Compare $2^{2016}$ and $10^{605}$ without a calculator

Question

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?

Answer 1

$$2^{2016}=(2^{10})^{201}\cdot2^6=1024^{201}\cdot64$$

$$10^{605}=(10^3)^{201}\cdot10^2=1000^{201}\cdot100$$

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 $$\boxed{2^{2016}>10^{605}}$$

Answer 2

$\log_{10}2^{2016}=2016\log_{10}2\approx 606.88>605$

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*}

Answer 3

\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

Answer 4

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 $1.024^{201} \cdot 0.64 > 1$.