Prove that $O(3^{2n}) \subseteq O(2^{3n})$

Question

I need to prove that $O(3^{2n}) \subseteq O(2^{3n})$. So far I have made this solution:

I)Lets assume that this is true, and that there $ \exists \space c \in \mathbb{R}^+ $ such as $$3^{2n} \leq 2^{3n} * c$$

II) therefore this fraction must be greater or equal as $1$

$$\frac{c*2^{3n}}{3^{2n}} \geq 1$$

III) lets look on $$\lim_{n\to\infty} \frac{c*2^{3n}}{3^{2n}}=(\frac{8}{9})^n=0$$ since the limit is equal to $0$ there occurs an conflict that shows us that II) can not ever be greater or equal as $1$. Is my solution correct? Any help is appreciated.

Answer 1

This is false.

Suppose it holds: then, since obviously $3^{2n} \in O(3^{2n}) \subseteq O(2^{3n}) \implies 3^{2n} \in O(2^{3n})$, $\lim_{n \to \infty}\, \sup |\frac{9}{8}|^n < \infty$, which obviously does not hold.

Answer 2

More generally, when is $a^{bn} \subseteq c^{dn}$?

$a^{bn} =(e^{\ln(a)})^{bn} =e^{\ln(a)b n} =(e^n)^{\ln(a)b } =(e^n)^{\ln(a^b) } $.

Therefore $a^{bn} \subseteq c^{dn} \iff \ln(a^b) \le \ln(c^d) \iff a^b \le c^d $.

Putting $a=3, b=2, c=2, d=3$, this is true iff $3^2 \le 2^3$, which is false.

For the case $c=b, d=a$, i.e., when is $a^{bn} \subseteq b^{an} $, this is the old problem of when $a^b \le b^a$, or $a^{1/a} \le b^{1/b}$.

It is well known that this is true when $e \le b \lt a$.

The case $a < e < b$ is trickier. I have a result that shows this is decided if $ab > e^2$ but I can't find it right now. I'll add to this if I find it.