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Solving logarithm problem for Y

  1. Aug 23, 2008 #1
    (For those values of x for which a solution exists), solve the following equation for y

    3e3y-6 = 2x2-1

    What does it mean by the values of x exists??
    Last edited: Aug 23, 2008
  2. jcsd
  3. Aug 23, 2008 #2


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    Re: logarithm

    Because of the nature of the domain of the logarithm one can not evaluate [itex]y=\ln x[/itex] for all x. Specifically the domain of the logarithm is all positive numbers. Therefore, we say that no solution exists for [itex]y=\ln x[/itex] in the domain [itex]x\in\left(-\infty, 0\right][/itex].

    So the question is asking you to solve the equation for y, for all values of x which exist. I hope that makes sense.
    Last edited by a moderator: Aug 23, 2008
  4. Aug 23, 2008 #3
    Re: logarithm

    Say you have [tex]ln(x)=y[/tex].
    This could be seen as [tex]e^y=x[/tex]
    It is impossible to raise any real number to any power and have it equal 0 or any number below that (feel free to try, in fact, I encourage it). Since it can't exist as [tex]x\in(-\infty,0)[/tex], there are only certain numbers it can exist as.
  5. Aug 23, 2008 #4


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    Hi fr33pl4gu3! :smile:

    Simple answer:

    3e3y-6 can only be positive.

    So the equation doesn't work if 2x2-1 is negative or zero. :smile:
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