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solve the equation log(y)/log(2) + log(y)/log(4) = 6

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asked by dan
  1. logy/log2 = log_2(y)
    logy/log4 = log_4(y)

    since 2^2 = 4, log_2(y) = 2log_4(y) = log_4(y^2)

    so, now we have

    log_4(y^2*y) = 6
    log_4(y^3) = 6
    y^3 = 4^6
    y = 4^2 = 16

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    posted by Steve

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