log5[log4(log3 81)]-搜答案-智慧作业帮

log4(27)=3/2log2(3)=a,既log2(3)=2a/3.既lg2/lg3=2a/3,log5(2)=log2(2)/log2(5)=1/[-1+1+log2(5)]=1/[-1+log

/>2007*log5(1)+log4(27)*log3(4)=2007*0+(lg27/lg4)*(lg3/lg4)=lg27/lg3=lg3³/lg3=3lg3/lg3=3

换成以10为底的对数,lg3/lg2*lg4/lg3*lg5/lg4*lg2/lg5=1

好复杂哟,不过小可有方法设log5(3x+4x)=log4(5x-3x）=a则有5^a=7x,4^a=2x两式相除可得1.25^a=3.5a=log1.25(3.5)你可用计算器算出a的值,代入方程5

(log25+log41/5)(log52+log251/2)=(log25-log45)(log52-log252)=(lg5/lg2-lg5/lg4)(lg2/lg5-lg2/lg25)=1/2l

log2(3)*log3(4)*log4(5)*log5(2)=log2(3)*log3(2^2)*log2^2(5)*log5(2)=log2(3)*2log3(2)*1/2*log2(5)*log

1、logaC×logcA=logaC×(1/logaC)=12、log23×log34×log45×log52=[1/(log32)]×2log32×[1/(log54)]×log52=2×[1/2

原式=lg25/lg4-2lg10/lg4+lg9/lg2×lg√5/lg3×lg2/lg5=(2lg5)/(2lg2)-2/(2lg2)+(2lg3)/lg2×(½lg5)/lg3×lg2

log5^41所以log4^5>log5^4log5^3再问：是不是打太快了？log5^41是什么？我想知道详细点的再答：哪里有log5^41首先只有log4^5>1所以log4^5最大无悬念然后1>

log4(x)和log5(x)都是增函数所以P>log4(4)=11

log5^a这个的意思应该是以5为底的loga吧……应该不是5的a次方吧~要不就没有底数了……〔log4（log5^a）〕=〔log3（log5^b）〕=1log5^a=4,log5^b=3,上面两个

log4(log5a)=1log5a=4a=5^4=625同理,b=5^3所以a/b=5

log5{log4(log3x)}=0,{log4(log3x)}=1(log3x)=4x=81

log4\8=log4(2*4)=log4\2+1log5\10=log5(2*5)=log5\2+1log4\2>log5\2所以log4\8>log5\10

∵y=log5(x)是增函数,y=log5(5)=1.∴0

1、log23*log34*log45*log52=lg2/lg3xlg3/lg4xlg4/lg5xlg5/lg2（换底公式）=1（约分）2.(log43+log83)*(log32+log92)中间

log5[log4(log3 81)]=0

求值log5[log4(log3x)]=0解方程log4(log3x)=5^0=1log₃x=4¹=4x=3⁴=81证明：log5[log4(log381)]log5

log4(5)>10

2^[log2(3)*log3(4)*log4(5)*log5(6)*log6(7)*log7(8)]=(2^log2(3))*log3(4)*log4(5)*log5(6)*log6(7)*log7