求不等式log1/2(x+1)≥log2(2x+1)的解集
问题描述:
求不等式log1/2(x+1)≥log2(2x+1)的解集
答
log1/2(x+1)
=[log2(x+1)]/log2(1/2)
=[log2(x+1)]/log2(2^-1)
=-log2(x+1)
=log2(x+1)^(-1)
=log2[1/(x+1)]
∴log1/2(x+1)≥log2(2x+1)
=>log2[1/(x+1)]≥log2(2x+1)
∴
{1/(x+1)≥2x+1
{x+1>0
{2x+1>0
解出-1/2