TITLE: Prove that $3^n$ is not $O(2^n)$
QUESTION [9 upvotes]: I have this question in my assignment. I need to prove, using only the definition of $O(\cdot)$, that $3^n$ is not $O(2^n)$. It is obviously true for any $n \geq 1$.
To prove  $3^n \in O(2^n)$, we must find $n_0$, $c$ such that $f(n) \leq c \cdot g(n)$ for all $n \geq n_0$.
$$
\begin{aligned}
3^n &\leq c \cdot 2^n\\
\left(\frac{3}{2}\right)^n& \leq c
\end{aligned}
$$
I am stuck here....
I could use a log here, but I don't see the use
Any hint?
UPDATE
$$
\begin{aligned}
3^n &\geq c \cdot 2^n\\
\left(\frac{3}{2}\right)^n& \geq c\\
n \log(3/2) &\geq \frac{\log c}{\log 3/2}\\
n &\geq \log \frac{2c}{3}
\end{aligned}
$$
For every $n \geq \log \frac{2c}{3}$, $3^n \geq 2^n$. Therefore, $3^n$ is not $O(2^n)$.

REPLY [2 votes]: Contradiction:   Assume: $${3^n} \in O\;({2^n})
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therefore:
$${3^n} \le c{2^n} \to \frac{{{3^n}}}{{{2^n}}} \le c \to {(\frac{3}{2})^n} \le c \to \log {(\frac{3}{2})^n} \le \log c \to n \le \frac{{\log c}}{{\log 1.5}}
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$$
Since 'c' is a positive constant,
$$\log c = Const. \to \frac{{\log c}}{{\log 1.5}} = Const. \to n \le Const.
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But we have to find:
$${n_0}\;{\rm{such}}\;{\rm{that}},{\rm{for}}\;any\;n \ge {n_0}\;(n \to \infty )\;{\rm{satisfies}}\;{\rm{our}}\;{\rm{inequality}}\;{\rm{(}}n \le Const.{\rm{)}}
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$${\rm{Since}}\;{\rm{for}}\;{\rm{any}}\;{n_0},\;n \to \infty \;\;{\rm{conflicts}}\;{\rm{with}}\;\;n \le Const.,\;{\rm{we}}\;{\rm{can}}\;{\rm{say:}}\;{{\rm{3}}^n} \notin O\;({2^n})
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$$