Cho x > 0,,,x ne 1 thỏa mãn biểu thức frac1log _2x + frac1log _3x + ... +

$\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + ... + \frac{1}{{{{\log }_{1993}}x}} = {\log _x}2 + {\log _x}3 + ... + {\log _x}1993 = {\log _x}\left( {2.3....1993} \right) = M \Leftrightarrow {x^M} = 1993! \Leftrightarrow x = \sqrt[M]{{1993!}}.$