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Math Help - [SOLVED] Checking my x solutions?

  1. #1
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    [SOLVED] Checking my x solutions?

    Hi again all,

    I've got:

    log(base3) (x) + log(base3) (x + 2) = log(base3) (8)

    I think I've solved it by turning into

    x^2 + 2x - 8 = 0

    so I get x = 2 and x = -4

    The x = -4 doesn't work in the original equation but it does work in subsequent steps in the equation (for example in log(base3) (x(x + 2)) = log(base3) (8))

    Does this matter or is this still not a solution?
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  2. #2
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    Quote Originally Posted by DannyMath View Post
    Hi again all,

    I've got:

    log(base3) (x) + log(base3) (x + 2) = log(base3) (8)

    I think I've solved it by turning into

    x^2 + 2x - 8 = 0

    so I get x = 2 and x = -4

    The x = -4 doesn't work in the original equation but it does work in subsequent steps in the equation (for example in log(base3) (x(x + 2)) = log(base3) (8))

    Does this matter or is this still not a solution?
    The domain is x > 0 because a negative log gives a complex number. Hence -4 is not in the domain so it cannot be a real solution
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