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Math Help - Indices past paper question help

  1. #1
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    Indices past paper question help

    By using y=2^x or otherwise, solve

    4^x - 3 (2^(x+1)) + 8= o

    so i understand to convert the 4^x to 2^x(2) but how to proceed ??
    sorry cant type equations properly dont know what language this forums use and again sorry.
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  2. #2
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    Re: Indices past paper question help

    Quote Originally Posted by mathkid12 View Post
    By using y=2^x or otherwise, solve

    4^x - 3 (2^(x+1)) + 8= o

    so i understand to convert the 4^x to 2^x(2) but how to proceed ??
    sorry cant type equations properly dont know what language this forums use and again sorry.
    \displaystyle \begin{align*} 4^x - 3 \cdot 2^{x + 1}  + 8 &= 0 \\ \left( 2^2 \right) ^x - 3 \cdot 2 \cdot 2^x + 8 &= 0 \\ 2^{2x} - 6 \cdot 2^x + 8 &= 0 \\ \left( 2^x \right) ^2 - 6 \cdot 2^x + 8 &= 0 \\ X^2 - 6X + 8 &= 0 \textrm{ after making the substitution } X = 2^x \end{align*}

    You should now be able to solve.
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  3. #3
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    Re: Indices past paper question help

    thanks, thank you very very much!!!!!!!!1
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