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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.