Compare $2^{2016}$ and $10^{605}$ without a calculator So, I am supposed to compare $2^{2016}$ and $10^{605}$ without using a calculator, I have tried  division by $2$ on both sides  and then comparing $2^{1411}$ and $5^{605}$, and then substituting with $8,16,10$ and then raising to powers and trying to prove that but that did not go anywhere, I have also tried taking $\log$ of both sides, but did not help either. Also is there a more general approach to these kind of problems?
 A: $$2^{2016}=(2^{10})^{201}\cdot2^6=1024^{201}\cdot64$$
$$10^{605}=(10^3)^{201}\cdot10^2=1000^{201}\cdot100$$
Hence by Bernoulli's Inequality, $$\frac{2^{2016}}{10^{605}}=\left(\frac{1024}{1000}\right)^{201}\cdot\frac{64}{100}=1.024^{201}\cdot0.64>(1+201\cdot0.024)\cdot0.64>5.8\cdot0.64>1$$ so $$\boxed{2^{2016}>10^{605}}$$
A: $\log_{10}2^{2016}=2016\log_{10}2\approx 606.88>605$
If calculators are not allowed, we have
\begin{align*}
2^{2016}&=64(1000+24)^{201}\\
&>64\left[1000^{201}+\binom{201}{1}1000^{200}(24)\right]\\
&=64(10^{603})(5.824)\\
&>100(10^{603})\\
&=10^{605}
\end{align*}
A: \begin{aligned}
2 ^ {2016} > 10 ^ {605} & \iff \\
e ^ {\ln (2 ^ {2016}) } > e ^ {\ln (10 ^ {605}) } & \iff \\
\ln (2 ^ {2016}) > \ln (10 ^ {605}) & \iff \\
2016 \cdot \ln (2) > 605 \cdot \ln (10) & \iff \\
\frac{2016}{605} > \frac{\ln 10}{\ln 2} & \iff \\
\frac{2016}{605} > \frac{\log 10}{\log 2} & \iff \\
\frac{2016}{605} > \frac{1}{\log 2} & \iff \\
3.33 \ldots > \frac{1}{0.301 \ldots} & \iff \\
3.33 \ldots > 3.32 \ldots
\end{aligned}
PS: the two boxed expressions (i.e $\boxed{ \frac{2016}{605} }$ and $\boxed{ \log 2}$ ) can be derived by hand: one can divide on paper and (as far as I can remember) during tests you can bring logarithmic tables
A: Since $2^{2016}=1024^{201}\cdot64$ and $10^{605} = 1000^{201}\cdot 100$, the key issue is to determine if the number $1.024^{201}\cdot 0.64$ is greater than 1.
To estimate the value of $1.024^{201}$, we define function
$$
f(x)=(1+x)^{201}.
$$
Then, we have the first and second order derivatives as:
$$
f'(x) = 201\cdot(1+x)^{200}
$$
and
$$
f''(x)=40200\cdot (1+x)^{199} > 0.
$$
Thus, f(x) is a convex function. Using the property of Taylor's polynomials we have
$$
1.024^{201} = f(0.024) > f(0) + f'(0)(x - 0) = 1 + 201\cdot 0.024 = 5.824
$$
It is certain $1.024^{201} \cdot 0.64 > 1$.
