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Math Help - Solving for x in an exponential equation

  1. #1
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    Solving for x in an exponential equation

    Hello, one question has proved ​ insurmountable to me. Here it is:
    (3/4)^x =(27/64)^(2x-3)

    Here is what I have tried:



    (3)^x/(4)^x=(27)^(2x-3)/(64)^(2x-3)

    Then i cross multiplied to get...

    (192)^(2x^2-3x)=(108)^(2x^2-3x)

    I know that I need to make the bases the same, but i can't see how to do that at this point. Some guidance would be greatly appreciated.
    Last edited by takaj; February 14th 2013 at 06:55 PM.
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  2. #2
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    Re: Solving for x in an exponential equation

    Quote Originally Posted by takaj View Post
    Hello, one question has proved ​ insurmountable to me. Here it is:
    (3/4)^x =(27/64)^(2x-3)

    \left(\frac{27}{64}\right)^{2x-3}=\left(\frac{3}{4}\right)^{6x-9}
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  3. #3
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    Re: Solving for x in an exponential equation

    Quote Originally Posted by Plato View Post
    \left(\frac{27}{64}\right)^{2x-3}=\left(\frac{3}{4}\right)^{6x-9}
    ?
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  4. #4
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    Re: Solving for x in an exponential equation

    Quote Originally Posted by takaj View Post
    ?

    The point is if \left(\frac{3}{4}\right)^a=\left(\frac{3}{4}\right  )^b then a=b.

    I showed how the bases are the same.
    Thanks from takaj
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  5. #5
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    Re: Solving for x in an exponential equation

    Thank you. I got it now.
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