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I need some math help

Discussion in 'Members' Lounge' started by Dustin_m, Dec 19, 2007.

  1. Dustin_m

    Dustin_m Active Member

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    I'm stuck on a problem here. It's not complicated, but i cant figure it out

    I need to solve this:

    e^(2x)+24=11e^x

    The thing that screws me up is the 24 in there. if i move that to the right side and the 11e^x to the left, is that correct? I would have e^(2x)-11e^x=-24 . Can i LN both sides of that?
     
  2. Citizen_Insane

    Citizen_Insane Senior Member

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    Ok, so first look at it like a regular trinomial. Remember that e^(2x) = (e^x)^2. Therefore:

    (e^x)^2-11e^x+24=0. Subsitute y = e^x:

    y^2-11y+24=0
    (y-8)(y-3)=0

    Your roots are 8 and 3. So, replacing y with e^x:

    e^x=8, and e^x=3
    x = ln(8) and x=ln(3)
     
  3. Dustin_m

    Dustin_m Active Member

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    Ahh, ok. I got it now.

    Thanks
     
  4. Citizen_Insane

    Citizen_Insane Senior Member

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