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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?

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    $\begingroup$ $10^{605}/2\neq 5^{605}$ $\endgroup$ – marwalix Jan 24 '18 at 7:54
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    $\begingroup$ One standard approach is to look at powers of both numbers which are close. For instance, $2^{10}$ is almost equal to $10^3$. Using this, you can try to compare $$ 2^6(2^{10})^{201}\text{ and } 10^2(10^3)^{201} $$ $\endgroup$ – Arthur Jan 24 '18 at 7:55
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    $\begingroup$ @marwalix He clearly divided by $2^{605}$, if you look at what happened to the exponent of $2$. $\endgroup$ – Arthur Jan 24 '18 at 7:56
  • $\begingroup$ We used to be able to know that $\log_{10} 2 \approx 0.30103$ (and just less than this value) so that $0.301\times 2016 \lt \log_{10} 2 \lt 0.30103 \times 2016$. The calculation will inevitably be quite delicate because $0.3\times 2016 \lt 605 \lt 0.301*2016$. So the approximation $2^10\approx 10^3$ is not good enough, and effectively you need an approximation as good as $0.301$. $\endgroup$ – Mark Bennet Jan 24 '18 at 8:40
  • $\begingroup$ @Arthur Yes, you can try. $\endgroup$ – uniquesolution Jan 24 '18 at 9:01
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$$2^{2016}=(2^{10})^{201}\cdot2^6=1024^{201}\cdot64$$

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

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

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    $\begingroup$ A small note on how you can handle $1.024^{201}$ rigorously without a calculator would be nice. $\endgroup$ – Arthur Jan 24 '18 at 7:58
  • $\begingroup$ @Arthur: via $(1+x)^{201}\geq 1+201 x$, I guess. $\endgroup$ – Jack D'Aurizio Jan 24 '18 at 11:07
  • $\begingroup$ @JackD'Aurizio Truncating the binomial theorem works, and it was what I had in mind. $\endgroup$ – Arthur Jan 24 '18 at 11:09
  • $\begingroup$ @Arthur Please see edit. Apologies for delay. $\endgroup$ – TheSimpliFire Jan 24 '18 at 19:02
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$\log_{10}2^{2016}=2016\log_{10}2\approx 606.88>605$

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*}

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    $\begingroup$ He specifically says in the comments that he does not have access to a calculator... $\endgroup$ – Arthur Jan 24 '18 at 7:57
  • $\begingroup$ @Arthur Then how on earth did he post the message if not on computer? $\endgroup$ – ploosu2 Jan 24 '18 at 7:58
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    $\begingroup$ You are aware of the concept of practicing for tests, right? Or coming here after a test wondering what you could've done? That is one possible reason. If we didn't allow technical restrictions on problems, 95% of the questions on this site would automatically be solved by a link to WolframAlpha. $\endgroup$ – Arthur Jan 24 '18 at 7:59
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\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

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

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