求log(1-2x)(3x 2)中x的取值范围求
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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)