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Question

Sum of integers satisfying log2x112log2(x3)+2>0 is

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Solution

log2x112log2(x3)+2>0x>0 and log2x10x>0 and x2

log2x132log2x+2>0
log2x132(log2x1)+12>0
Let
log2x1=t0
t32t2+12>0
3t22t1<0(3t+1)(t1)<0
13<t<1
So,
0t<1

0log2x1<1
0log2x1<1
1log2x<2
2x<4
Hence, the integral values are 2,3
Sum =5




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