Prove that $10^{340} < \frac{5^{496}}{1985}$ 
Prove that $10^{340} < \dfrac{5^{496}}{1985}$.

I said since $2^{13} < 10^{4}$, we see that $5 = \dfrac{10}{2} > 10^{\frac{9}{13}}$ and so $10^{340} < \dfrac{10^{343.38}}{1985} <\dfrac{5^{496}}{1985}$. But then I have to estimate  $\log_{10}{2} < 0.31$, which is pretty computational. Is there an easier way?
 A: $$2^{340}\cdot 5^{340} \lt \dfrac{5^{496}}{1985}$$
is equivalent to
$$2^{340}\lt \frac{5^{496-340}}{5\times 397}=\frac{5^{155}}{397}$$
which is equivalent to 
$$397\cdot 2^{340}\lt 5^{155}$$
Since $397\lt 1024=2^{10}$, it is sufficient to prove that
$$2^{350}\lt 5^{155}$$
which is equivalent to
$$2^{70}\lt 5\cdot 5^{30}$$
which is equivalent to
$$5\gt \left(\frac{2^7}{5^3}\right)^{10}=(1.024)^{10}$$
Now, $(1.024)^{10}\lt (1.1)^{10}=(1.21)^5\lt 1.3\times (1.3)^4\lt 1.3\times (1.7)^2=3.757\lt 5$.
A: $625=5^4>2^9=512$
$496-340 = 156 = 4\times 39 $
So $5^{156}>2^{351}$ and $5^{496}>10^{340}\cdot2^{11}>10^{340}\cdot 1985$
So $10^{340} < \dfrac{5^{496}}{1985}$ as required.
A: You can check with a calculator that 
$$
\frac{344}{153}<\frac{\ln{5}}{\ln{2}}
$$
then it follows:
$$
\begin{align*}
\frac{344}{153}&<\frac{\ln{5}}{\ln{2}}\\
344\ln{2}&<153\ln{5}\\
2^{344}&<5^{153}\\
2^{344} \cdot 5^{343}&<5^{496}\\
2000 \cdot 10^{340}&<5^{496}\\
10^{340}&<\frac{5^{496}}{2000}<\frac{5^{496}}{1985}.\\
\end{align*}
$$
A: Starting with
$$21^3=9261\lt10^4$$
and allowing ourselves $2^{11}=2048$, we have
$$2^{33}=2048^3\lt2100^3=21^3\cdot10^6\lt10^4\cdot10^6=10^{10}$$
which implies $10^{33}\lt5^{33}\cdot10^{10}$, or
$$10^{23}\lt5^{33}$$
From this we get
$$10^{345}=(10^{23})^{15}\lt(5^{33})^{15}=5^{495}$$
and thus
$$10^{340}\lt{5^{495}\over10^5}={5^{496}\over5\cdot10^5}\lt{5^{496}\over1985}$$
since $1985\lt5\cdot10^5=500{,}000$.
Remark:  If you want to keep everything checkable by eye, you can start instead with
$$21^3=3^3\cdot7^3=27\cdot49\cdot7\lt28\cdot50\cdot7=14\cdot100\cdot7=9800\lt10^4$$
