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Math Help - exponents and e problem

  1. #1
    jut
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    exponents and e problem

    Here's the problem:

    7e^(-2t) - 5e^(-5t) + 4e^(-6t) = 0

    Solve for t.

    I've tried to solve it a few ways (taking the natural log) but each leads to a dead end.

    Please help!
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  2. #2
    Math Engineering Student
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    Factorise by e^{-6t}

    Then you tell us...
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  3. #3
    jut
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    I factored out a e^{-6t} but that didn't seem to help.

    I am left with :

    e^{-6t}*(7e^{4t}-e^{t}-4)=0


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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jut View Post
    I factored out a e^{-6t} but that didn't seem to help.

    I am left with :

    e^{-6t}*(7e^{4t}-e^{t}-4)=0

    Well as e^z \ne 0 \mbox{ for any z } , this reduces to:

    7e^{4t}-5 e^{t}-4=0

    putting y=e^t gives:

    7 y^4-5 y-4=0

    which by Descartes rule of signs we know has exactly one positive root, which is what
    we want if we are looking for a real solution to the original equation.

    RonL
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