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Math Help - Monster problem

  1. #1
    Junior Member
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    Monster problem

    16^x - 4^(2x-2) + 2^(3x) = 4


    I've gotten to:

    2^4x - 2^2(2x-2) + 2^3x = 2^2
    I've gotten beyond lost after here...

    I've also tried dividing by 2^2 to get....


    2^(4x-2) - 2^(4x-6) + 2^(3x-2) = 1 but I'm not sure what I can do with this
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Intrusion View Post
    16^x - 4^(2x-2) + 2^(3x) = 4


    I've gotten to:

    2^4x - 2^2(2x-2) + 2^3x = 2^2
    I've gotten beyond lost after here...

    Very good! You can reduce that a little more
    2^{4x}- 2^{2(2x-2)}+ 2^{3x}= (2^x)^4- (2^{-4})(2^x)^4+ (2^x)^3= 4

    Now, let y= 2^x and the equation becomes
    y^4- \frac{1}{16}y^4+ y^3= 4
    or 17y^4+ 16y^3= 4

    but, frankly, I don't see any simple solution to that.

    I've also tried dividing by 2^2 to get....


    2^(4x-2) - 2^(4x-6) + 2^(3x-2) = 1 but I'm not sure what I can do with this
    Follow Math Help Forum on Facebook and Google+

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