Which of the numbers $99^{100}$ & $100^{99}$ is the larger? Solve without using logarithms.
Which of the numbers $99^{100}$ and $100^{99}$ is the larger one?
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0I love this question, not because of the question, but the vast variety of answers you give (well you all give the same answer) but many of you have completely different methods, and I just love that about math. – 2014-09-09
8 Answers
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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1@Henry: my comment was to point out that the fact that \lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e<3 does not give a clue as to $\max\limits_{n\in\mathbb{N}}\left(1+\frac1n\right)^n$. – 2012-02-07
$99^{100} - 100^{99}$ is:
3560323412732295049306160265725173861897 1207663892369140595737269931704475072474 8187196543510026950400661569100652843274 7182356968017994158571053544917075742738 9035006098270837114978219916760849490001
Since this number is positive, $99^{100}$ is the bigger number.
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6"...Since this number is positive...": as Orwell could have said, many numbers are positive but some are more positive than others... – 2012-01-29
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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1@Doug: The ratio between $(x+1)^x$ and $x^{x+1}$ is $\varrho(x)/x$, hence, between $2/x$ and $\mathrm e/x$ for every $x\geqslant1$, and even between $9/(4x)$ and $\mathrm e/x$ for every $x\geqslant2$. For example, this proves that the ratio $100^{99}/99^{100}$ is between $0.02272727$ and $0.02745739$ without knowing the values of $100^{99}$ and $99^{100}$. This even suggests that this ratio is much closer to the latter value than to the former (to wit, the actual ratio is $0.02731999$). – 2012-01-30
$x^{x+1}=x x^x$ 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
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1@robjohn I've now upvoted your answer, since the details can help here, while the older version didn't help much. – 2012-01-29
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 $e$.
${99^{100}\over 100^{99}} = 99\left({99\over 100}\right)^{99}\approx {99\over e} > 1.$
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3Starting 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... – 2012-01-30
$100^{99}$=$(10*10)^{99}$=$(10^{99})(10^{99})$=$10^{198}$ 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.
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0@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 – 2012-01-30