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A log question (probably easy)

  1. Sep 29, 2006 #1
    find the solution(s) to 2e^2x = e^x

    its on a practise exam im doing. i did with my calculator, but how could this be done without a calculator?
     
    Last edited: Sep 29, 2006
  2. jcsd
  3. Sep 29, 2006 #2

    Hurkyl

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    You certainly seem to be aware that logarithms ought to be useful -- so use them.
     
  4. Sep 29, 2006 #3
    ohh yeah like this?

    x=loge(2e^2x)
    x = loge2 + loge(e^2x)
    x = loge2 + 2xloge(e )
    x = loge2 +2x
    -x = loge2
    x= -loge2
     
  5. Sep 30, 2006 #4

    Hurkyl

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    That looks right.

    Now, there's another way to solve this problem too. :smile: It comes up often enough that you should know about it (or at least will know about it). e^(2x) is just (e^x)^2 -- so your equation is a quadratic equation in e^x... and you know how to solve quadratic equations.
     
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