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

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

## closed as off-topic by user21820, Qwerty, Watson, ervx, user91500Aug 15 '16 at 13:03

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• Why do I have to solve it? Even more, why do I have to solve it without using logarithms? – 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 $y > x$.

• 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$$for all x>0. – Jonas Meyer Jan 29 '12 at 3:59 • @Doug: At least for natural x, we can show monotonicity simply with the binomial theorem. I have added the details. – robjohn Jan 29 '12 at 14:37 • @robjohn I've now upvoted your answer, since the details can help here, while the older version didn't help much. – Doug Spoonwood Jan 29 '12 at 16:44 Proof that x^y > y^x for all y > x > e: Raising both sides to the {1 \over xy} power, this is equivalent to x^{1 \over x} > y^{1 \over y}. The derivative of x^{1 \over x} with respect to x is {1 - \ln(x) \over x^2} x^{1 \over x}}, which is negative whenever \ln(x) > 1 i.e. when x > e. Thus x^{1 \over x} is a decreasing function of x for x > e. Yeah I know, I used logarithms. But someone needed to say this ;) I cheat and use a basic fact about e.$${99^{100}\over 100^{99}} = 99\left({99\over 100}\right)^{99}\approx {99\over e} > 1.$$• Nice cheat, what is the general formula for this fact about e please? Thanks. – NoChance Jan 29 '12 at 15:09 • For large n, (1 + \lambda/n)^n \sim e.$$ – 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.

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.

• 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