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Solving Exponential Equations

  1. Sep 21, 2008 #1
    Can anyone show me how to solve the following equations:



    I would greatly appreciate your assistance.
  2. jcsd
  3. Sep 21, 2008 #2


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    It should be a bit easier now.
  4. Sep 21, 2008 #3
    Here is a general way to solve things like this
    1. [tex]2^{x-1}-2^{x} = 2^{-3}[/tex]
    Now factor out a [tex]2^{x-1}[/tex]
    [tex]2^{x-1}(1 - 2) = 2^{3} \rightarrow 2^{x - 1} = -\frac{1}{8}[/tex]
    You can take the logarithm at this point or you could notice that [tex]-\frac{1}{8}
    [/tex] is out of the range of [tex]2^{x}[/tex]

    Similar approach
    2. [tex]3^{x+1} + 3^{x} = 36 \rightarrow
    3^{x}(3 + 1) = 36 \rightarrow
    3^{x} = 9 \rightarrow
    x = 2 [/tex]
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