How can you prove that $1.05^{50} < 100$ without a calculator?
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?
$\endgroup$ 45 Answers
$\begingroup$$$ 1.05^{50} = \left(1+\frac{5}{100}\right)^{50} = \sqrt{\left(1+\frac{5}{100}\right)^{100}} < \sqrt{e^5} = \sqrt{e}^5 < 2^5 = 32 $$
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$$ \begin{split} \left(\frac{2^{10}}{10^3}\right)^{100} &= \left(\frac{1024}{1000}\right)^{100} = \left(\frac{1}{1-\frac{24}{1024}}\right)^{100} \geq e^{100\cdot\frac{24}{1024}} = e^{75/32} \\ &= e^{2+11/32} > e^{2+11/33} = \exp(2)\exp(1/3) > \exp_5(2)\exp_2(1/3) \\ &= \left(1+2+2+\frac43+\frac23+\frac4{15}\right)\left(1+\frac13+\frac1{18}\right) \\ &= \frac{109}{15} \cdot \frac{25}{18} = \frac{545}{54} > 10. \end{split} $$
$\endgroup$ 10 $\begingroup$$\log(1.05^{50}) = 50\log(1.05) < 50·0.05 = 2.5 < 3$. Then$1.05^{50} < e^3 < 3^3 = 27$.
$\endgroup$ 4 $\begingroup$Here is an elementary way using GM-HM.
First note
- $1.05^{50} < 100 \Leftrightarrow \boxed{1.05 < \sqrt[50]{100}}$
$$\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}$$
$\endgroup$ 1 $\begingroup$We have to prove $\left(\frac{21}{20}\right)^{50} < 100$, which simplifies to:$$ 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*}
$\endgroup$ $\begingroup$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.
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