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...
 A: $$2\log_{2x+3}x$$$$=\log_{2x+3}x^2$$$$=\frac{\log x^2}{\log (2x+3)}$$
So ${\log x^2\over\log(2x+3)}\lt1$ so we get $\log x^2\lt\log(2x+3)$
$\log$ is an increasing function so this implies $x^2\lt2x+3$
This should be simple to solve.
A: Use the formula
$$
\log_ba=\frac{\log a}{\log b}
$$
where in the right-hand side the base of the logarithm is irrelevant. I assume natural logarithms.
There are also some conditions to be satisfied at the outset:

*

*$x>0$

*$2x+3>0$

*$2x+3\ne1$
but they just boil down to $x>0$. Now your inequality is
$$
\dfrac{\log(x^2)}{\log(2x+3)}<1
$$
Since $x>0$, we have $2x+3>3$ and so $\log(2x+3)>0$. Hence you can safely remove the denominator getting
$$
\log(x^2)<\log(2x+3)
$$
that's the same as
$$
x^2<2x+3
$$
Since $x^2-2x-3=(x+1)(x-3)$ and $x+1>0$, your inequality is satisfied for $x<3$, so finally
$$
0<x<3
$$
A: 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:
$$
\begin{align}&x^2-2x-3<0\wedge x>0\\
\implies &x>0\wedge (x+1)(x-3)<0\\
\implies &x\in(0,3).\end{align}
$$
A: Working over ${\bf R}^{*}_{+}$, first we can re-write the LHS as $\displaystyle \frac{2\log(x)}{\log(2x+3)}$, since the fact $\log_{b}(a)=\frac{\log_{c}(a)}{\log_{c}(b)}$ for the change of basis in logarithms.
Then, the inequality can be written as $$\forall x\in {\bf R}^{*}_{+}:\quad 2\log_{2x+3}(x)<1\iff \frac{2\log(x)}{\log(2x+3)}<1.$$
Since $\log(2x+3)>0$ for $x>-1$ and we are assuming that $x\in {\bf R}_{+}^{*}$ so not problem with the denominator and it is positive with that assumption.  Thus, via $x\mapsto e^{x}$ we have
the inequality over ${\bf R}_{+}^{*}$ given by $$x^{2}<2x+3.$$
Remember that we are working over the interval $I:=\,]0,+\infty[$. But, the above inequality $x^{2}<2x+3$ can be written as $x^{2}-2x-3<0$ that is $(x-3)(x+1)<0$ with set solutions given by interval $J:=\,]-1,3[$.
Thus, the original inequality $2\log_{2x+3}(x)<1$ holds over the interval $I\cap J$, that is, the interval $]0,3[$.
