Simple looking inequality I would to find the smallest possible constant $c$ that satisfies
$$\frac{3^{3k}e\sqrt{3}}{\pi\sqrt{k}2^{3/2+2k}} \leq 2^{ck}$$
assuming $k\geq 1$ is an integer.   
I tried taking logs base $2$ of both sides but that does not seem to be the right approach.  How can you solve this problem?
When you do take log of both sides base $2$ you end up with  this I think.
$$c \geq \frac{2+6k\ln{3} - 4k\ln{2} - 2\ln{\pi} + \ln{3} - 3\ln{2} - \ln{k}}{2k\ln{2}}$$
 A: Taking natural logarithm, your inequality is equivalent to
$$ck\log2\geq3k\log3+1+\frac12\log3-\log\pi-\frac12\log k-\biggl(\frac32+2k\biggr)\log2\,,$$
and so
$$c\geq\frac1{\log2}\Biggl[3\log3-2\log2+\frac{1+\frac{\log3}2-\log\pi-\frac{3\log2}2}k-\frac{\log k}{2k}\Biggr]\,.$$
Therefore 
$$\begin{align*}c=&\,\sup_{k\in\mathbb Z^+}\frac1{\log2}\Biggl[3\log3-2\log2+\frac{1+\frac{\log3}2-\log\pi-\frac{3\log2}2}k-\frac{\log k}{2k}\Biggr]\\
=&\,\frac1{\log2}\Biggl[3\log3-2\log2+\frac12\,\sup_{k\in\mathbb Z^+}\biggl(\frac{a-\log k}k\biggr)\Biggr]\,,
\end{align*}$$
where $a=2+\log3-2\log\pi-3\log2=\log\Bigl(\frac{3e^2}{8\pi^2}\Bigr)<0$.
If $f(x)=\frac{a-\log x}x$ for $x>0$, then $f'(x)=\frac{\log x-(a+1)}{x^2}$. Therefore 
$$f'(x)>0\iff\log x>a+1\iff x>e\cdot e^a=\frac{3e^3}{8\pi^2}\,.$$
Since $\frac{3e^3}{8\pi^2}<1$, it follows that $f$ is increasing on the interval $[1,\infty)$. Therefore
$$\sup_{k\in\mathbb Z^+}\biggl(\frac{a-\log k}k\biggr)=\lim_{k\to\infty}\biggl(\frac{a-\log k}k\biggr)=0\,,$$
so your required value of $c$ is
$$c=-2+\frac{3\log3}{\log2}\,.$$
A: Making use of the Maple command $$sol := solve(3^{3*k}*exp(1)*sqrt(3)/(Pi*sqrt(k)*2^{3/2+2*k}) = 2^{c*k}, c) ,$$ I obtain $$sol=1/2\,{\frac {-4\,\ln  \left( 2 \right) +6\,\ln  \left( 3 \right) }{
\ln  \left( 2 \right) }}+
 $$ $$1/2\,{\frac {2+\ln  \left( 3\,{\pi }^{-2} \right) +\ln  \left( {k}^{-1
} \right) -3\,\ln  \left( 2 \right) }{\ln  \left( 2 \right) k}}.
 $$
In view of $$2+\ln  \left( 3\,{\pi }^{-2} \right) -3\,\ln  \left( 2 \right)=-1.270289025 $$ and $$1/2\,{\frac {-4\,\ln  \left( 2 \right) +6\,\ln  \left( 3 \right) }{
\ln  \left( 2 \right) }}=2.754887504 $$ and taking into account that $\ln(\frac 1 k) \le0$ if $k \ge 1,$ I draw the conclusion that $c$ can be taken $2.754887504$.
