Any ideas with this logarithms system? Find $n\in\mathbb N$ for which exist $x,y\in\mathbb R$ such that $\log_2 x+\log_3(x-y)=n+1$ and $\log_2 y+\log_3(2x+y)=n$.
I have found the solution, but I don't know how to show that it is the only one:

 A: In fact, if $x,y\in\mathbb{R}$, then any $n\in\mathbb{N}$ can be obtained. If $x,y$ are only allowed to be in $\mathbb{Q}$, then $n=2$ is the only solution.
Subtract the two equations (this will be a necessary condition for the system to be solved):
$$\tag{1}\log_2{x}-\log_2{y}+\log_{3}{(x-y)}-\log_3{(2x+y)}=1$$
Thus:
$$\log_2{\left(\frac{x}{y}\right)}+\log_3{\left(\frac{x-y}{2x+y}\right)}=1$$
Re-write the second logarithm:
$$\log_2{\left(\frac{x}{y}\right)}+\log_3{\left(\frac{\frac{x}{y}-1}{\frac{2x}{y}+1}\right)}=1$$
Notice that this shows that if some $(x_1,y_1)\in\mathbb{R}^2$ satisfy equation $(1)$, then so will any $(x_2,y_2)$ with $x_1/y_1=x_2/y_2$. This suggests that we use the substitution $a=x/y$:
$$\log_2{a}+\log_3{\left(\frac{a-1}{2a+1}\right)}=1$$
(Observe that $a=4$ satisfies this equation.) Re-write the second logarithm:
$$\log_2{a}+\log_3{\left(\frac{1}{2}-\frac{3/2}{2a+1}\right)}=1$$
Now it is apparent that both logarithms are strictly-increasing, so the sum is strictly-increasing; thus, $a=4$ is in fact the only solution to the equation.
Let's plug in $x=4y$ into the original equations:
\begin{align}
\log_2{(4y)}+\log_3{(4y-y)}=3+\log_2{y}+\log_3{y}&=n+1 \\
\log_2{y}+\log_3{(8x+y)}=2+\log_2{y}+\log_3{y}&=n
\end{align}
Notice that your solution, $n=2$, corresponds to $y=1$ (and $x=4$).
Suppose that we want the $y$ that corresponds to a general $n$. Since $y$ is allowed to be any (positive) real number, we just need to solve the equation:
$$\log_2{y}+\log_3{y}=n-2$$
But $\log_2{y}$ and $\log_3{y}$ are strictly-increasing and range from $-\infty$ to $\infty$, so their sum does as well. Thus, there exists exactly one solution to this equation (though it is not only irrational but also transcendental).
