quad(x.*log(1 x),0,1)什么意思

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 09:14:44
函数log(x-1)的定义域?

真数大于0x-1>0x>1所以是(1,+∞)

log(2x)+log(3y)-log(2z) 怎么化简? log3(x^2-2x-6)=2 log(3x+6)=1+l

log(2x)+log(3y)-log(2z)=log(12xyz)log3(x^2-2x-6)=2=log3(9)x^2-2x-6=9x=-3x=5x=5使得x^2-2x-6

y=log 7 x+1

换底公式y=log7(x+1)=lg(x+1)/lg7=lg1001/lg7计算器或者查表所以原式=3.55

log(x)(100)-lgx+1=0

首先,把log(x)(100)变形为1/{log(10^2)(X)}PS:^2是平方的意思把平方提出来为1/1/2{log(10)(X)}=1/[1/2lgX]=2/lgX所以原式=2/lgx-lgx

对数的运算问题-log a 1/x=-(log a 1-log a x)=log a x

∵log(a)(m/n)=log(a)(m)-log(a)(n)∴-loga1/x=-(loga1-logax)∵log(a)(1)=0∴-(loga1-logax)=logax明教为您解答,如若满意

log(1/2) |x-1|>0

答:log1/2|x-1|>0则有:0

log a(1/2次方) x等于2log a

是的loga(1/2次方)x=logax^2=2logax

解不等式:log½(x²-1)>log½(2x²-x-3)

∴0<x²-1<2x²-x-3;x²-1>0;x²>1;x>1或x<-1;x²-x-2>0;(x-2)(x+1)<0;x>2或x<-1;∴x>2或x<

已知函数f(x)=log.(1-x)+log.(x+3)(0

a是底数吧?由题得f(x)=loga(1-x)(x+3)则得定义域为(-3,1)因为0

f(x)=log^2(x+1)/(x-1)+log^2(x-1)+log^2(p-x)的值域(负无穷,log^(p+1)

f(x)=log^2(x+1)/(x-1)+log^2(x-1)+log^2(p-x)定义域:(x+1)/(x-1)>0,且x-1>0,且p-x>0x<-1或x>1,且x>1,且x<p∴1<x<pf(

log(X+5)+log(X+2)=1

㏒(X+5)(X+2)=1,X^2+7X+10=10X=0或X=-7,当X=0时,符合题意,当X=-7时,㏒(X+5)无意义,舍去.∴X=0.

化简:1/log(3)x 1/log(4)x 1/log(5)x=……(答案是1/log(60)x

我没有算出来,但我肯定,一定不是1/log(60)x

log₂log₃log₄ X=log₃log₄logS

log₂log₃log₄X=0log₃log₄X=1log₄X=3X=64log₃log₄logS

已知log(1/7)[log(3)(log(2)x)]=0

log1/7[log3(log2x)]=0=log1/7(1)所以log3(log2x)=1log3(log2x)=log3(3)log2(x)=3x=2³x=8

f(x)=|log(a)(x)-1|+|2log(a)(x)|,求使f(x)<2的x范围,

令b=log(a)(x)则x=a^b则f(x)=|b-1|+2|b|当f(x)

y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x

因为:(1-2x+x^2)=(1-x)^2所以:log(4)(1-2x+x^2)=log(4)[(1-x)^2]对其使用换底公式(log(a)b=[log(c)b]/[log(c)a]),将以4为底,