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Math Help - Isolating for x

  1. #1
    Junior Member NAPA55's Avatar
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    Isolating for x

    How would I go about isolating for x in this case?

    1 = -ln(x+1)
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  2. #2
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    TheEmptySet's Avatar
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    Quote Originally Posted by NAPA55 View Post
    How would I go about isolating for x in this case?

    1 = -ln(x+1)

    Write in expo form..

    1 =-ln(x+1) \iff -1=ln(x+1)

    e^{-1}=x+1 \iff e^{-1}-1 = x
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  3. #3
    Junior Member NAPA55's Avatar
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    Hmm... that's what I was doing and I got the same answer, but that's not what it's supposed to be. It should be 1.7 instead of -0.6.

    EDIT: I see why I'm wrong. I checked my previous work and I made a mistake. The ln should NOT be negative. Now it works and I get the answer I want.

    1 =ln(x+1)

    e^{1}=x+1

    e^{1}-1 = x

    1.7 = x
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