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Math Help - exponential problem

  1. #1
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    exponential problem

    Solve the following for x.
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  2. #2
    Senior Member Peritus's Avatar
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    \begin{gathered}<br />
  3^{4x} 3^{ - x - 2}  = 3^3 3^{ - 2x}  \hfill \\<br />
  3^{3x - 2}  = 3^{3 - 2x}  \hfill \\<br />
   \Leftrightarrow 3x - 2 = 3 - 2x \hfill \\<br />
   \Leftrightarrow x = 1 \hfill \\ <br />
\end{gathered} <br />
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  3. #3
    Member
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    Solution

    I think I got it:

    Rewrite all of the terms as base 3:

    9^(2x) = (3^2)^(2x) = 3^(4x)
    (1/3)^(x+2) = 3^(-x-2)
    27 = 3^3
    (3^x)^(-2) = 3^(-2x)

    Now rewrite the original equation:
    (3^(4x)) * (3^(-x-2)) = (3^3) * 3^(-2x)

    Now simplify by adding exponents as follows:
    3^(3x-2) = 3^(3-2x)

    Now, take the log of both sides to yield:

    (3x-2)*log3 = (3-2x)*log3

    Cancel log 3 from both sides. You are left with:
    3x-2 = 3-2x
    5x=5
    x=1

    If you substitute x = 1 back in it should check out.
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