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Math Help - [SOLVED] Log equation problem.

  1. #1
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    [SOLVED] Log equation problem.

    Hi guys,

    I'm doing exercises with exponential and logarithmic equations and I got to this one:

    e^x - 12e^-x - 1 = 0

    I was tempted to factor a e^x from the first two terms but guess I can't do that? I don't recognize this as a quadratic and frankly I'm not sure how to go about it. Any help is appreciated!
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  2. #2
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    Quote Originally Posted by DannyMath View Post
    Hi guys,

    I'm doing exercises with exponential and logarithmic equations and I got to this one:

    e^x - 12e^-x - 1 = 0

    I was tempted to factor a e^x from the first two terms but guess I can't do that? I don't recognize this as a quadratic and frankly I'm not sure how to go about it. Any help is appreciated!
    You are almost right. You should factorise  e^{-x} instead.
     e^{-x} (e^{2x} - e^x - 12) = 0

    Then substitute  a = e^x to get a quadratic
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  3. #3
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    Haha I've never factorised a negative exponent in such a manner! Ingenious! Thanks.
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  4. #4
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    Another way to do that is to multiply both sides of e^x- 12e^{-x}- 1= 0 by e^x to get (e^x)^2- 12- e^x= 0. Let a= e^x and the equation becomes the quadratic a^2- a- 12= (a- 4)(a+ 3).

    Now you have a= e^x= 4 so x= ln(4) and a= e^x= -3 which is not true for any x.
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  5. #5
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    Oh yes, that seems like a straightforward solution, thanks!
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