【高二数学】解不等式log2 (x+1)+log1/4 (x-1)>log4(2x-1)
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【高二数学】解不等式log2 (x+1)+log1/4 (x-1)>log4(2x-1)
解不等式log2 (x+1)+log1/4 (x-1)>log4(2x-1)
说明:log2 (x+1)中底数为2,真数为x+1
log1/4 (x-1)中底数为1/4,真数为x-1
log4(2x-1)中底数为4,真数为2x-1
答案是{x|1
还有一题变式:
解不等式log2(x+1)+log1/2(x-1)>2
答案是{x|1
解不等式log2 (x+1)+log1/4 (x-1)>log4(2x-1)
说明:log2 (x+1)中底数为2,真数为x+1
log1/4 (x-1)中底数为1/4,真数为x-1
log4(2x-1)中底数为4,真数为2x-1
答案是{x|1
还有一题变式:
解不等式log2(x+1)+log1/2(x-1)>2
答案是{x|1
log2(x+1)+log1/4(x-1)〉log4(2x-1)
log4(x+1)^2-log4(x-1)〉log4(2x-1)
(x+1)^2/(x-1)>2x-1
x+1>0
x-1>0
2x-1>0
log4(x+1)^2-log4(x-1)〉log4(2x-1)
(x+1)^2/(x-1)>2x-1
x+1>0
x-1>0
2x-1>0
【高二数学】解不等式log2 (x+1)+log1/4 (x-1)>log4(2x-1)
解不等式log2(x+1)+log1/4(x-1)〉log4(2x-1)
解不等式log1/2^(x+1) +log2^[(1/(6-x)] 大于log1/2^12
解关于x的不等式 log2(x-1)>log4[a(x-2)+1] (a>1).
解不等式:[log4(3^x-1)]×[log1/4(3^x-1)/16]≤3/4
log4(x+3)+log1/4(x-3)=log1/4(x-1)+log4(2x+1),说明:
不等式log2(x+1)≤log4(1-x)的解集
log1/4(x+3)-log4(x-1)=log1/4(x+9)?
求不等式log1/2(x+1)≥log2(2x+1)的解集
解不等式log1/2(3-x)>=log2(x+3)-1
解方程log4(2-x)=log2(x-1)-1
解方程log1/2[log4(x^2-1)]=1