3log32-2(log34)(log827)
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原式=log252•log322•log532=2log25•2log32•2log53=8log25•log32•log53=8×lg5lg2×lg2lg3×lg3lg5=8.再问:但全部相约定之后
[3^(1/2)+1]^2=4+2(3)^(1/2),2+3^(1/2)=[3^(1/2)+1]^2/2,[2+3^(1/2)]^(1/2)=[3^(1/2)+1]/2^(1/2).[3^(1/2)-
8=log33^8就这么算.
【log2(3)+log4(9)+log8(27)+log16(81)+log32(243)】-5log2(3/2)=【log2(3)+log2^2(3^2)+log2^3(3^3)+log2^4(3
解题思路:考查对数式的化简解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/read
∵log32=m,log35=n,∴lg2lg3=m,lg5lg3=1−lg2lg3=n,1-lg2=nlg3,∴lg2=mlg3,∴1-mlg3=nlg3,∴lg3=1m+n,lg5=nlg3=n×
:(log43+log83)(log32+log92)=(lg3lg4+lg3lg8)•(lg2lg3+lg2lg9)=(lg32lg2+lg33lg2)•(lg2lg3+lg22lg3)=3lg3+
(lg5)^2+2lg2+(lg2)^2+log2(3)log3(4)=(lg5+lg2)^2+log2(4)=1+2=3
a=log32
(1)原式=(lg5)2+2lg2-(lg2)2+2log23•2log32=(lg5)2+2lg2-(lg2)2+4=(lg5+lg2)(lg5-lg2)+2lg2+4=lg5-lg2+2lg2+4
a=30.5>30=1,0=log31<b=log32<log33=1,c=cos23π=-cosπ3<0,∴c<b<a.故答案为:c<b<a.
(log23+log89)(log34+log98+log32)=(log827+log89)(log916+log98+log94)=log8243•log9512=lg35lg8×lg83lg32
(log32+log92)•(log43+log83)=(log32+12log32)•(12log23+13log23)=log3232•log2356=32×56=54故答案为:54
(log43+log83)(log32+log92)=(log23/log24+log23/log28)(log32+log32/log39)=(log23/2+log23/3)(log32+log3
-2(log34-1)=-3LOG33/4所以等于答案为(3/4)的-2次这步不对噢,-2(log34-1)=-2(log34-log33)=-2(log34/3)=log3(4/3)负二次方与前面的
log34与log43互为倒数log43=1/log34=log34的负一次=-log34log34+log43=log34-log34=0最后答案为0
因为log89=log2^33^2=2/3log23,log32*log23=1那么,log32×log89=2/3