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Question

Number of integers less than or equal to 10 satisfying the inequality log1/2(x1)131logx2x8 is

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Solution

log1/2(x1)131logx2x8
x1>0x>1,x2x>0 and x2x1x(x1)>0x<0 or x>1x2x1x2x10x1±52
x(1,){1+52}

Now,
log1/2(x1)131logx2x8log2(x1)13log2(x2x)33log2(x1)1log2(x2x)log2(x2x)log2(x1)31log2(x2x)(x1)31
log2x(x1)21[x>1]
As the base of the log is greater than 1, so
x(x1)22x2x24x+22x25x+20(2x1)(x2)0x12 or x2

Therefore, the required solution is,
x2
Hence, the required number of intergral solution is 9

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