log2的5次方加log4的125次方
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原式=log2(12)-(1/2)log2(9)=log2(12)-log2(3)=log2(4)=2
原式=[log2底5+(log2底0.2)/2]*[1/2+(log2底0.4)/(log2底25)]=(log2底5)/2*[(log2底5)/(log2底5²)+(log2底0.4)/(
a=log45b=log22.2c=log0.50.3a=lg5/lg4=lg5/2lg2=1/2*lg5/lg2=lg√5/lg2√5>2.2所以lg√5/lg2>lg2.2/lg2=b所以a>bc
log2(12)-log4(9)=log2(4*3)-log2(3)=log2(4)+log2(3)-log2(3)=2log2(2)=2
(log2(5)+log4(125))(log3(2)/log√3(5))=(log2(5)+3/2log2(5))(log3(2)/2log3(5))=5/2log2(5)*1/2log5(2)=5
1、log2(4*5)-(2/2)log2(5)=log2(4)+log2(5)-log2(5)=2;2、换底公式:(lg3/lg2)*(lg5/lg2)*(lg5/lg3)=1;3、log2(5-l
log(2)5+log(2)1/10=log(2)(5×(1/10))=log(2)2=1
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
定义域x-1>0,x>1a(x-2)+1>0x-2>-1/ax>2-1/aa>10log2[a(x-2)+1](x-1)^2>a(x-2)+1x^2-2x+1>ax-2a+1x^2-(a+2)+2a>
先确定定义域-1再问:log4(1-x)=log2(1-x)^2这一步是这样吗?为什么我学的是log4(1-x)=1/2log2(1-x)那就应该等于log2(1-x)^(1/2)难道我公式记错了?再
(1)log(2)3.6,log(4)3.2,log(4)3.6log(2)3.6=log(4)12.96∵4>1∴log(2)3.6>log(4)3.6>log(4)3.2(2)6的0.7次方,0.
∵log2(x+1)-log4(x+4)=1,∴log4x+4(x+1)2=1,∴x+1>0x+4>0(x+1)2x+4=4,解得x=5,或x=-3(舍).故答案为:5.
∵log2(x+1)2+log4(x+1)=5,∴log4(x+1)4+log4(x+1)=5,∴log4(x+1)5=5,∴(x+1)5=45,∴x=3.故答案为:3.
(1/2)^(log2底5)=[2^(-1)]^(log2底5)=2^(-log2底5)=2^(log2底1/5)=1/52的(log2底5)次方]+3=5+3=87的(1-log2底3)次方————
log23·log34·log45·log52=lg3/lg2*lg4/lg3*lg5/lg4*lg2/lg5=1再问:用换底公式?再答:嗯,都换成以10为底的log23=lg3/lg2用e为底的ln
log2(x+1)+log0.25(x-1)>log4(2x-1)log4(x+4)/log4(2)+log4(x-1)/log4(1/4)>log4(2x-1)log4(x+1)^2-log4(x-
y=log4(2)+log4(√x)log4(√x)=y-1/2log2(x)=4y-2x=2^(4y-2)y=2^(4x-2)
log76log47所以log23>log47>log76
首先,不是log2的x次方,而是以2为底,X的对数