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Math Help - Extraneous solutions, absolute value

  1. #1
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    Thumbs up Extraneous solutions, absolute value

    Hi Forum!
    I've just found something strange

    The closest integral of sum of the solutions to 27^{5x+3}=81^{|x+2|-7} is

    OK, calculating, we get two solutions

    \frac{-29}{11} and \frac{-45}{19}

    But the solution says that \frac{-29}{11} should be disconsidered.
    Why?
    The sum of them would give something about -5.
    Can someone explain?
    Does x=\frac{-29}{11} makes the function diverge?

    Thanks
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  2. #2
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    Re: Extraneous solutions, absolute value

    Does \displaystyle 27^{5\left(-\frac{29}{11}\right)+3} equal \displaystyle 81^{\left|-\frac{29}{11} + 2\right| - 7}?
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  3. #3
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    Re: Extraneous solutions, absolute value

    Quote Originally Posted by Prove It View Post
    Does \displaystyle 27^{5\left(-\frac{29}{11}\right)+3} equal \displaystyle 81^{\left|-\frac{29}{11} + 2\right| - 7}?
    Hi Prove It
    I get your point, then it is necessary to test the values.
    I thought there was something else to it; this is really simple.
    Thanks for your reply.

    All the best.
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