio of the logarithm of any same base,
  
  logn 3 / logn 2
  = log2 3
  
  Great, you have discovered a logarithm concept that is not found in the usual textbooks, guidebooks and ten years series ! Congratulation !
  
  This forum has not only helped the students to do their homework, it can also help to generate more ideas, new ways and methods, short-cut methods to solve maths questions. Please continue to contribute more to the forum.
  
  Thank you for your kind attention.
  
  Regards,
  
  ahm97sic

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