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Find the solution if it exists.

  1. May 29, 2013 #1
    1. ln(5x+6)+ ln(x-2)=1



    2. Relevant equations: log(xy)= log(x)+ log(y)



    3. The attempt at a solution:

    ln(5x+6)+ ln(x-2)=1
    ln((5x+6)(x-2))=1
    ln(5(x^2)-14x-12)=1
    e^(ln(5(x^2)-14x-12)) = e^1
    5(x^2)-4x-12=e
    5(x^2)-4x-(e-12)=0

    use quadratic formula to find x and I get x=2.1617

    I don't think I am doing this right. Do you see where my concepts went wrong ?
     
    Last edited: May 29, 2013
  2. jcsd
  3. May 29, 2013 #2

    jedishrfu

    Staff: Mentor

    Your logic looks good, but the quadratic expansion should be 5x^2 + 6x - 10x - 12 = e

    Also when you brought e to the lefthand side that term should be - (12 + e)

    Giving 5x^2 - 4x - (12+e) = 0

    Now you can use the quadratic formula to find both x values.
     
  4. May 29, 2013 #3

    Sorry thats what I meant. It was a typo. So my x is a long decimal answer. Is there any way of finding a preciser and a more clear answer ?
     
  5. May 29, 2013 #4

    jedishrfu

    Staff: Mentor

    Well if you redefine e to be zero and change all of mathematics then maybe but as it stands now nope.
     
  6. May 29, 2013 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your 3rd and 4th lines are wrong, your 5th line is right, your 6th line is wrong. Magically, you still manage to get the correct answer! You really do need to be more careful. Mistakes like than on an exam can be very costly.

    BTW: an "exact" answer would be
    [tex] x = \frac{e + 16 + 2\sqrt{5e+64}}{8 + \sqrt{5e+64}},[/tex]
    where e is the base of the natural logarithms: e = 2.7182818284590452354 ... .
     
    Last edited: May 29, 2013
  7. May 29, 2013 #6
    sorry it was a typo. instead of 14x i meant to write 4x...

    thanks.
     
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