求log(1-2x)(3x 2)中x的取值范围求-搜答案-智慧作业帮

1.两边取对数,以根号2减1为底数x=log(根号2减1）22.可知：log2x+log2y=log2(xy)=2则xy=4,因为x,y≥0而（x+y)≥2根号（xy）=43.1+1/2lg9-lg2

设T＝x²－3x无限小到2／3为减2／3到无限大为增Y＝㏒1／3T0到无限大为减x²－3x＞0x＜0x＞3X减减得增0到无限小为增减增得减3到无限大为减

log(1/2,x+1)-log(1/2,9x^2-1)+11/4(x+1)/(9x^2-1)-1/4>0(4x+4-9x^2+1)/[4(9x^2-1)]>0(9x^2-4x-5)/(9x^2-1)

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

真数大于0-x2+2x+3>0x2-2x-3

不等式log（x-1）（x2-5x+10）＞2即log（x-1）（x2-5x+10）＞log(x-1)(x-1)²（i）当0＜x－1＜1即1＜x＜2时,得x²－5x+10

log(0.5)(x²-5x-6)>log2(1/2(x+6))log2[1/(x²-5x-6)]>log2[1/2(x+6)]1/(x²-5x-6)>1/2(x+6)x

[log(底数为a)x]=[1\2log(底数为a)b]-[log(底数为a)c][log(底数为a)x]=【log(底数为a)根号b】-[log(底数为a)c][log(底数为a)x]=log(底数

log(3x)²=log4(6-x)所以9x²=24-4x9x²+4x-24=0x=(-2±2√55)/9真数3x>0,6-x>0所以x=(-2+2√55)/9再问：lo

据题意得：1-2a>03a-1>0=>1/3x）因此可得：1-2a>3a-1=>a

log∨3∧（log∨2∧x）＝1log∨2∧x＝3x=2^3x^1/2=6^1/2

首先,令f(x)=x²-5x+6,则必须f(x)＞0,可得x＜2或x＞3所以f(x)在（-∞,2）单减,（3,+∞）单增,而y=log（1/2)f(x)的底数为1/2,所以y)在（-∞,2）

log(0.5)(1/根号5次2）=1/5log(0.2)1/5=1log(0.1)1=0,所以x=0.

本题是复合函数求单调性和值域问题令u=3+2x-x^23+2x-x^2>0解得-1

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

2(log(1/2)x)^2-14log(4)x+3≤02(log(2)x）^2-7log(2)x+3≤0=>1/2≤log(2)x≤2log(2)√2≤log(2)x≤log(2)8∴√2≤x≤8f

x-2x+3=(x-1)+2≥2定义域为x∈R--------------------------值域f(x)=log((x-1)+2)≥log（2）值域：[log2,+∞)(x-1)+2在（-∞,1

1.由底数x+3>0且不等于1；x2-4x+3>0得定义域为：(-3,-2))u(-2,1)u(3,+无穷)2.f(x)