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](/uploads/image/z/1475882-26-2.jpg?t=%E6%B1%82%E4%B8%8D%E7%AD%89%E5%BC%8Flog1%2F2%28x%2B1%29%E2%89%A5log2%282x%2B1%29%E7%9A%84%E8%A7%A3%E9%9B%86)
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
=[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
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