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?
Best Answer
$$ 1.05^{50} = \left(1+\frac{5}{100}\right)^{50} = \sqrt{\left(1+\frac{5}{100}\right)^{100}} < \sqrt{e^5} = \sqrt{e}^5 < 2^5 = 32 $$
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 $$ \begin{split} \left(\frac{2^{10}}{10^3}\right)^{100} &= \left(\frac{1024}{1000}\right)^{100} = \left(\frac{1}{1-\frac{24}{1024}}\right)^{100} \geq e^{100\cdot\frac{24}{1024}} = e^{75/32} \\ &= e^{2+11/32} > e^{2+11/33} = \exp(2)\exp(1/3) > \exp_5(2)\exp_2(1/3) \\ &= \left(1+2+2+\frac43+\frac23+\frac4{15}\right)\left(1+\frac13+\frac1{18}\right) \\ &= \frac{109}{15} \cdot \frac{25}{18} = \frac{545}{54} > 10. \end{split} $$