wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
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




flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon