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Math Help - Finding the roots of this equation

  1. #1
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    Finding the roots of this equation

    \displaystyle y=e^{-3x} - e^{2x}

    I factored it to: \displaystyle y=(e^{-3x})(e^{2x}-1)

    \displaystyle e^{-3x} Never equals zero (right?)

    How would I solve for \displaystyle e^{2x}-1=0?

    Edit: I know the answer is x=0 by looking at it, but how would I solve a similar equation?
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  2. #2
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    Quote Originally Posted by IanCarney View Post
    \displaystyle y=e^{-3x} - e^{2x}
    I factored it to: \displaystyle y=(e^{-3x})(e^{2x}-1)
    The correct factorization is e^{-3x}\left(1-e^{5x}\right).
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  3. #3
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    Quote Originally Posted by Plato View Post
    The correct factorization is e^{-3x}\left(1-e^{5x}\right).
    Ah, yes. I wrote down the factored form of its derivative by accident >.<

    Here's what I'm trying to solve:

    e^{2x}=3

    Would I take the log of both sides? Because I get a different answer.
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  4. #4
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    Factor in the usual way. \left(e^{x}+1\right)\cdot\left(e^{x}-1\right)=0

    You'll have trouble with the fist piece, for reasons similar to that 3x you already threw out.

    You'll need a logarithm or an emperical rule for the last piece. Very informally, Number^{What} = 1?
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  5. #5
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    Quote Originally Posted by IanCarney View Post

    e^{2x}=3

    Would I take the log of both sides? Because I get a different answer.
    Natural Log: 2x = \ln(3)
    Last edited by e^(i*pi); April 11th 2011 at 01:28 PM. Reason: quoting relevant post
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  6. #6
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    Quote Originally Posted by IanCarney View Post
    \displaystyle y=e^{-3x} - e^{2x}

    I factored it to: \displaystyle y=(e^{-3x})(e^{2x}-1)

    \displaystyle e^{-3x} Never equals zero (right?)

    How would I solve for \displaystyle e^{2x}-1=0?

    Edit: I know the answer is x=0 by looking at it, but how would I solve a similar equation?
    y = e^(-3x) - e^(2x) = e^(-3x) (1 - e^(5x)) = 0
    1 - e^(5x) = 0
    1 = e^(5x)
    ln 1 = ln e^(5x) = 5x
    ln 1 = 0
    ∴ x = 0/5 = 0



    Moderator edit: Quote of OP has been added. This post answers the original question posted. However, it should be noted that in post #3 the original poster gave a new question.
    Last edited by mr fantastic; April 12th 2011 at 02:28 PM.
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