Make each side of the equation an exponent of (2x+3):
x^2<(2x+3)^1 is the same as x^2<2x+3 or x^2-2x-3<0.
(x-3)(x+1)<0. Therefore x-3<0 and x+1>0; or x-3>0 and x+1<0.
The first case is x<3 and x>-1 so -1<x<3 (x is between -1 and 3).
The second case makes x>3 and x<-1, which is impossible.
The solution is -1<x<3. But we have to eliminate x=0 because we can't have log0. Therefore the answer is:
-1<x<0 or 0<x<3. What about the base? Base=2x+3 so 1<base<3 or 3<base<9.
CHECK
Try base=2, so 2x+3=2, x=-1/2: log(x^2)=log(1/4)=log(2^-2)=-2 which is less than 1. So the inequality holds true.
Try base=5, so 2x+3=5, x=1: log(x^2)=log(1)=0 which is less than 1. So the inequality holds.
Try base=9, so 2x+3=9, x=3. We would not expect the inequality to hold because 9 is out of range. Log(x^2)=log(9)=1, which is not less than 1, and the inequality fails as expected.