Which number is bigger?

Question

Which number is bigger? $1.01^{101}$ or $2$? and how about $e^{\pi}$ or $\pi^e$?

Tried some algebraic manipulations to no end, so would love some suggestions or some different ways to approach those kind of problem

Answer 1

Hint for the first: With $x = 0.01$ and $n = 101$, note that $$ (1 + x)^n = 1 + n\,x + \binom n2 x^2 + \cdots \geq 1 + n\,x $$

Hint for the second: It is equivalent to show that $e^{1/e} > \pi^{1/\pi}$. In order to do so, consider the function $f(x) = x^{1/x}$ and use calculus.

Answer 2

Consider $$A=1.01^{101}$$ So $$\log(A)=101\log(1.01)$$ Now use the very fast convergent expansion $$\log\Big(\frac{1+x}{1-x}\Big)=2\,\Big(\frac {x}{1}+\frac {x^3}{3}+\frac {x^5}{5}+\cdots\Big)$$ and make $\frac{1+x}{1-x}=\frac{101}{100}$ that is to say $x=\frac{1}{201}$. The first term (to which only positive terms will be added) is already $$\log(A)=101\log(1.01)=2\times 101\times \frac{1}{201}=\frac{202}{201}>1$$ and then $A>e>2$

Answer 3

$$x \to 0 , (1+x)^n \approx 1+nx \\(1.01)^101=(1+\frac{1}{100})^101 \approx 1+ 101(\frac{1}{100}) >2$$ for the second one Consider this function $$x^{\frac{1}{x}}$$. $f'=x^{\frac{1}{x}}(\frac{1}{x^2})(1-\ln x)$, function has global maximum at $x=e$.

so $e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}} \to $, and it is clear that the inequality is strict,$$(e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}})^{e\pi} \to $$ so $$e^{\pi}>\pi^{e}$$

Answer 4

HINT: we get $$e^{\pi}>\pi^{e}$$ use the function $$f(x)=\sqrt[x]{x}$$