a=log3(2),b=log3(4)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/09 20:04:35
log a b=log b a,则:ab=?

应该有个a≠b吧,ab=1logab=lnb/lnalogba=lna/lnb因为logab=logba所以lnb/lna=lna/lnb(lnb)^2=(lna)^2lnb=lna或者lnb=-ln

我是这样做的log2(3)*log3(5)*log5(49)*log7(64)=log2(3)*log3(5)*2log

不知你的连锁公式是指什么此处可以用换底公式log2(3)*log3(5)*log5(49)*log7(64)=log2(3)*log3(5)*2log5(7)*6log7(2)=12*log2(3)*

log 243÷log9 =log3^5÷log3^2 =log3^5-log3^2 然后怎么写

log243÷log9=log3^5÷log3^2=5log3÷(2log3)=5/2=2.5

已知log3(2)=a,3^b=5,则用a,b表示log(3)根30为

=log3(5)原式=1/2*log3(30)=1/2*log3(2*3*5)=1/2*[log3(2)+log3(3)+log3(5)]=(a+b+1)/2

对数函数题 log a - log b = 3 log (c/2) 求 a = ______________

loga-logb=3log(c/2)loga/b=log(c/2)^3a/b=(c/2)^3a=b(c/2)^3=bc^3/8

若log3 2=a,log3 5=b,则log3 40=?若lg2=a,则lg5=?lg8=?lg25=?lg0.2=?

/>因为:log32=a,log35=b,所以:log340=log38+log35=3log32+log35=3a+b因为:lg2=a,所以:lg5=lg(10/2)=lg10-lg2=1-alg8

a=(4/3)^(0.5),b=log以3/4为底(log3 4)的对数 ,C=log以2为底sin(2Pi/5)的对数

A=2根号3/3A属于(2/3,1)B=log(3)*(log(3)*4)/log(4)*(log(3)*4)=2.8864663987286C=log2(sin72°)因为sin72°小于1所以lo

设a=log3π,b=log3根号3,c=log3根号2比较abc

点击图片就可以看清楚了,加油!

a=log3²,那么log3八次方-2log3六次方用a表示是?

log3八次方-2log3六次方=4log3²-2×3log3²=-2log3²=-2a

公式:log(a)(b)*log(b)(a)=?

log(a)(b)*log(b)(a)=1

log3(2)=a,3^b=5,用a,b表示log3(根号30)

因为3^b=5所以b=log(3)5所以log3(根号30)=1/2log3(30)=1/2log3(2*5*3)=1/2[log(3)2+log(3)5+log3(3)]=1/2(a+b+1)

log(8)a+log(4)b^2=5,log(8)b+log(4)a^2=7,求log(2)ab的值

因为log(8)a+log(4)b^2=5所以1/3log(2)a+log(2)b=5①因为log(8)b+log(4)a^2=7所以1/3log(2)b+log(2)a=7②由①②将log(2)a与

LOG3底2=A 3的B次方=5 ,试用A.B来表示LOG3底根号30

log3(2)=A3^B=5log3(5)=Blog3(根号30)=log3[30^(1/2)]=1/2*log3(30)=1/2*log3(3*10)=1/2*[log3(3)+log3(10)]=

已知a=2b>0,(1)化简log3(a-b)-1/2log3(a^2+5b^2)

很高兴您解答.图片可能不是很清晰.

已知a=2b>0,(1)化简log3(a-b)-1/2log3(a2+5b^2)

∵集合A={x|log3(x2-3x+3)=0}={1,2},B={x|mx-2=0}={2m},A∩B=B,∴B=∅,或B={1},或B={2}.当B=∅时,2m不存在,∴m

log(ab)=log(a)+ log(b)吗

对,一样正确再问:好快的回答。。那log(a)(X)=clog(b)(X)=d用c、d表示log(ab)(X)怎么做呢。。

2log(b)x=log(a)x+log(c)x

2log(b)x=log(a)x+log(c)x2lgx/lgb=lgx/lga+lgx/lgc把lgx约分2/lgb=1/lga+1/lgc=(lga+lgc)/(lgalgc)=lg(ac)/(l

log a b*log b c=log b b*log a c成立吗

用换底公式,左边=logab*logbc=(lgb/lga)*(lgc/lgb)=lgc/lga=logac右边=logbb*logac=1*logac=logac左边=右边所以,logab*logb