Which of the numbers $99^{100}$ and $100^{99}$ is the larger one? [closed]

Solution 1:

Note that $$\begin{align} 99^{100} > 100^{99} &\iff 99 \cdot 99^{99} > 100^{99} \\ &\iff 99 > (100/99)^{99} \\ &\iff 99 > \left( 1 + \frac{1}{99}\right)^{99} \end{align}$$

Since $(1 + \frac{1}{n})^n < 3$ for all integers $n$, the above inequalities are all true. Thus, $99^{100} > 100^{99}$. In general, you should expect that $x^y > y^x$, whenever $y > x$.

Solution 2:

$99^{100} - 100^{99}$ is:

3560323412732295049306160265725173861897
1207663892369140595737269931704475072474
8187196543510026950400661569100652843274
7182356968017994158571053544917075742738
9035006098270837114978219916760849490001

Since this number is positive, $99^{100}$ is the bigger number.

Solution 3:

A purely math solution: Using AM-GM inequality:

$$(x+1)^x\times \frac{x}{2} \times \frac{x}{2} < \left(\frac{x(x+1)+x}{x+2}\right)^{x+2}=x^{x+2}.$$

Therefore

$$(x+1)^x < 4x^x$$

and easily we see that $(x+1)^x< x^{x+1}$ for any $x\ge 4$.

Solution 4:

$x^{x+1}=x x^x$ while for large $x$, $(x+1)^x\sim e x^x$. Since $99>e$, I would say that $99^{100}>100^{99}$.

More Detail:

To show that $(x+1)^x=\left(1+\frac1x\right)^xx^x<ex^x$, without just saying so and without using logarithms, consider the binomial expansion $$ \left(1+\frac1x\right)^x=1+1+\frac12\frac{x-1}{x}+\frac16\frac{(x-1)(x-2)}{x^2}+\frac{1}{24}\frac{(x-1)(x-2)(x-3)}{x^3}+\dots $$ and note that, at least for $x\in\mathbb{N}$, each term is monotonically increasing. Thus, $\left(1+\frac1x\right)^x$ monotonically increases to $e=\sum\limits_{k=0}^\infty\frac{1}{k!}$.