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math

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Solve
4^y -7(2^y) = 8

  • math -

    4^y = (2^2)^y = (2^y)^2
    so, we have a quadratic in 2^y:
    (2^y)^2 - 7*2^y - 8 = 0
    (2^y-8)(2^y+1) = 0
    y=3
    2^y is never negative, so 2^y+1=0 has no real solution.

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