求不等式log1/2(x+1)≥log2(2x+1)的解集

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求不等式log1/2(x+1)≥log2(2x+1)的解集求不等式log1/2(x+1)≥log2(2x+1)的解集求不等式log1/2(x+1)≥log2(2x+1)的解集log1/2(x+1)=[

求不等式log1/2(x+1)≥log2(2x+1)的解集
求不等式log1/2(x+1)≥log2(2x+1)的解集

求不等式log1/2(x+1)≥log2(2x+1)的解集
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

求不等式log1/2(x+1)≥log2(2x+1)的解集。

log1/2(x+1) ≥ log2(2x+1)
log2[(x+1) - log2 (1/2)] ≥ log2(2x+1)
log2 [2(x+1)] ≥ log2 (2x+1)

2(x+1) ≥ 2x+1
2x+2 > 0 ,x > - 1
2x+1 > 0,x > - 1/2

解集为: (-1/2 ,+∞) 。