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Math Help - indices problem

  1. #1
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    indices problem

    solve the equation 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}

    anyone help!!
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  2. #2
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    First step

    Square both sides to get rid of the radical.
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  3. #3
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    dont get the answer
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  4. #4
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    Quote Originally Posted by mastermin346 View Post
    solve the equation 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}

    anyone help!!
    8^{2x-3}=\frac{1}{\sqrt{4^x+2}}

    \implies 8^{2x-3}=(4^x+2)^{-\frac{1}{2}}

    \implies 8^{-4x+6}=4^x+2

    \implies 8^{-4x+6}-4^x = 2

    Now, convince your self that 8^n = 4^{\frac{3}{2}n} \forall n.

    \implies 4^{\frac{3}{2}(-4x+6)}-4^x = 2

    \implies 4^{(-6x+9)}-4^x = 2

    Hmmmm, well here is the answer...

    Code:
    >> syms x;
    >> solve('8^(2*x-3) = (4^x + 2)^(-1/2)')
     
    ans =
     
    ((log(2)*log(2, z))/2 + pi*i*k)/log(2)
    ...and it's an ugly complex number, what do you know.
    Last edited by Anonymous1; May 8th 2010 at 06:22 PM.
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