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A. Maths Questions - surds and logarithms

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  • weewee
    weewee's Avatar
    406 posts since Aug '03
    • 28 Jun `08, 10:01PM

      I need help on these two questions.

      Q1) If a^(2x-1) = b^(1-3y) and a^(3x-1) = b^(2y-2), show that 13xy = 7x + 5y -3.

      Q2) Given that log5 (x) = 4 logx (5), calculate the possible values of x.

       

      For question 2, the first term is log with base 5  and the second term is log with base x.

  • jiaxing2
    jiaxing2's Avatar
    15 posts since Jun '08
    • 28 Jun `08, 10:46PM

      for qn 1

      multiply log base 10 on both sides of both eqn

      u will get (2x-1)/(1-3y) = lg b / lg a----(1)

                        (3x-1)/(2y-2) = lg b / lg a----(2)

      (1) = (2)

      thus                        4xy + 2 - 4x- 2y = 3x + 3y- 1- 9xy

      Hence 13xy = 5y + 7x - 3

       

      Qn 2)

      Re-express the eqn with lg base 10

      lg x / lg 5 = lg 625 / lg x   (5^4=625)

      (lg x)^2 = lg 5 ( lg 625 )

      solve for values of x using ur own calculator

       

  • weewee
    weewee's Avatar
    406 posts since Aug '03
  • secretliker
    secretliker's Avatar
    174 posts since Dec '06
    • 29 Jun `08, 2:35AM

      Q1) If a^(2x-1) = b^(1-3y) and a^(3x-1) = b^(2y-2), show that 13xy = 7x + 5y -3.

      (2x-1)(2y-2) = (1-3y)(3x-1)

      4xy-2y-4x+2 = 3x-9xy-1+3y

      13xy = 7x + 5y - 3

      Q2) Given that log5 (x) = 4 logx (5), calculate the possible values of x.

      log5(x) / log5(5) = 4 log5(5) / log5(x)

      [ log5(x) ] ^2 =  4 [ log5(5) ] ^2

      [ log5(x) ] = ±(2)

      x=5^(-2) or 5^2

      x=1/25 or 25

      Edited by secretliker 29 Jun `08, 2:38AM
  • weewee
    weewee's Avatar
    406 posts since Aug '03
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