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Homework Help: Calculus Natural Log

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data
    ln(3x+1) = 3-lnx


    2. Relevant equations
    Solve for x


    3. The attempt at a solution
    Well I put the ln on the left side
    ln(3x+1)+ln(x) = 3

    Then I combine them
    ln ((3x+1)(x)) = 3

    So I take e

    e^ln(x(3x+1)) = e^3

    I get x(3x+1) = e^3

    So I divide the x

    (3x+1)/(x) = (e^3)/(x)

    Multiply x by both sides
    3x+1 = e^3

    Subtract the 1
    3x = (e^3)-1

    Divide by 3
    x = ((e^3)-1)/3

    Can someone tell me if I made a mistake or did it totally wrong ?
     
  2. jcsd
  3. Sep 10, 2010 #2

    jgens

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    Gold Member

    When you divide by x, you should get 3x+1 = e3x-1. This is different than what you show and it should have an impact on the final answer.
     
  4. Sep 10, 2010 #3
    So I did what you said.
    3x+1 = e^3 x^-1

    3x-x^(-1) +1 =e^3

    factor out x(3-1^(-1))=e^3

    Divide both sides by 3-1^(-1)

    ans: (e^3)/ 2 But I still get it wrong
     
  5. Sep 10, 2010 #4

    jgens

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    Gold Member

    How did you get this? It doesn't follow from the equality you had before this.
     
  6. Sep 10, 2010 #5
    I subtracted the x^-1
     
  7. Sep 10, 2010 #6

    jgens

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    Gold Member

    You can't subtract it out like that. Remember that it's multiplied to e3. If you're not familiar with this, you should probably review your elementary algebra.

    Anyway, before this goes further, the approach that you're taking (dividing through by x) doesn't really work. You should be able to get the equation 3x2+x-e3 = 0 and then solve for x.
     
  8. Sep 10, 2010 #7
    I suppose I use the quadratic formula but I'm only suppose to get one answer

    I get (sqrt (12e^3))/(6) and -(sqrt (12e^3))/(3)
     
  9. Sep 10, 2010 #8

    jgens

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    Gold Member

    Think about which of those values make sense for log(x). The domain of the functions involved is very important.
     
  10. Sep 10, 2010 #9

    Mark44

    Staff: Mentor

    Yes, use the quadratic formula. Since your first equation has ln(3x + 1) and ln(x), x must be positive, which will probably eliminate one of the solutions you get from the quadratic formula.
     
  11. Sep 10, 2010 #10
    Nope the positive value is wrong too
     
  12. Sep 10, 2010 #11

    jgens

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    Gold Member

    That would be because you applied the quadratic formula incorrectly.
     
  13. Sep 10, 2010 #12
    Just to be sure c would be -e^3 right?
     
  14. Sep 10, 2010 #13

    jgens

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    Gold Member

    That's correct.
     
  15. Sep 10, 2010 #14
    Heres what i Have

    3x^2+x-e^3=0

    -b+-√((b^2) - 4ac)/2a

    -1+-√(1^2)-4(3)(-e^3))/2(3)

    -1+-√(1+12e^3)/6

    Then two possibilities but only need the positive value

    1st possibility = -1+1+√(12e^3)/6 = √(12e^3) / 6
     
  16. Sep 10, 2010 #15

    Mark44

    Staff: Mentor

    You still have a mistake. If 3x^2 + x - e^3 = 0,
    then
    [tex]x = \frac{-1 \pm \sqrt{1 + 12e^3}}{6}[/tex]

    One of these values is positive and the other is negative.
     
  17. Sep 10, 2010 #16
    The second value is (-1/3)+√(12e^3)/(6) so should I find the lcd
     
  18. Sep 10, 2010 #17

    Mark44

    Staff: Mentor

    How are you getting that?

    You have 1 + 12e^3 in the radical, and there's no way you can simplify that.
     
  19. Sep 10, 2010 #18
    I thought you can take the radical one

    -1-√(1+12e^3)/6

    I still get it wrong though
     
  20. Sep 10, 2010 #19

    Mark44

    Staff: Mentor

    The two solutions to the quadratic equation are
    x1 = [-1 + √(1+12e^3)]/6
    and x = [-1 - √(1+12e^3)]/6

    The first one above is positive; the second one is negative.

    I don't know what you mean by this:
     
  21. Sep 10, 2010 #20
    I thought I could separate √(1) and √12e^3.

    Those x-values do not give me the right answer unless I am suppose to do something after that. However, if I already solved for x it should be the answer though right?
     
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