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

Question

Which of the numbers $99^{100}$ & $100^{99}$ is the larger? Solve without using logarithms.

Answer 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 $y > x$.

Answer 2

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

3560323412732295049306160265725173861897
1207663892369140595737269931704475072474
8187196543510026950400661569100652843274
7182356968017994158571053544917075742738
9035006098270837114978219916760849490001

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

Answer 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 $x\ge 4$.

Answer 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 $e=\sum\limits_{k=0}^\infty\frac{1}{k!}$.

Answer 5

Proof that $x^y > y^x$ for all $y > x > e$: Raising both sides to the ${1 \over xy}$ power, this is equivalent to $x^{1 \over x} > y^{1 \over y}$. The derivative of $x^{1 \over x}$ with respect to $x$ is ${\displaystyle {1 - \ln(x) \over x^2} x^{1 \over x}}$, which is negative whenever $\ln(x) > 1$ i.e. when $x > e$. Thus $x^{1 \over x}$ is a decreasing function of $x$ for $x > e$.

Yeah I know, I used logarithms. But someone needed to say this ;)

Answer 6

I cheat and use a basic fact about $e$.

$${99^{100}\over 100^{99}} = 99\left({99\over 100}\right)^{99}\approx {99\over e} > 1.$$

Answer 7

$100^{99}$=$(10*10)^{99}$=$(10^{99})(10^{99})$=$10^{198}$ exactly.

$99^{100}$=$(9*11)^{100}$=$(9^{100})(11^{100})$ exactly. My "hand" calculator approximates $9^{100}$ as about $(2.656)(10^{95})$.

11=(2)(2)(2.75).

$2^{100}$ equals about (1.267)($10^{30}$), $2.75^{100}$ equals about (8.575)($10^{43}$). Dropping the coefficients here we can thus approximate ($11^{100}$) by a lower bound of ($10^{30}$)($10^{30}$)($10^{43}$)=$10^{103}$.

Keeping the coefficients on the approximation of $9^{100}$ we have a lower bound for $99^{100}$ as $(2.656)((10^{95}$)($10^{103}$))=(2.656)($10^{198}$) which comes as greater than $10^{198}$.

So, $99^{100}$>$100^{99}$.

Note that if we kept the coefficients in here, we would also have more of an idea as to how much greater $99^{100}$ is than $100^{99}$. Some of the other answers do this, some don't. This doesn't necessarily make this answer better though, since such information might come as extraneous to the problem.

Answer 8

From experimenting with small numbers:

scala> (0 to 5).map (x=> (math.pow (x, x+1) - math.pow (x+1, x))).mkString ("; ")
res18: String = -1.0; -1.0; -1.0; 17.0; 399.0; 7849.0

scala> (0 to 5).map (x=> (math.pow (x, x+1), math.pow (x+1, x))).mkString ("; ")
res19: String = (0.0,1.0); (1.0,2.0); (8.0,9.0); (81.0,64.0); (1024.0,625.0); (15625.0,7776.0)

you can conclude, that the first one is growing faster than the second. Of course this is only an indication.