Question

(1)log_0.27和log_0.29；(2)log₃5和log₆5；(3)(lgm)^1.9和(lgm)^2.1(m＞1)；(4)log₈₅和lg4.

Answer 1

(1)log_0.27＞log_0.29；(2)log₃5＞log₆5；
(3)m＞10時，(lgm)^1.9＜(lgm)^2.1，m=10時，lgm=1，(lgm)^1.9=(lgm)^2.1，1＜m＜10時，(lgm)^1.9＞(lgm)^2.1；(4)log₈₅＞lg4.

  (1)log_0.27和log_0.29可看作是函數(shù)y=log_0.2x當x=7和x=9時對應的兩函數(shù)值，由y=log_0.2x在(0，+∞)上單調遞減，得log_0.27＞log_0.29.
(2)考察函數(shù)y=log_ax底數(shù)a＞1的底數(shù)變化規(guī)律，函數(shù)y=log₃x(x＞1)的圖象在函數(shù)y=log₆x(x＞1)的上方，故log₃5＞log₆5.
(3)把lgm看作指數(shù)函數(shù)的底數(shù)，要比較兩數(shù)的大小，關鍵是比較底數(shù)lgm與1的關系.若lgm＞1即m＞10，則(lgm)^x在R上單調遞增，故(lgm)^1.9＜(lgm)^2.1.若0＜lgm＜1即1＜m＜10，則(lgm)^x在R上單調遞減，故(lgm)^1.9＞(lgm)^2.1.若lgm=1即m=10，則(lgm)^1.9=(lgm)^2.1.
(4)因為底數(shù)8、10均大于1，且10＞8，
所以log₈₅＞lg5＞lg4，即log₈₅＞lg4.