Logarithms, HELP!

• Nathpoop

  7 Sep 09, 22:51

  How do i solve this two questions. THANKS!

  (a) e^x(2e^x - 1) = 10       [Ans:0.916] (SOLVED!)

  (b) log5(8y-6) - log5(y-5) = log4(16), the BOLDED numbers are the base [Ans: 7]

• wee_ws

  7 Sep 09, 23:23

  Hi,

    For (b):

    log_5 [(8y - 6)/(y - 5)] = log_4 (4^2) = 2.

    Then (8y - 6)/(y - 5) = 5^2, from which y can be determined.

    Thanks!

  Cheers,
  Wen Shih

• Nathpoop

  8 Sep 09, 01:41

  Another question, 3log_5 (y) - log_25 (y) = 10 [Ans: 625] How do i solve thanks!
• wee_ws

  8 Sep 09, 07:34

  Hi,

    Apply a change of base, e.g.

    log_25 (y) = (log_5 (y)) / (log_5 (25)),

    so that all terms on LHS follow the same base.

    Thanks!

  Cheers,
  Wen Shih