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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
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Python: 99**100 > 100*99 == True –  nightcracker Jan 29 '12 at 12:02
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-1 This question has showed absolutely no effort whatsoever. @TheChaz I don't get the upvotes either. –  user38268 Jan 29 '12 at 14:53
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@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
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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

8 Answers 8

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
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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
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@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
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@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
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@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.

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Nice reminder that the world has changed since I first did mathematics. –  André Nicolas Jan 29 '12 at 1:57
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@AndréNicolas I suppose it doesn't use logarithms... –  Matt Jan 29 '12 at 2:06
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@Matt, technically correct is the best kind of correct. –  Hammerite Jan 29 '12 at 2:24
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So, Hammerite, is $999999^{1000000}$ larger than $1000000^{999999}$? –  Myself Jan 29 '12 at 2:48
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@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 $x\ge 4$.

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10  
"purely math solution" as opposed to? –  Najib Idrissi Jan 29 '12 at 7:42
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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
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@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
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@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
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@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 $e=\sum\limits_{k=0}^\infty\frac{1}{k!}$.

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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
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$$\frac{(x+1)^x}{x^x}=\left(1+\frac{1}{x}\right)^x<e<99$$ for all $x>0$. –  Jonas Meyer Jan 29 '12 at 3:59
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@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
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@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 ;)

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

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

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Nice cheat, what is the general formula for this fact about $e$ please? Thanks. –  Emmad Kareem Jan 29 '12 at 15:09
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For large $n$, $(1 + \lambda/n)^n \sim e.$$ –  ncmathsadist Jan 29 '12 at 15:18
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@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
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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
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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

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