problem: Describe the set of all real numbers x which satisfy 2 {\log_{{2x+3}^x}} <1.

solution: 2 {\log_{{2x+3}^x}} < 1

Now x< 0 is not in the domain of logarithm.

So x> 0.

Now as x>0           2x+3 > 1.

So {(2x+3)^a} > {(2x+3)^b} for a>b
So 2 {\log_{{2x+3}^x}} < 1 or {(2x+3)^{\frac{1}{2}}} > x
or 2x+3 > {x^2}
or {x^2} -2x -3 < 0
or -1 < x < 3

But x > 0 so set of all real number

x is 0 < x < 3.