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Can you solve this equation? watch

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    • Thread Starter
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    hi everyone

    i'm having difficulty in solving this equation. It's Q1b from C3 June 2007

    e^x + 3e^-x = 4

    The solutions are 0 and ln 3. How do you get this?

    I can get 0 by logic by saying that if 1 + 3 = 4, so the power that e is raised to should equal 1, hence the solution being 0 (e^0 = 1).
    But how would you solve this in a way to achieve ln 3 as another solution??

    Thanks soo much in advance
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    what would you multiply the equation by inorder to get a quadratic....
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    Multiply all terms by e^x, shift all terms over to one side of the equation, and you'll have yourself a lovely quadratic in disguise to solve for x.
    • Wiki Support Team
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    Wiki Support Team
    Multiply both sides by e^x. (You're allowed to do this because e^x is never 0.) Then substitute u = e^x, and you'll get a quadratic in u.
    • Thread Starter
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    thanks all of you so much , and thanks for all replying so quickly
    • Thread Starter
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    actually, i read your answers and thought 'oh that's easy' but im trying to do it and im getting stcuk. can someone give me a step or two please?

    my attempt:
    (e^2x) + 3e^-2x - 4e^x = 0

    let u = e^x

    u^2 - 3u^2 - 4u = 0
    -2u^2 - 4y = 0
    -2u^2 = 4u
    (divide by 2) -u^2 = 2u

    is this wrong already?
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    (Original post by sheena18)
    actually, i read your answers and thought 'oh that's easy' but im trying to do it and im getting stcuk. can someone give me a step or two please?

    my attempt:
    (e^2x) + 3e^-2x - 4e^x = 0

    let u = e^x

    u^2 + 3....?

    is this wrong already?
    mistake
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    X multiplied by e^-x doesn't give e^-2x it gives 1...
    • Thread Starter
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    oh yeah samson89 e^-x multiplied by e^-x is e^-x+x
    = e^0
    = 1
    gotcha (gosh im dumb)
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    thanks again (a final thanks! i've got the answers now yay)
    have a good evening!
 
 
 
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