比较log3[2]和log4[3]大小
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换成以10为底的对数,lg3/lg2*lg4/lg3*lg5/lg4*lg2/lg5=1
log3+log16-log4-log12=log[(3*16)/(4*12)]=log1=0不需要底数,1对于任何底的对数都是0
由logab=1/(logba)得log31/5=1/log1/53log41/5=1/(log1/54)因为3log1/54即1/(log1/53)
当底数大于1时,真数大的对数大,则log32log412当底数在0到1之间时,真数大的对数小,则log0.52>log0.53当真数相同1时,底数在0到1之间时,底数大的对数大,则log0.52
alog23<b=log32<c=log46再问:为什么c>b再答:哦,开始以为同底呢,应该这样:b=log32<1a=log23>1c=log46>1a=log23=lg3/lg2c=log46=l
log3(2)/log4(3)=lg2*lg4/lg3*lg3=2(lg2)²/(lg3)²因为(lg2)²/(lg3)²=(log3(2))²>1所
log(a)(b)表示以a为底的b的对数.所谓的换底公式就是log(a)(b)=log(n)(b)/log(n)(a).log3^2*log4^9=(1g2/1g3)*(1g9/1g4)=(1g2/1
log1/3(5)=-log3(5)log3(1/4)=-log3(4)log4(1/3)=-log4(3)=-1/log3(4)log5(4)1所以有:2^0.7>log5(4)>log4(1/3)
[log4(3)+log8(3)]×[log3(2)+log9(2)]=[1/2log2(3)+1/3log2(3)]×[log3(2)+1/2log3(2)]=[5/6log2(3)]×[3/2lo
换底:log3(2)=lg2/lg3log4(2)=lg2/lg4lg1=0
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/
六分之五再问:过程,谢啦再答:log4(3)=1/2log2(3)log8(3)=1/3log2(3)故原式=5/6log2(3)Xlog3(2)=5/6
(log43+log83)*(log32+log42)→(1/2log23+1/3log23)*(log32+1/2log22)→[5/6log23]*[log32+1/2]→[5/6log23*lo
log3(4)>log3(3)=1log4(3)<log4(4)=1∴log3(4)>log4(3)
log1/2(1/3)=log2(3)=log2^2(3^2)=log4(9)>log4(8)
log6[log4(log3^81)]=log6[log4(log3^3^4)]=log6[log4(4)]=log6(1)=0
作差:M-N=log(3)[4]-log(4)[5]=[lg4/lg3-lg5/lg4]【换底公式】=[lg²4-lg3lg5]/[lg3lg4]因为lg3>0、lg4>0而:lg3+lg5
log4(20)=log4(5)+1,log3(12)=log3(4)+1,很明显log3(4)与log4(5)比较大小即可.log4(5)/log3(4)=log4(5)*log4(3)
log2为底0.4,log3为底0.4,log4为底0.4分别等于lg0.4/lg2,lg0.4/lg3,lg0.4/lg4,因为lg0.4小于0,lg2小于lg3小于lg4,lg2大于0,所以log