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(shí)，(lgm)^1.9＜(lgm)^2.1，m=10時(shí)，lgm=1，(lgm)^1.9=(lgm)^2.1，1＜m＜10時(shí)，(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(dāng)x=7和x=9時(shí)對應(yīng)的兩函數(shù)值，由y=log_0.2x在(0，+∞)上單調(diào)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(guān)鍵是比較底數(shù)lgm與1的關(guān)系.若lgm＞1即m＞10，則(lgm)^x在R上單調(diào)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(diào)遞減，故(lgm)^1.9＞(lgm)^2.1.若lgm=1即m=10，則(lgm)^1.9=(lgm)^2.1.
(4)因?yàn)榈讛?shù)8、10均大于1，且10＞8，
所以log₈₅＞lg5＞lg4，即log₈₅＞lg4