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Math Help - quadratic equation

  1. #1
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    quadratic equation

    hello i am trying to solve this equation for the values of x. can i get some help solving this one.
     <br />
3^{2x-1} -4(3^x) + 1= 0<br />
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sigma1 View Post
    hello i am trying to solve this equation for the values of x. can i get some help solving this one.
     <br />
3^{2x-1} -4(3^x) + 1= 0<br />
    Hi sigma1,

     3^{2x-1} -4(3^x) + 1= 0

     3^{-1}3^{2x} -4(3^x) + 1= 0

     \left(\frac{1}{3}\right)\left(3^{x}\right)^2 -4(3^x) + 1= 0

     \left(\frac{1}{3}\right)y^2 -4y + 1=  0

    y^2-12y+3=0

    where y = 3^x.
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  3. #3
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    Quote Originally Posted by undefined View Post
    Hi sigma1,

     3^{2x-1} -4(3^x) + 1= 0

     3^{-1}3^{2x} -4(3^x) + 1= 0

     \left(\frac{1}{3}\right)\left(3^{x}\right)^2 -4(3^x) + 1= 0

     \left(\frac{1}{3}\right)y^2 -4y + 1= 0

    y^2-12y+3=0

    where y = 3^x.
    thanks alot.. i finally have some perspective on this one
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