log23乘log34乘log45
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/30 17:06:05
log(2)5=log(3)5/log(3)2log(5)3=log(3)3/log(3)5=1/log(3)5log(2)25乘log(3)4乘log(5)9=(2*log(2)5)(2*log(3
再问:第二步到第三步是怎么求的?再答:
log(2)9×log(3)4=2log(2)3×2log(3)2=4
log(14)56=[log3(56)]/[log3(14)]=[3log3(2)+log3(7)]/[log3(2)+log3(7)]=[(3/a)+b]/[(1/a)+b]=[ab+3]/[ab+
lg8×log2(5)×log5(4)换底公式:loga(b)=[logc(b)]/[logc(a)]=lg2^3×(lg5/lg2)×(lg4/lg5)=3lg2×(lg4/lg2)=3lg2×(l
125*³√5=5³*³√5=5^(3+1/3)=5^(10/3)所以原式=10/3
(log43+log83)*(log32+log92)=((1/2)*log23+(1/3)*log23)*(log32+(1/3)*log32)=(5/6)log23*(4/3)log32=(5/6
2*3+5*9行不?
(lg5)^2+2lg2+(lg2)^2+log2(3)log3(4)=(lg5+lg2)^2+log2(4)=1+2=3
换底公式我喜欢ln所以log2(9)=ln9/ln2=ln(3^2)/ln2=2*ln3/ln2log3(4)=ln4/ln3=ln(2^2)/ln3=2*ln2/ln3log2(9)*log3(4)
f(x)=log2(x/2)*log2(√x/2)=1/2[log2(x/2)]^2=1/2[log2(x)-1]^2=1/2[log2(x)]^2-log2(x)+1/2
8再问:是不是换成分数形式可以互相约掉再答:log2(25)*log3(4)*log5(9)=lg25*lg4*lg9/lg2*lg3*lg4=log2(4)*log3(9)*log5(25)=2*2
(Log(4)3+log(8)3)(log(3)2+log(9)2)=[(1/2)log2[3]+(1/3)log2[3]][log3[2]+(1/2)log3[2]]=(5/6)log2[3]*(3
具体过程都在这儿了,自己看吧!
(log23+log89)(log34+log98+log32)=(log827+log89)(log916+log98+log94)=log8243•log9512=lg35lg8×lg83lg32
你确定你的题目对吗,要补充好我才能回答哦再问:再答:2log310=log3100log3100+log30.27=log3(100*0.27)=log327=3
解题思路:考查幂的运算解题过程:解:(1)第一次撕成2片,即21;第二次撕成4片,即22;....所以第n次撕成2n,则2n=64所以n=6所以要把这张纸撕成64片需要6次(2)撕7次后这张纸变成27