Four real roots (Tomato subjective 67)

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.

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