Is there a way to prove that $1.05^{50} < 100$ without a calculator?
I have tried this...
$$1.05^{50}<10^2$$ $$(\frac{105}{100})^{50}<10^2$$ $$(\frac{21}{20})^{50}<10^2$$ $$\frac{21^{50}}{20^{50}}<10^2$$ $$\frac{21^{50}}{2^{50}*10^{50}}<10^2$$ $$\frac{21^{50}}{2^{50}}<10^{52}$$ $$10.5^{50}<10^{52}$$
...but I don't know where to go. Can someone assist me (alternate methods are fine)?
EDIT: Can anyone help me prove that $2^{1000}<10^{302}$ without a calculator?
$$ 1.05^{50} = \left(1+\frac{5}{100}\right)^{50} = \sqrt{\left(1+\frac{5}{100}\right)^{100}} < \sqrt{e^5} = \sqrt{e}^5 < 2^5 = 32 $$
Edit: for you second inequality, $2^{1000}<10^{302}$ is equivalent to $(2^{10})^{100}<10^2(10^3)^{100}$, which is equivalent to $$ \left(\frac{2^{10}}{10^3}\right)^{100} = \left(1+\frac{24}{1000}\right)^{100} < 100. $$ From $a\log(1+t)\leq at$ for $a>0$ we deduce $(1+t)^a\leq e^{at}$. Therefore $$ \left(1+\frac{24}{1000}\right)^{100} < e^{\frac{24}{1000}\cdot100} <e^{5/2} < 32 < 100. $$
Addendum. What is trickier, is proving that $2^{1000}>10^{301}$. Can you do that?
Here is how I go about it. Maybe someone else can find a simpler derivation.
Define $$ \exp_n(x) = \sum_{k=0}^n \frac{x^k}{k!} < \exp(x) . $$
We want $\left(\frac{2^{10}}{10^3}\right)^{100}>10$. From $(1-t)^a\leq e^{-at}$ you deduce $\left(\frac1{1-t}\right)^a\geq e^{at}$. So $$ \begin{split} \left(\frac{2^{10}}{10^3}\right)^{100} &= \left(\frac{1024}{1000}\right)^{100} = \left(\frac{1}{1-\frac{24}{1024}}\right)^{100} \geq e^{100\cdot\frac{24}{1024}} = e^{75/32} \\ &= e^{2+11/32} > e^{2+11/33} = \exp(2)\exp(1/3) > \exp_5(2)\exp_2(1/3) \\ &= \left(1+2+2+\frac43+\frac23+\frac4{15}\right)\left(1+\frac13+\frac1{18}\right) \\ &= \frac{109}{15} \cdot \frac{25}{18} = \frac{545}{54} > 10. \end{split} $$
$\log(1.05^{50}) = 50\log(1.05) < 50·0.05 = 2.5 < 3$. Then $1.05^{50} < e^3 < 3^3 = 27$.
Here is an elementary way using GM-HM.
First note
$$\color{blue}{\sqrt[50]{100}} = \sqrt[50]{2^2\cdot 5^2 \cdot 1^{46}} \color{blue}{\stackrel{\mbox{GM-HM}}{>}}\frac{50}{\frac{2}{2}+\frac{2}{5}+46}=\frac{250}{237}=1+\frac{13}{237}> \color{blue}{1.05}$$
Use $$\left(1+\frac1n\right)^n<\left(1+\frac1n\right)^{n+1}\leq4,$$ because $\{\left(1+\frac1n\right)^{n+1}\}$ is a decreasing sequence, we have $$1.05^{50}=\left(\left(1+\frac1{20}\right)^{20}\right)^{\frac52} <4^{\frac52}=2^5=32<100.$$
$(1+x)^r\leq e^{rx}$. If $x=0.05$ and $r=50$, then $rx=50\times \frac{5}{100}=\frac{250}{100}=2.5$. Therefore $(1.05)^{50}\leq e^{2.5}$. Since $2\lt e\lt 3$, we have $2^{2.5}\lt e^{2.5}\lt 3^{2.5}$. We also have $3^{2.5}\lt 3^3$, and since $3^3=27$, we may say that $(1.05)^{50}\lt 27\lt 100$, proving the claim. No calculators required!
We have to prove $\left(\frac{21}{20}\right)^{50} < 100$, which simplifies to: $$ 3^{50}7^{50} < 2^{102}5^{52}. $$
We have $3^5 = 243 < 256 = 2^8$, and $7^2 = 49 < 50 = 2\cdot5^2$, therefore: $$ 3^{50}7^{50} = (3^5)^{10}(7^2)^{25} < (2^8)^{10}(2\cdot5^2)^{25} = 2^{105}5^{50} = 2^{102}\cdot8\cdot5^{50} < 2^{102}5^{52}. $$
(I didn't notice that a second, unrelated question had been tacked onto the end - naughty, that!)
From \begin{equation} \tag{$*$}\label{ineq:crux} 2^{10} = 1024 < 1025 = 5^2\cdot41, \end{equation} we get $$ \frac{2^{30}}{5^6} = \left(\frac{2^{10}}{5^2}\right)^3< 41^3 = 68921 < 69000 = 69\cdot2^3\cdot5^3, $$ and from this, together with \eqref{ineq:crux} again, we get $$ \frac{2^{37}}{5^{11}} = \left(\frac{2^{27}}{5^{9}}\right)\left(\frac{2^{10}}{5^{2}}\right) < 69\cdot41 = 2829 < 3125 = 5^5, $$ whence $$ 2^{37} < 5^{16}. $$ One more slog, and we're done: \begin{gather*} \left(2^{349}\right)^{16} = 2^{5584} < 2^{5587} = \left(2^{37}\right)^{151} < \left(5^{16}\right)^{151} = \left(5^{151}\right)^{16}, \\ \therefore\ 2^{1000} = 2^{302}\left(2^{349}\right)^2 < 2^{302}\left(5^{151}\right)^2 = 2^{302}5^{302} = 10^{302}. \end{gather*}
I'll use this CW answer to gather together some deductions from the power series expansion $$ \frac12\log\frac{1+x}{1-x} = x+\frac{x^3}3+\frac{x^5}5+\cdots. $$ This can be truncated to give lower bounds, and its remainder can be replaced by a geometric series with common ratio $x^2$ to give upper bounds.
Taking the first question first: \begin{multline*} 50\log\frac{21}{20} = 100\cdot\frac12\log\frac{1+\tfrac1{41}}{1-\tfrac1{41}} < 100\cdot\frac1{41}\left(1-\frac1{41^2}\right)^{-1} \\ = \frac{100\cdot41}{41^2-1} = \frac{100\cdot41}{40\cdot42} = \frac{205}{84} < \frac52, \end{multline*} whence $$ \left(\frac{21}{20}\right)^{50} < e^{5/2} < 3^{5/2} = \sqrt{243} < \sqrt{256} = 16 < 100. \quad \square $$
It doesn't take much more work to get a sharper upper bound: $$ \left(\frac{21}{20}\right)^{50} < 12. $$ Proof: $$ 50\log\frac{21}{20} < \frac{205}{84} < \frac{49}{20} < \log{12}. $$ The penultimate inequality follows from $$ 20\cdot205 = 4100 < 4116 = 4200 - 84 = 50\cdot84 - 84 = 49\cdot84. $$ The final inequality is proved as follows: \begin{gather*} \log3 = 2\cdot\frac12\log\frac{1+\tfrac12}{1-\tfrac12} > 2\left(\frac12 + \frac1{3\cdot2^3}\right) = \frac{13}{12}; \\ \log2 = 2\cdot\frac12\log\frac{1+\tfrac13}{1-\tfrac13} > 2\left(\frac13 + \frac1{3^4}\right) = \frac{56}{81}; \\ \log{12} = \log3 + 2\log2 > \frac{13}{12} + \frac{112}{81} = \frac{351+448}{324} = \frac{799}{324} > \frac{800}{325} = \frac{32}{13} > \frac{49}{20}, \end{gather*} because $20\cdot32 = 640 > 637 = 650-13 = 50\cdot13-13 = 49\cdot13. \quad \square$
An answer to the OP's second question can be deduced from the inequality in the first question. As @Federico notes in his answer, the second question reduces to $$ \left(\frac{2^{10}}{10^3}\right)^{100} < 100. $$ Therefore, it is enough to prove $$ \left(\frac{2^{10}}{10^3}\right)^2 = \left(1 + \frac3{125}\right)^2 < 1 + \frac1{20} = \frac{21}{20}; $$ and the central inequality does hold, because $$ \frac1{20} - \left(\frac6{125} + \frac9{125^2}\right) = \left(\frac6{120} - \frac6{125}\right) - \frac9{125^2} = \frac1{125}\left(\frac14 - \frac9{125}\right) > 0. \quad \square $$ (With hindsight, it would have been simpler just to write $1.024^2 = 1.048576 < 1.05$!)
Alternatively, we can answer the second question directly, using the same power series again: \begin{multline*} 100\log\frac{2^{10}}{10^3} = 100\log\left(1+\frac6{250}\right) = 200\cdot\frac12\log\frac{1+\tfrac3{253}}{1-\tfrac3{253}} \\ < \frac{600}{253}\left(1-\left(\frac3{253}\right)^2\right)^{-1}\!\! = \frac{600\cdot253}{253^2-3^2} = \frac{600\cdot253}{250\cdot256} = \frac{3\cdot253}{5\cdot64} = \frac{759}{320} < \frac52, \end{multline*} whence, exactly as with the first question, $$ \left(\frac{2^{10}}{10^3}\right)^{100} < e^{5/2} < 3^{5/2} = \sqrt{243} < \sqrt{256} = 16. \quad \square $$
Also in his answer, @Federico poses (and solves) the problem of proving $$ \left(\frac{2^{10}}{10^3}\right)^{100} > 10. $$ We can prove this by using the power series again. To begin with, $$ 100\log\frac{2^{10}}{10^3} = 100\log\left(1+\frac6{250}\right) = 200\cdot\frac12\log\frac{1+\tfrac3{253}}{1-\tfrac3{253}} > \frac{600}{253} > \frac73. $$
@Federico gives a short and clear proof that $e^{7/3} > 10$ by using the power series for $e^x,$ and this is probably to be preferred; but I'll give another proof, using the same power series I've been using throughout. (There are several other reasonable-looking ways to use the same power series to prove the result, and I make no claim that this is the neatest.) $$ \log3 = 2\cdot\frac12\log\frac{1+\tfrac12}{1-\tfrac12} < 2\left(\frac12 + \frac1{3\cdot2^3}\left( 1-\frac1{2^2}\right)^{-1}\right) = 2\left(\frac12+\frac1{18}\right) = \frac{10}9; $$ \begin{multline*} \log\frac{10}9 = 2\log\frac{1+\tfrac1{19}}{1-\tfrac1{19}} < 2\left(\frac1{19} + \frac1{3\cdot{19}^3}\left( 1-\frac1{{19}^2}\right)^{-1}\right) = \\ 2\left(\frac1{19} + \frac1{3\cdot19\cdot(19^2-1)}\right) = 2\left(\frac1{19} + \frac1{3\cdot18\cdot19\cdot20}\right) < 2\left(\frac1{19} + \frac1{18\cdot19}\right) = \frac19; \end{multline*} $$ \log{10} = 2\log3 + \log\frac{10}9 < \frac{20}9 + \frac19 = \frac73. \quad \square $$
That will have to be enough for today, but it would be fun to give similar proofs of the two middle inequalities here: $$ 10 < \left(\frac{2^{10}}{10^3}\right)^{100}\!\! < 11 < \left(\frac{21}{20}\right)^{50}\! < 12, $$ and as this is a CW answer, anyone should feel free to append such proofs.
By writing 1.05 as (1 + 0.05) and expanding by using binomial Theorem, you can clearly see that in all terms there is positive power in 0.05 ( except the first one) which is making it smaller. So it can never be greater than 100 or even leave 100 it can never be greater than 50.
