log4的3次方×log9的2次方 log2的4倍根号下32

log2√2+log927+4^log413=log22^(1/2)+log(3^2)(3^3)+13=1/2+3/2+13=15

log(2^n)(3^n)=n/nlog23=log23log(9)8^(1/n)=1/n*log98原式=(log23+log23+...+log23)*1/n*log98=nlog23*1/n*l

2^log4(2-√3)+3^log9(2+√3)=2^log2^2(2-√3)+3^log3^3(2+√3)=2^(1/2[log2(2-√3)])+3^(1/2[log9(2+√3)])=2^1/

=lg√2/lg9(lg3/lg4+lg3/lg8)=lg2/4lg3/(lg3/2lg2+lg3/3lg2)=1/4(1/2+1/3)=5/24

验证(log4＾3和log8＾3的关系,用log2^4和log2^8比较,好像是平方倍关系吧!后一个括号同理,然后两log相乘,就是前面乘前面,后面乘后面!再问：能写一下过程和结果吗？谢谢。再答：推理

公式loga²(b)=1/2loga(b)   loga(b)×logb(a)=1

[log4（3）+log8（3）]×[log3（2）+log9（2）]=[1/2log2（3）+1/3log2（3）]×[log3（2）+1/2log3（2）]=[5/6log2（3）]×[3/2lo

log4(9)=2*1/2log2(3)=log2(3)log3(4)=2log3(2)log9(2)=1/2log3(2)[log2(3)+log4(9)][log3(4)+log9(2)]=2lo

4^a=3公式1过程两边取lg()的对数：lg4^a=lg3a=lg3/lg4（公式1）公式2过程两边取log4()的对数：log4(4^a)=log4(3),又因为log4(4^a)=a*log4(

原式=(1/2*log2^3+1/3*log2^3)(log3^2+1/2*log3^2)=(5/6*log2^3)*(3/2*log3^2)={5/6*(lg3)/(lg2)}*{3/2*(lg2/

(log4(3)+log8(3))×(log3(2)+log9(2))log4(3)这里的(3)立方?

(log4^3+log8^9)*(log3^2+log9^16)=(log2^3/log2^4+log2^9/log2^8)*(log3^2+log3^16/log3^9)=(1/2log2^3+1/

=3(log4+log8)*2(log3-log9)=3(log32)*2(log1/3)=15(log2)*(-2)(log3)=-30(log2*log3)

2^log4(2-√3)+3^log9(2+√3)＝(2-√3)^log4(2)+(2+√3)^log9(3)＝(2-√3)^(1/2)+(2+√3)^(1/2)＝√6公式:a^logb(c)=c^l

求图

我也在做这个,做出来是-44/3（该死的预习作业啊!）原式=2^log2为底12-3^log3为底27+5^log5为底1/3=12-27+1/3不知道对不对,因为只在书上看来了换底公式...理解无能

原式=(log3(2)+1/2log3(2))*(2log2(3)+3log2(3))=3/2*5*log2(3)*log3(2)=15/2

【log4(3)+log8(3)】×【log3(2)+log9(2)】=[(1/2)log2(3)+(1/3)log2(3)]×[log3(2)+(1/2)log3(2)]=(5/6)log2(3)×

首先解释一下ln4=2ln2底数也有类似规律log4^3=1/2（log2^3）所以(log3^2+log9^4)×(log4^3+log8^3)=(log3^2+log3^2)×(（1/2）log2

(log32+log92)*(log43+log83)=(lg2/lg3+lg2/lg9)*(lg3/lg4+lg3/lg8)=(lg2/lg3+lg2/2lg3)*(lg3/2lg2+lg3/3lg