prove that $\log_2(n+1)-\log_3(2n+1) \le c$ eventually How can I prove that there is $c>0$ such that starting from a specific $n$ the following is true:
$$
\log_2(n+1) - \log_3(2n+1) \le c
$$
 A: HINT
$$
\log_2(n+1) -\log_3(2n+1) \le \log_2(n+1)
$$
so suffices to prove that $\log_2(n+1) \le cn$ for some $c \in \mathbb{R}^+$, which is equivalent to proving
$$
n+1 = 2^{\log_2(n+1)} \le 2^{cn} = \left(2^c\right)^n = k^n
$$
for $k=2^c \in \mathbb{R}^+$.
Can you finish?

UPDATE
Now that you updated the question, it became a more interesting problem. One way is to consider the function
$$
f(x)
 = \log_2(x+1) - \log_3(2x+1)
 = \frac{\ln(x+1)}{\ln 2} - \frac{\ln(2x+1)}{\ln 3}
$$
over $x \ge 0$ and prove that the function is bounded by showing that it achieves a maximum or reaches a horizontal asymptote.
Achieving maximum (if true) can be shown by solving $f'(x)=0$ and computing $f''(x)$ to establish concavity with the 2nd derivative test.
Horizontal asymptotes are also part of standard calculus class.
Can you finish?

UPDATE 2
It appears this claim is not true. Note that as $x \to \infty$, we have
$$
\begin{split}
f(x)
 &\approx \log_2(x) - \log_3(2x) \\
 &= \log_2(x) - \log_3(x) - \log_3 2 \\
 &= \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 3} - \log_3 2 \\
 &= \ln x \left(\frac{1}{\ln 2} - \frac{1}{\ln 3}\right) - \log_3 2 \\
 &= a\ln(x) - b,
\end{split}
$$
where $a = \frac{1}{\ln 2} - \frac{1}{\ln 3} > 0$, so $f(x) \to \infty$ as $x \to \infty$. That means the difference between your logarithmic factors diverges with increasing $n$...
A: $\log_2(n+1)-\log_3(2n+1)=\dfrac{\log_3(n+1)}{\log_32}-\log_3(2n+1)=$
$=\log_23\cdot\log_3(n+1)-\log_3(2n+1)=$
$=\left[\log_2\left(\dfrac32\right)+1\right]\cdot\log_3(n+1)-\log_3(2n+1)=$
$=\log_2\left(\dfrac32\right)\cdot\log_3(n+1)+\log_3(n+1)-\log_3(2n+1)=$
$=\log_2\left(\dfrac32\right)\cdot\log_3(n+1)+\log_3\left(\dfrac{n+1}{2n+1}\right)>$
$>\log_2\left(\dfrac32\right)\cdot\log_3(n+1)+\log_3\left(\dfrac12\right)\to+\infty$
as $\;n\to+\infty\;.$
Hence, there does not exist any $\;c>0\;$ such that
$\log_2(n+1)-\log_3(2n+1)\le c\;$ eventually.
