MathematicsEasy178×since 2002Q5311If the sum of the first 20 terms of the series log(71/2)x+log(71/3)x+log(71/4)x+...{\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...log(71/2)x+log(71/3)x+log(71/4)x+... is 460, then x is equal to :Ae²B7^1/2C7²D7^46/21Check answerSkip