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

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

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Why do I have to solve it? Even more, why do I have to solve it without using logarithms? If this is a homework question please use the [homework] tag. –  Asaf Karagila Jan 28 '12 at 23:53
Python: 99**100 > 100*99 == True –  orlp Jan 29 '12 at 12:02
-1 This question has showed absolutely no effort whatsoever. @TheChaz I don't get the upvotes either. –  user38268 Jan 29 '12 at 14:53
@cardinal: I should explain why @nightcracker's Python returns False. Python allows chained comparisons (1 <= x < 9), so it was interpreting 99**100 > 100**99 == True as one of these. True has an integer comparison value of 1, so this is really 99**100 > 10**99 == 1, which is false. 99**100 > 100**99 and (99**100 > 10**99) == True both return True as you'd expect. –  DSM Jan 29 '12 at 15:05
No one has explicitly said that the question is simply rude. Michael, using the imperative when asking for a favor will not pay in the long run. –  user23211 Jan 29 '12 at 15:30

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 $y > x$.

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it holds when $x>2$ –  ratchet freak Jan 28 '12 at 23:53
Mr. Man, sir, can you please cite where $(1 + \frac{1}{n})^n < 3$ comes from, or prove it, or something? I'm Rusty on this sort of thing. –  Ed Staub Jan 29 '12 at 2:38
@EdStaub: The inequality comes from the calculation of e. e=2.7(ish), and is defined as the limit of that equasion, as n= infinity. –  PearsonArtPhoto Jan 29 '12 at 2:54
@EdStaub: Expanding $(1 + \frac{1}{n})^n$ using the binomial theorem, it is enough to show $\frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} < 1$. But this follows immediately since $n! \geq 2^{n-1}$ for positive integers $n$ (which you can prove by induction), and hence $$\sum_{k =2}^n \frac{1}{k!} \leq \sum_{k=2}^n \frac{1}{2^{n-1}} < 1. Adding 2 to both sides gives the result. There are many places where you can read more (by googling "limit definition of e"), say for example, here: physicsforums.com/showthread.php?t=176076 – JavaMan Jan 29 '12 at 3:25 @Pearsonartphoto: no it doesn't. What you get from the calculation of e=2.71828\dots is that there is an N so that for x>N, \left(1+\frac1x\right)^x<3. However, what happens for x\le N? – robjohn Jan 29 '12 at 22:39 99^{100} - 100^{99} is: 3560323412732295049306160265725173861897 1207663892369140595737269931704475072474 8187196543510026950400661569100652843274 7182356968017994158571053544917075742738 9035006098270837114978219916760849490001  Since this number is positive, 99^{100} is the bigger number. - Nice reminder that the world has changed since I first did mathematics. – André Nicolas Jan 29 '12 at 1:57 @AndréNicolas I suppose it doesn't use logarithms... – Matt Jan 29 '12 at 2:06 @Matt, technically correct is the best kind of correct. – Hammerite Jan 29 '12 at 2:24 So, Hammerite, is 999999^{1000000} larger than 1000000^{999999}? – Myself Jan 29 '12 at 2:48 @Myself - Yep, wolframalpha.com/input/?i=999999%5E1000000+%3E+1000000%5E999999. Might have to give it a few seconds ;) – DMan Jan 29 '12 at 4:18 A purely math solution: Using AM-GM inequality:$$(x+1)^x\times \frac{x}{2} \times \frac{x}{2} < (\frac{x(x+1)+x}{x+2})^{x+2}=x^{x+2}$$Therefore$$(x+1)^x < 4x^x$$And easily we see that (x+1)^x< x^{x+1} for any x\ge 4. - "purely math solution" as opposed to? – Najib Idrissi Jan 29 '12 at 7:42 In this case, it's no computation. In general, it's just personal sense. Since the question is pretty easy with calculus, I expected that an "elementary" solution (secondary-school) is a best fit. – hiro Jan 29 '12 at 13:48 @GeorgesElencwajg I don't see how using a hand calculator, as say I did in my answer, doesn't come as something check-able. Or for that matter if we calculate 99^{100} by multiplying 99s out on paper, it only involves 100 multiplications. Sure, that's several sheets of paper and probably takes a few hours to do, but we could all do that before we die if we had the persistance and desire to do so. – Doug Spoonwood Jan 29 '12 at 17:24 @Doug: But why would we want to do such a thing, since (a) this is most boring, (b) using our brain a few seconds would solve the question and would furthermore suggest easy generalizations? – Did Jan 29 '12 at 17:40 @Doug: If I didn't trust my calculating machines, and I wanted to know how much greater one of the numbers is than the other, I would much prefer to know that (x+1)^x=\varrho(x)x^x, where the function x\mapsto \varrho(x) is increasing on x\geqslant1 from \varrho(1)=2 to \varrho(+\infty)=\mathrm e. The proof is easier to check and the result is more informative. – Did Jan 30 '12 at 6:13 x^{x+1}=x x^x while for large x, (x+1)^x\sim e x^x. Since 99>e, I would say that 99^{100}>100^{99}. More Detail: To show that (x+1)^x=\left(1+\frac1x\right)^xx^x<ex^x, without just saying so and without using logarithms, consider the binomial expansion$$ \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 $$and note that, at least for x\in\mathbb{N}, each term is monotonically increasing. Thus, \left(1+\frac1x\right)^x monotonically increases to e=\sum\limits_{k=0}^\infty\frac{1}{k!}. - Is your approximation here necessarily an over or under approximation? If not, then how do you have anything more than a good guess? – Doug Spoonwood Jan 29 '12 at 3:48$$\frac{(x+1)^x}{x^x}=\left(1+\frac{1}{x}\right)^x<e<99$$for all x>0. – Jonas Meyer Jan 29 '12 at 3:59 @Doug: At least for natural x, we can show monotonicity simply with the binomial theorem. I have added the details. – robjohn Jan 29 '12 at 14:37 @robjohn I've now upvoted your answer, since the details can help here, while the older version didn't help much. – Doug Spoonwood Jan 29 '12 at 16:44 Proof that x^y > y^x for all y > x > e: Raising both sides to the {1 \over xy} power, this is equivalent to x^{1 \over x} > y^{1 \over y}. The derivative of x^{1 \over x} with respect to x is {\displaystyle {1 - \ln(x) \over x^2} x^{1 \over x}}, which is negative whenever \ln(x) > 1 i.e. when x > e. Thus x^{1 \over x} is a decreasing function of x for x > e. Yeah I know, I used logarithms. But someone needed to say this ;) - I cheat and use a basic fact about e.$${99^{100}\over 100^{99}} = 99\left({99\over 100}\right)^{99}\approx {99\over e} > 1.$$- Nice cheat, what is the general formula for this fact about e please? Thanks. – Emmad Kareem Jan 29 '12 at 15:09 For large n, (1 + \lambda/n)^n \sim e.$$ –  ncmathsadist Jan 29 '12 at 15:18
@ncmathsadist Your comment has a small typo. Surely, you meant to write either $(1 + \lambda/n)^n \sim e^{\lambda}$ or $(1 + 1/n)^n \sim e$. –  Srivatsan Jan 29 '12 at 17:27
Somebody should probably mention that such asymptotics, per se, can give NO information about the $n$th term of a sequence, even for $n=99$, hence the (true) basic fact used here CANNOT prove the desired inequality. –  Did Jan 29 '12 at 19:19
Starting from $2\le1+1/n\le3$, I doubt one can go far. And if you wish to use $(1+1/n)^n\le3$ for every $n$, then mention it... –  Did Jan 30 '12 at 16:29

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

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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.

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Do you have computer code in this answer? I simply don't know how to interpret "scala> (0 to 5).map" and the like. –  Doug Spoonwood Jan 30 '12 at 4:40
@DougSpoonwood: Yes, Scala code. (0 to 5) creates a Range, a collection of the numbers (0, 1, ..., 5). The map (x => takes each of them, and puts them, named x, into a function, math.pow (x, x+1) - ... so for the first element it is math.pow (0, 0+1) or 0^(1), then 1^2, 2^3 and so on minus 1^0, then 2^1, 3^2 and so on. mkString is only used for formatting the output a bit. (3^4-4^2) = 81.0-64.0 = 16 –  user unknown Jan 30 '12 at 14:50