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Solving Equation With Natural Log

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the Equation, Check for extraneous solutions.

    (ln x)^2 = ln (x^2)



    2. Relevant equations

    -None-

    3. The attempt at a solution

    I tried simplifying it down for myself but got stuck and don't know what to do next.

    (ln x)^2= ln x^2

    (loge x)^2 = loge x^2

    Please help! Thanks
     
  2. jcsd
  3. Jan 20, 2012 #2

    micromass

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    Use [itex]ln(x^2)=2ln(x)[/itex] to obtain a quadratic equation in ln(x).
     
  4. Jan 20, 2012 #3
    What do you mean quadratic equation?

    I know through what you are saying that I will get 0= ((ln x)^2)-2lnx right?

    would i plug it into the quadratic forumla or something like:

    x= (-(2ln) +/-√((2ln)^2 - 4(lnx)(0)))/2(lnx)

    ?
     
  5. Jan 20, 2012 #4

    Mark44

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    The equation is actually quadratic in ln(x). IOW, that means it's an equation in which ln(x) appears to, at most, the second power.
    You wouldn't be solving for x; you will be solving for ln(x), and you should get two values. For each value you can solve the resulting equation for x.

    Note that 2ln is meaningless, just as the symbol √ by itself is also meaningless.
     
  6. Jan 20, 2012 #5
    How do I solve for it? What would it be/look like?
     
  7. Jan 20, 2012 #6
    You had the right idea going, darshanpatel. If you follow Mark44's advice and instead of using the quadratic equation to find x, try to find ln(x). In your case, if you replace ln(x) = y or some other variable into your first equation in post #3, you would get:

    y^2 - 2y = 0

    If you solve for y using the Quadratic Equation, you should get two answers.

    After you get those two answers, replace back in ln(x) for y.

    Then solve for x with your two equations.

    Good luck!

    (I hope that made sense and uses correct logic. I am an engineering major, so I am trained toward using mathematics as a tool for design purposes. I love using math; I just am not trained in proofs or at in-depth mathematics period....at least, not yet)
     
  8. Jan 21, 2012 #7

    HallsofIvy

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    You surely don't need the quadratic formula to solve [itex]y^2- 2y= 0[/itex]! However, although the quadratic equation has two solutions, the original equation has only one solution.
     
  9. Jan 21, 2012 #8
    I get two perfectly viable solutions for the original equation from this process, HallsofIvy.
     
  10. Jan 21, 2012 #9
    You do get two solutions, HallsofIvy. You were correct, however, in saying that the Quadratic Formula is not necessary to solve the equation.
     
  11. Jan 21, 2012 #10
    so would it be like this:

    (ln x)^2=ln x^2

    (ln x)^2= 2ln x

    ln(x)=y

    y^2=2y

    y^2-2y=0

    y(y-2)=o

    y=0 and y=2

    as solutions?

    because if i plug this into the quadratic equation:

    y^2-2y=0

    where a:1 b:-2 and c: 0

    I still get: y=0 and y=2

    is that right?
     
  12. Jan 21, 2012 #11
    After I get y=0 and y=2, i plug it into lnx^2-2lnx=0 right?

    So plug in 0 and 2 for the x's?
     
  13. Jan 21, 2012 #12

    SammyS

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    You don't plug 0 and 4 in for x !

    You solve each equation for x: ln(x) = 2 and ln(x) = 0 .
     
  14. Jan 22, 2012 #13
    so would it be like for ln(x)=2

    e^ln(x)=e^2
    x=e^2
    x is about 7.389

    and ln(x)=0
    e^ln(x)=e^0
    x=1

    x= 1, 7.389

    right?
     
  15. Jan 22, 2012 #14
    Would that be the correct answer?
     
  16. Jan 22, 2012 #15
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