设log(3)4log(4)3log(3) m=log(4)16,则m的值为

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用换底公式简化[log(4)3+log(8)3][log(3)2+log(9)2]

[log(4)3+log(8)3][log(3)2+log(9)2]=[1/2log(2)3+1/3log(2)3]*[log(3)2+1/2log(3)2]=5/6log(2)3*3/2log(3)

log(3)4*log(4)8*log(8)M =log(2)16 M是多少

用换底公式log(3)4*log(4)8*log(8)M=(lg4/lg3)*(lg8/lg4)*(lgM/lg8)=lgM/lg3log(2)16=log(2)2^4=4lgM/lg3=4lgM=4

1.设log(3为底)4·log(4为底)8·log(8为底)m=log(4为底)16,那么m等于?

1.log(3为底)4·log(4为底)8·log(8为底)m运用公式log(m为底)n*log(n为底)m=1得到上述式子=log(3为底)m=log(4为底)16=2所以m=92.3^a=5^b=

设a=1/log(4)3+1/log(7)3,求证a属于(3,4)

log(4)3=lg3/lg4log(7)3=lg3/lg7所以上式可以写成a=lg4/lg3+lg7/lg3=(lg4+lg7)/lg3alg3=lg4+lg7lg(3^a)=lg4+lg7等式两边

比较log[2] (3)和log[3] (4)的大小

log(2)3=lg3/lg2log(3)4=lg4/lg3log(2)3-log(3)4=lg3/lg2-lg4/lg3=[(lg3)^2-lg2lg4]/(lg2lg3)因为分母lg2lg3大于0

log(2)9×log(3)4

log(2)9=2log(2)3log(3)4=2log(3)2=》4log(2)3*log(3)2=>4*(log(2)3*(1/(log(2)3)))=4//由换底公式 log(a)(b

对数:log(4)8-log(1/9)3 求值

log(4)8-log(1/9)3=lg8/lg4-lg3/lg1/9=lg2×4/2lg2-lg3/(-2×lg3)=4/2+1/2=5/2

log(2)9log(3)4等于

答案等于:4

log(2)3=a,则log(6)4=(

换底公式a=lg3/lg2lg3=alg2所以原式=lg4/lg6=2lg2/(lg2+lg3)=2lg2/(lg2+alg2)=2/(1+a)

已知log(2)[log(3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x

log(2)[log(3)(log(4)x)]=0log(3)(log(4)x)=1log(4)x=3x=64log(3)[log(4)(log(2)y)]=0log(4)(log(2)y)=1log

(log(4)3+log(8)3)(log(3)2+log(9)8) 怎么个思路

原式=[log(2的平方)3+log(2的三方)3][log(3)2+log(3的平方)2的三方]=[0.5log(2)3+三分之一log(2)3][log(3)2+二分之三(3)2]=[六分之五lo

log底数2 真数3×log底数3 真数4×log底数4 真数5×log底数5 真数2

再问:怎么等出1再答:分子和分母约分,分子和分母都是lg2×lg3×lg4×lg5

化简log(2)3 * log(3)4 * log(4)7 *log(7)16

原式=(lg3*lg4*lg7*lg16)/(lg2*lg3*lg4*lg7)=lg16/lg2=4

有关对数的计算题(log(4)3+log(8)3)(log(3)2+log(9)2)=?

(log(4)3+log(8)3)(log(3)2+log(9)2)=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg

log下标2 3乘以log下标3 4

原式=(lg3/lg2)×(lg4/lg3)=lg下标24=2运用换底公式不懂可以向我们团队追问哦

(log(4)3+log(8)3)(log(3)2+log(9)2) 怎么个思路

用换底公式logab=lnb/lna(log(4)3+log(8)3)(log(3)2+log(9)2)=(ln3/(2ln2)+ln3/(3ln2))(ln2/ln3+ln2/(2ln3))=5/6

log(27)16/log(3)4=

使用换底公式log(27)16/log(3)4=(lg16/lg27)/(lg4/lg3)=(lg16/lg27)*(lg3/lg4)=(2lg4/3lg3)*(lg3/lg4)=2/3

对数计算1.{log(4)(3)+ log(8) (3)} X {log(3)(2)+log(9)(2)}2.{log(

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

4^log(3)(4)+4^-log(3)(4)怎么算,

log(3)(4)是真数为3,底为4吗?原式=3+4^log4(3^-1)=3+3^(-1)=10/3