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log.23与log34比大小

解:log(2)3×log(3)4 =log(2)3×log(2)4/log(2)3 =log(2)4 =log(2)2² =2log(2)2 =2

应用对数换底公式得到哦 log2 3•log3 4•log4 5•log5 6•log6 7•log7 8 =(lg3/lg2)×(lg4/lg3)×(lg5/lg4)×(lg6/lg5)×(lg7/lg6)×(lg8/lg7) =log2 8 =log2 2³ =3

log23?log34?log45?log52=lg3?lg4?lg5?lg2lg2?lg3?lg4?lg5=1.故答案为:1.

∵log23?log34?log4m=log327,∴lg3lg2?2lg2lg3?lgm2lg2=32lg3lg3,化为lgm=32lg2,∴m=22.故答案为:22.

(log23+lOg427)(log34+log98) =(lg3/lg2+lg27/lg4)(lg4/lg3+lg8/lg9) =(lg3/lg2+3lg3/2lg2)(2lg2/lg3+3lg2/2lg3) =(5lg3/2lg2)(7lg2/2lg3) =(5/2)(7/2) =35/4

(log23+log29)×(log34+log38)=(log94+log92)(log6427+log649)=log9(4×2)log64(27×9)=lg8lg9?lg(27×9)lg64=3lg22lg3?5lg36lg2=54,故答案为:54.

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(1)原式=33×23-1×log22?3+lg3lg2×2lg2lg3=9-(-3)+2=14.(2)∵0<x<1,且x+x-1=3,∴(x12?x?12)2=x+x-1-2=1,x12<x?12,∴x 12-x ?12=-1.

(1)(2764)?13+(214)12+12log26=(6427)13+(94)12+12log26=43+32+16=3(2)(log23+log89)?(log34+log98+log32)=(log23+23log23)?(2log32+ 32log32+log32)=53log23?92log32=152

(18)?23+(log29)(log34)=382+(2log23)(2log32)=4+4=8.故答案为8.

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