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Thread: Exponential equation with modulus

  1. #1
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    Exponential equation with modulus

    How to solve something like this?
    $\displaystyle |3x-2|^{\sqrt{2x-3x^2}}=1$
    $\displaystyle 2x-3x^2=0$ 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 $\displaystyle \pm 1$.

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


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


    edit: Not sure why 0 cannot be a solution because $\displaystyle f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1$
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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 $\displaystyle x=2/3$ isn't working is that this gives you $\displaystyle 0^0=1$, 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:

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

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

    Take logs of both, starting with:

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

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

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

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

    Interestingly, with my approach, the $\displaystyle x=0$ 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 $\displaystyle lg$?
    I also don't understand why x can't be = 0. Because it is included in the domain of the equation. $\displaystyle x\in[0;\frac{2}{3}]$
    And as e^(i*pi) wrote $\displaystyle f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1$
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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 $\displaystyle lg$?
    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 $\displaystyle f(0) = |3(0)-2|^{\sqrt{0-0}} = (2)^0 = 1$
    I also don't understand why x can't be = 0. Because it is included in the domain of the equation. $\displaystyle x\in[0;\frac{2}{3}]$
    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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