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Solving exponential equations by logs

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data

    2^(2x) + 2 ^(x) - 12 = 0

    2. Relevant equations

    none really

    3. The attempt at a solution

    so I think what you have to do is factor it
    so it would be like

    (2^x- )(2^x + )

    then you set the factor equal to zero and solve for x but I'm not sure how to factor it.
  2. jcsd
  3. Nov 7, 2007 #2
    let u=2^x

    now can you factor it?
  4. Nov 10, 2007 #3
    Also notice that 2^2x = (2^x)^2
    as previously mentioned u = 2^x, (u > 0, because the exponential function is always positive) so you get u^2 + u - 12 = 0, which is easy to solve. Just don't forget that u > 0.
  5. Nov 10, 2007 #4
    also, it would serve you well to remember the graph of "e" and "ln"
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