Stampboy123's maths thread

• Stampboy123

  12 Apr 11, 23:48

  For each of the following, find y in terms of x

  2 lg y = x-2

  ln(x+y)-4x = 0

   

  I appreciate if you can show the working.

• FireIce

  12 Apr 11, 23:50

  since u always have qns to ask

  u can have a dedicated thread for yourself to ask

  dun always open a new thread

• Stampboy123

  12 Apr 11, 23:58

  So how to edit the topic?

• Darkness_hacker99

  12 Apr 11, 23:58

  edited for you already 

• wee_ws

  13 Apr 11, 09:27

  Hi,

    Consider these useful results:

    1. lg y = x => y = 10^x

    2. ln y = x => y = e^x

    3. lg (a^n) = n lg a

    Thanks.

  Cheers,
  Wen Shih

• Stampboy123

  14 Apr 11, 16:50

  Given that � is the root of the equation x^2=2x+1, show that

  A) ð�œ¶^3 = 3ð�œ¶-2                                        B) ð�œ¶^4 - ð�œ¶^2 = 2ð�œ¶ - 2                 

  Please show working for the qn cause i need to learn through example. Thanks

• wee_ws

  14 Apr 11, 17:38

  Hi,

    What do the boxes represent? Thanks for clarifying.

  Cheers,
  Wen Shih

• Stampboy123

  14 Apr 11, 17:51

  it is alpha.

• wee_ws

  14 Apr 11, 23:15

  Hi,

    I believe something is not right about the two results.

   Given x^2 = 2x + 1, one can solve for alpha easily. alpha is either 1 + sqrt(2) or 1 - sqrt(2).

    One can verify via substitution that the two results do not hold.

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  14 Apr 11, 23:30

  Hi,

    Since α is a root to the equation, then α^2 = 2α + 1, which is useful for our later proof.

    One (correct) result we can show is α^3 = 5α + 2.

    Proof:

    LHS = α^3

    = α. α^2

    = α (2α + 1)

    = 2α^2 + α

    = 2(2α + 1) + α

    = 5α + 2

    = RHS.

    Thanks.

  Cheers,
  Wen Shih

• Stampboy123

  15 Apr 11, 01:09

  thanks wee_ws

  x^2+ax < b is only satisfied by -2<x<4

  Find value a and b

   

   

• wee_ws

  15 Apr 11, 08:27

  Hi,

    Now x^2 + ax - b < 0 is satisfied by (x + 2)(x - 4) < 0, so a and b can be found easily.

    Thanks.

  Cheers,
  Wen Shih