log23和log35
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 02:09:05
log(14)56=[log3(56)]/[log3(14)]=[3log3(2)+log3(7)]/[log3(2)+log3(7)]=[(3/a)+b]/[(1/a)+b]=[ab+3]/[ab+
利用换底公式.全部换成lg为底的.lg3/lg2*lg5/lg3*lg8/lg5=lg8/lg2=3
log35】9=1/log9]35log9]35=log9]7+log9]5log9]7=2log3]7=2n所以log35】9=1/(2n+m)
原式=(lg9/lg8)/(lg3/lg2)=(2lg3/3lg2)/(lg3/lg2)=2/3
根据函数单调性,(lg23)^x-(lg53)^x是减函数,所以x>-y
∵log32=m,log35=n,∴lg2lg3=m,lg5lg3=1−lg2lg3=n,1-lg2=nlg3,∴lg2=mlg3,∴1-mlg3=nlg3,∴lg3=1m+n,lg5=nlg3=n×
log9^5=a,log3^7=bLOG(3,7)=2LOG(9,7)=B==>LOG(9,7)=B/2lOG(35,9)=1/LOG(9,35)=1/(LOG(9,7)+LOG(9,5))=1/(B
(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].
log37=log27/log23所以log27=log23log37log37=log27/log23所以log27=log23log37公式logab=logcb/logca
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×
换底lg9/lg8=alg5/lg3=b相乘得ab=2lg5/3lg2则2lg5=3lg2xab1=lg10=lg2x5=lg2+lg52(1-lg2)=3lg2Xablg2=2/(2+3ab)
(log23+log89)(log34+log98+log32)=(log827+log89)(log916+log98+log94)=log8243•log9512=lg35lg8×lg83lg32
由log(3)7=a,log(23)=blog(3)7=log(2)7/log(2)3=a,则log(2)7=a*
log37^½=log33^(½b)=½b,不知这样表达有没有问题
log35^2=log35^14-log35^7=1/(a+b)-1/log7^35=1/(a+b)-1/(1+b/a)=(1-a)/(a+b)原式=(1+1-a)/(a+b)=(2-a)/(a+b)
lg3=alg2lg5=blg3lg20=xlg152lg2+lg5=x(lg3+lg5)2lg2+blg3=x(lg3+blg3)2lg2+ablg2=x(alg2+ablg2)2+ab=x(a+a
log23=lg3/lg2=b/a选A