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

Question

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?

Answer 1

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

Answer 2

You can check with a calculator that $$ \frac{344}{153}<\frac{\ln{5}}{\ln{2}} $$ then it follows: $$ \begin{align*} \frac{344}{153}&<\frac{\ln{5}}{\ln{2}}\\ 344\ln{2}&<153\ln{5}\\ 2^{344}&<5^{153}\\ 2^{344} \cdot 5^{343}&<5^{496}\\ 2000 \cdot 10^{340}&<5^{496}\\ 10^{340}&<\frac{5^{496}}{2000}<\frac{5^{496}}{1985}.\\ \end{align*} $$

Answer 3

$625=5^4>2^9=512$

$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.

Answer 4

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

$$21^3=3^3\cdot7^3=27\cdot49\cdot7\lt28\cdot50\cdot7=14\cdot100\cdot7=9800\lt10^4$$