* 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.