SOLVElog(x+4)/2{log2(2x-1)/(3+x)}<0

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

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