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```        SOLVE
log(x+4)/2{log2(2x-1)/(3+x)}<0```
7 years ago

147 Points
```										Dear

case 1    (x+4)/2 >1
x> -2
log(x+4)/2{log2(2x-1)/(3+x)}<0
so   0  <  log2(2x-1)/(3+x) <1
1  <  (2x-1)/(3+x) <  2
1  <  (2x-1)/(3+x)                           and         (2x-1)/(3+x) <2
x>4                                                    and         for all x
so solution is x>4

case 2    0<(x+4)/2 <1
-4 <x<-2

log(x+4)/2{log2(2x-1)/(3+x)}<0
so    log2(2x-1)/(3+x) >1
(2x-1)/(3+x) >  2
now again let  x+3 > 0
2x-1 > 6+2x
no solution
x+3 <0
x<-3
so   2x-1 < 6+2x
all x
so final solution is    -4<x<-3

so solution for x is     -4<x<-3   and x>4
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin

```
7 years ago
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