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Math Help - Solving Exponential Equations...

  1. #1
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    Solving Exponential Equations...

    Guys and Gals,

    I'm going over my old textbooks in preparation for going back to university, and one question has me totally stumped.

    I understand when encountering problems like $3^2^x-9(3^x)=0 that you substitute for the exponential e.g. $y=3^x and solve from there, but i've now been presented with:

    2^2^x-2^x^+^1+1=0

    Can anyone advise me how I go about solving this? I'll no doubt kick myself when I see it, but I just can't figure out what to do!
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  2. #2
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    Re: Solving Exponential Equations...

    Hello, cac2008!

    We have to "hammer" the equation into the proper form.


    Solve for x\!:\;2^{2x}-2^{x+1}+1\:=\:0

    We have: . 2^{2x} - 2\cdot2^x + 1 \:=\:0

    Let y \,=\,2^x\!:\;\;y^2 - 2y + 1 \:=\:0

    Got it?
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  3. #3
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    Re: Solving Exponential Equations...

    That was embarrasingly simple! I knew I was missing something elementary.

    Many thanks Soroban!
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