03 Dec, 10:21AM in sunny Singapore!
Home → Homework Forum →

Logarithm -maths

Subscribe to Logarithm -maths 2 posts

Please Login or Signup to reply.
  • bonkysleuth
    bonkysleuth's Avatar
    314 posts since Mar '07
    • 01 Sep `08, 3:47PM

      here's a logarithm question which i am not able to solve. i need the help of you guys out there.

       

      express the following in the form ln x = ax + b and find a and b

       

      (xe^x)^2 = 30e^-x

       

      answer's a = -1.5, b = 1.7

       

      here's what did...

       

      2x ln xe = -x ln 30e

      ln x + ln e = -x (ln 30 - ln e) / 2x

      ln x + 1 = -x (ln 30 - 1) / 2x

       

      after that i got stuck. ><

       

      P.S the above workings may not be correct for i'm unsure of whether i should bring down the -x on the right hand side.

      Edited by bonkysleuth 01 Sep `08, 3:48PM
  • Moderator
    eagle
    eagle's Avatar
    17,967 posts since Aug '01
    • 01 Sep `08, 3:51PM

      You went wrong on your first part

      "2x ln xe = -x ln 30e"

      That's not how logarthims work.

       

      Ans:

      (xe^x)^2 = 30e^-x
      2 ln(xe^x) = ln (30e^-x)
      2 ln x +2 ln (e^x) = ln 30 + ln (e^-x)
      2 ln x + 2x = ln 30 - x

      so
      2 ln x = -3x + ln 30
      ln x = -1.5x + 0.5 ln 30

      so a = -1.5, b = 0.5 ln30 = 1.7

       

      Regards,
      Eagle (Strategic Tuition)

Please Login or Signup to reply.