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?

$$2^{340}\cdot 5^{340} \lt \dfrac{5^{496}}{1985}$$
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, $(1.024)^{10}\lt (1.1)^{10}=(1.21)^5\lt 1.3\times (1.3)^4\lt 1.3\times (1.7)^2=3.757\lt 5$.
