Describe the set of real numbers $x$ which satisfy $2\log_{2x+3}x<1$. Describe the set of real numbers $x$ which satisfy $2\log_{2x+3}x<1$.
Here, I am really stuck . In this problem, I tried to compute by writing this expression as $(2x+3)^a=x^2$, where $a<1$.  But, I don't have a clue what to do ? I am not quite getting this...

General approach.
Notice that, the logarithm $\log_{2x+3}x$ is defined only by the following condition:
$$x>0\wedge 2x+3>0\wedge 2x+3≠1$$
This is equivalent to the condition $x>0$.
Remember that,

If $a>1$ and $b,c>0$ then the inequality $\log_a b<\log_a c$ is equivalent to
$b<c$ and if $0<a<1$ and $b,c>0$ then the inequality $\log_a b<\log_a c$ is equivalent to $b>c.$

Therefore, if $2x+3>1\wedge x>0$, then we have:
$$\begin{align}&\log_{2x+3}x^2<\log_{2x+3}(2x+3)\\
\iff &x^2<2x+3\end{align}$$
If $0<2x+3<1\wedge x>0$, then we have:
$$\begin{align}&\log_{2x+3}x^2<\log_{2x+3}(2x+3)\\
\iff &x^2>2x+3.\end{align}$$
However, we see that the inequality $0<2x+3<1\wedge x>0$ implies $x\in\emptyset$, therefore in this case we doesn't have a solution.
Thus, the original problem is reduced to the inequality $x^2<2x+3$ under the restriction $x>0$.
Finally we conclude:
$$
\begin{align}&x^2-2x-3<0\wedge x>0\\
\implies &x>0\wedge (x+1)(x-3)<0\\
\implies &x\in(0,3).\end{align}
$$
