C-04-Q131mediumsingle_mcqlogaMn=?\log_a \sqrt[n]{M} = ?loganM=?a1nlogaM\dfrac{1}{n} \log_a Mn1logaMbnlogaMn \log_a MnlogaMclogaM/n2\log_a M / n^2logaM/n2dlogaMn\log_a M^nlogaMnব্যাখ্যাThe nnn-th root is the power 1n\tfrac{1}{n}n1, so Mn=M1/n\sqrt[n]{M} = M^{1/n}nM=M1/n, and the power rule gives logaM1/n=1nlogaM\log_a M^{1/n} = \dfrac{1}{n}\log_a MlogaM1/n=n1logaM.