对数函数测试题一、选择题1.若5logxyz,则()A.5zyxB.5zyxC.5zyxD.5xyz答案:B2.对于0a,1a,下列命题中,正确命题的个数是()①若MN,则loglogaaMN;②若loglogaaMN,则MN;③若22loglogaaMN,则MN;④若MN,则22loglogaaMNA.0B.1C.2D.3答案:B3.若实数a满足4log15a,则a的取值范围是()A.40(1)5,,B.405,C.(01),D.(1),答案:A4.已知11loglog033ab,则ab,的关系是()A.1baB.1abC.01abD.01ba答案:D5.设函数1211()lg1xxfxxx,,,,≥若0()1fx,则0x的取值范围是()A.(010),B.(10),C.(2)(10),,D.(0)(10),,答案:B6.已知5()lgfxx,则(2)f()A.lg2B.lg32C.1lg32D.1lg25答案:D二、填空题7.已知lg2a,lg3b,那么3log6______.答案:abb8.计算522log253log648ln1______.答案:229.方程lg(42)lg2lg3xxx的解是______.答案:1x10.函数0.21xy的反函数是______.答案:0.2log(1)(1)yxx11.函数20.5log(43)yxx的定义域为______.答案:130144,,12.已知函数()lgfxx,则14f,13f,(2)f的大小关系是______.答案:11(2)43fff三、解答题13.已知1m,试比较0.9(lg)m与0.8(lg)m的大小.解:当lg1m,即10m时,0.90.8(lg)(lg)mm;当lg1m,即10m时,0.90.8(lg)(lg)mm;当0lg1m,即110m时,0.90.8(lg)(lg)mm.14.已知函数()yfx,12x,,若(2)sfs,求()fx的解析式.解:12x,,122s≤≤.设2st,则12t≤≤,且2logst.代入(2)sfs,得2()logftt.2()logfxx,12x,.15.已知22log(4)log(1)log5log(21)(01)aaaaxyxyaa,且,求8logyx的值.解:由已知,得22(4)(1)5(21)xyxy,即222244105xyxyxy,即2222(69)(44)0xyxyxyxy,即22(3)(2)0xyxy.3020xyxy,,故12yx.8811loglog23yx.16.已知()log(1)(01)xafxaaa,且,(1)求其定义域;(2)解方程1(2)()fxfx.解:(1)由已知条件,知10xa,即1xa.故当1a时,0x,当01a时,0x.即当1a时,函数的定义域为(0),∞,当01a时,函数的定义域为(0),∞.(2)令log(1)xaya,同1yxaa,log(1)yaxa,即1()log(1)xafxa.1(2)()fxfx∵,2log(1)log(1)xxaaaa∴,即211xxaa.2()20xxaa∴.2xa∴,或1xa(舍去).log2ax∴.