6log32*3log23
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log2(3)=alog3(2)=1/alog3(7)=blog3根号7(2根号21)=log根号63(根号84)=log3(84)/log3(63)={1+log3(7)+2log3(2)}/{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
利用换底公式.全部换成lg为底的.lg3/lg2*lg5/lg3*lg8/lg5=lg8/lg2=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
答案在图片上,用手机照的,效果差了点,将就着看吧
原式=(lg9/lg8)/(lg3/lg2)=(2lg3/3lg2)/(lg3/lg2)=2/3
(lg5)^2+2lg2+(lg2)^2+log2(3)log3(4)=(lg5+lg2)^2+log2(4)=1+2=3
函数y=log23(3x-2)的定义域为:{x|3x-2>0log23(3x-2)>0},解得{x|23<x≤1},故答案为:(23,1].
换底公式可得log3=alog2,log7=blog3,因此log7=ablog2.对log4256也运用换底公式可得log4256=log56/log42=log(2×2×2×7)/log(2×3×
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
log37^½=log33^(½b)=½b,不知这样表达有没有问题
∵1=log33>a=log32>log31=0,b=log52<log32=a,c=log23>log22=1,∴c>a>b.故答案为:c>a>b.
因为log89=log2^33^2=2/3log23,log32*log23=1那么,log32×log89=2/3