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2^x to Log base 2

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    The solution of 2^(2x+3) = 2^(x+1) + 3 can be expressed in the form of: a + log2b where a, b belong to the set of complex numbers.

    3. The attempt at a solution

    (2^3)(2^x)^2 = 2^(x+1) + 3

    [8((2^x)^2)] - [2(2^x)] - 3 = 0

    Solving the above quadratic for 2^x, I found that 2^x = 3/4 and -1/2. I can solve for x, but for simplicity's sake, I haven't yet.

    From here, I need to put x into the above log form.
     
  2. jcsd
  3. Sep 16, 2009 #2

    gabbagabbahey

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    Looks good so far, but unless x is allowed to be a complex number, you can throw one of these solutions away.....which one?:wink:

    In order to get 'x' into the log form, just take log base 2 of both sides of your equation for 2^x.... [itex]\log_2(2^x)=x[/itex]
     
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