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March 26, 2017

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Simplify exactly

log_2 1/16

  • calculus (math) - ,

    base^log x = x

    2^z = 1/16

    where z is log_2 (1/16)

    but we know that
    2^0 = 1
    2^-1 = 1/2
    2^-2 = 1/4
    2^-3 = 1/8
    2^-4 = 1/16 ah ha

    so 2^(-4) = 1/16
    z = -4
    and log_2 (1/16 ) = -4

    check:
    b^log_b(x) = x
    2^-4 = 1/16 ???
    1/2^4 = 1/16 sure enough

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