A.Maths Qn (Indices & Log)

• wishboy

  29 Sep 08, 19:27

  this is a difficult qn (to me) i encountered when doing past yr papers
  can someone pls help me with this?

  
  Solve for x.

• skythewood

  29 Sep 08, 20:16

  bring one of the term over.

  cross multiply.

  you will get numbers to the power of the log thing one side,===> change of base

  x to the power of the of a log addition subtraction the other side. ===> solve it

   

  maybe you will get the answer. just doing this from visual, no have software to show the maths.

   

• Ahm97sic

  29 Sep 08, 20:16

  Hi Wishboy,

  This question is fun ie it makes use of 2 interesting (but not so commonly used)concepts to solve the question.

  The answer is x = 6.

  Thank you or your kind attention.

  Regards,

  ahm97sic

• wishboy

  29 Sep 08, 20:28

  Originally posted by skythewood:

  bring one of the term over.

  cross multiply.

  you will get numbers to the power of the log thing one side,===> change of base

  x to the power of the of a log addition subtraction the other side. ===> solve it

  maybe you will get the answer. just doing this from visual, no have software to show the maths.

  i got until

  

  mind guiding me on how to continue?

• skythewood

  29 Sep 08, 20:57

  i see you change the log from 4 to 2, 9 to 3.

  that got me thinking... possible to use A^2 - B^2 = (A+ B)(A-B) ?

  than you will have 2 answers

• skythewood

  29 Sep 08, 21:03

  no idea
• wishboy

  29 Sep 08, 21:15

  

  Ahm97sic, can u help me continue? T.T

• skythewood

  29 Sep 08, 21:21

  Originally posted by wishboy:

  

  Ahm97sic, can u help me continue? T.T

  
  can change log n 2 = (log 10 2)/(log 10 n) ? is this legal change of base?
• wishboy

  29 Sep 08, 21:28

  ya can change like tat
  but it is still a power, only way to bring down is to make the base the same as the number

• skythewood

  29 Sep 08, 21:31

   

• skythewood

  29 Sep 08, 21:40

  last resort:

  bring the power from the x side to the number side,

  change on base on the number side.

  not sure if this is legal though

• Ahm97sic

  29 Sep 08, 21:53

  Hi Wishboy,

  I do not know how to type the mathematical symbols eg the log n 4 in the forum.

  Do you have any email account that can receive attached file ? I have typed out the answer using Mathtype in Microsoft Word 2003. I can send you an email with the answer in the attached file.

  Or I can send the attached file to the moderator Eagle and he will kindly type out the answer in the forum.

  Thank you for your kind attention.

  Regards,

  ahm97sic

   

• Mikethm

  29 Sep 08, 21:57

  Mathtype expressions can be saved as gif i mean.

• Ahm97sic

  29 Sep 08, 22:21

  Hi Mikethm,

  Thank you for the information.

  Regards,

  ahm97sic

   

   

• wishboy

  29 Sep 08, 22:21

  ok i have received the file

  so the main step is to bring over the power

  thanks everyone for ur help

• Ahm97sic

  29 Sep 08, 22:22

  Hi Wishboy,

  I have sent the answer in an email with the attached file to your email account.

  I like this question. It is fun. Please post more.

  Thank you for your kind attention.

  Regards,

  ahm97sic

• Uncertain

  30 Sep 08, 14:10

  Equate (2/x) ^ log = (3/x) ^ log

  Then take log base n on both side.

  Simplify the equations to arrive at log n 6 (log n 2/3) = log n x (log n 2/3)

  Ans: x = 6

• Mega wolf005

  14 Nov 08, 01:42

  Can you send me the answer too cause i really don't get how to do it, my email is [email protected]    thank you very much

• secretliker

  14 Nov 08, 03:11

  (2/x) ^ logn(4) = (3/x) ^ logn(9)

   

  Square root both sides, not sure if this is legal.

   

  (2/x) ^ logn(2) = (3/x) ^ logn(3)

   

  Change the base at the powers to power 2.

   

  (2/x) ^ [log2(2)/log2(n)] = (3/x) ^ [log2(3)/log2(n)]

   

  Take power log2(n) to remove the denominator.

   

  (2/x) = (3/x) ^ log2(3)

   

  Take log2 on both sides.

   

  1 - log2(x) = log2(3) [log2(3) - log2(x)]

   

  1 - log2(x) = [log2(3)]^2 - log2(3) * log2(x)

   

  1 - [log2(3)]^2 = log2(x) [1 - log2(3)]

   

  Use the identity (A-B)(A+B) = A^2 - B^2

   

  1 + log2(3) = log2(x)

   

  log2(6) = log2(x)

   

  x=6

• Mikethm

  15 Nov 08, 15:49

  Just wondering... why no one want to take the easy way out?

  (2/x)^logn4 = (3/x)^logn9

  (2/x) = (3/x)^(logn9/logn4)

  (2/x) = (3/x)^1.58496

  (x^1.58496)/x = (3^1.58496)/2

  x^0.58496 = 2.85225

  x = 2.85225^(1/0.58496)

  x = 6.0000

  x = 6.00 (at worse minus 1 mark for unnecessary 3sf but who cares about 1 mark?)

   

• secretliker

  15 Nov 08, 19:13

  Because we want to solve it without using calculator. It's good practice to know more methods nevertheless.

• Mikethm

  16 Nov 08, 14:53

   

  Oh ok... then you can have this simple use-only-simple-rules-no-calculator solution then :)

  (2/x)^2logn2 = (3/x)^2logn3

  Taking log to base n on both sides

  logn(2/x)^2logn2 = (3/x)^2logn3

  2logn2[logn2-lognx] = 2logn3[logn3-lognx]

  logn2[logn2-lognx] = logn3[logn3-lognx]

  (logn2)^2 -(logn2)(lognx) = (logn3)^2 - (logn3)(lognx)

  (logn3)(lognx)-(logn2)(lognx) = (logn3)^2-(logn2)^2

  lognx[logn3-logn2] = (logn3-logn2)(logn3+logn2)

  lognx = logn3 + logn2

  lognx = logn6

  x =6

  
  Same method as uncertain's probably. But just for the benefit of those who need it step by step. Does not make use of uncommon concepts. Only make use of 2 simple concepts.

  1. Must have same base.

  2. Making whatever contain x the subject. (E Maths)