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Math Help - Exponential equation with modulus

  1. #1
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    Exponential equation with modulus

    How to solve something like this?
    |3x-2|^{\sqrt{2x-3x^2}}=1
    2x-3x^2=0 doesn't work, because here x=2/3 and x=0, and the answer is 1/3...
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  2. #2
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    e^(i*pi)'s Avatar
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    The exponent will be 0 OR the base will be \pm 1.

    2x-3x^2 = 0 which is x=0. x cannot be 2/3 because it would give the undefined 0^0


    You can also solve |3x-2| = 1 - one of the solutions is 1/3


    edit: Not sure why 0 cannot be a solution because f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1
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  3. #3
    Forum Admin topsquark's Avatar
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    e^(i pi) beat me to it. 8]
    -Dan
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  4. #4
    Super Member Quacky's Avatar
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    The reason x=2/3 isn't working is that this gives you 0^0=1, which I don't think is defined. Anyway, here is my very poor and untidy method, which does give the right answer in a 'round-the-houses' approach:

    |3x-2|^{\sqrt{2x-3x^2}}=1
    Split it into:

    1) (3x-2)^{\sqrt{2x-3x^2}}=1
    and
    2) (2-3x)^{\sqrt{2x-3x^2}}=1

    Take logs of both, starting with:

    1) (3x-2)^{\sqrt{2x-3x^2}}=1

    \sqrt{2x-3x^2}\log(3x-2)=0

    Either 3x-2 = 1 (this gives log(1) = 0)
    In which case x = 1
    Or as we've already established, \sqrt{2x-3x^2}=0
    However neither work! The x=1 solution, when resubstituted, gives a complex value and is thus omitted. We've already gathered that the \sqrt{2x-3x^2}=0 doesn't work either. So then, we have to look at

    2) (2-3x)^{\sqrt{2x-3x^2}}=1
    Taking logs again:
    \sqrt{2x-3x^2}\log(2-3x)=Log(1)
    The only untested possibility is 2-3x=1
    In which case, -1=-3x
    x=\frac{1}{3}

    Interestingly, with my approach, the x=0 root indeed doesn't work - it gives negative values inside the logarithms, which are undefined, but that wasn't stated initially by the question. I assume whoever made the question took a similar approach as I did, or did something else, in which the 0 solution was lost along the way.

    I know my method isn't sound, but it does explain the textbook's answer.
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  5. #5
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    Quacky, I'm not sure what is the base for you logarithms? Do you mean lg?
    I also don't understand why x can't be = 0. Because it is included in the domain of the equation. x\in[0;\frac{2}{3}]
    And as e^(i*pi) wrote f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1
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  6. #6
    Super Member Quacky's Avatar
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    Quote Originally Posted by Evaldas View Post
    Quacky, I'm not sure what is the base for you logarithms? Do you mean lg?
    Yes, that's what I mean, although the base of the log doesn't matter in this case, as long as it's consistent.

    Quote Originally Posted by Evaldas View Post
    And as e^(i*pi) wrote f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1
    I also don't understand why x can't be = 0. Because it is included in the domain of the equation. x\in[0;\frac{2}{3}]
    I'm not sure either, as I said above, I supposed that someone followed a similar method as I did in which the 0 solution was accidentally dropped. Perhaps I didn't communicate my point very well.
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