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Using logarithms, it doesn't seem all too hard to figure out that $99^{(n+1)}$>$100^n$ when n<457.21 approximately. How does one figure out when $99^{(n+1)}$>$100^n$ without using a single logarithm?

Doug Spoonwood
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    Presumably, you're not looking for "by using a calculator or computer to graph or to make a table of values." – Isaac Feb 06 '12 at 06:56
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    The exact question you are asking is for which $n$ is $99> \left(1+\frac{1}{99}\right)^n$. This occurs precisely when $\frac{\log 99}{\log\left(1+\frac{1}{99}\right)}>n$. Presumably you can prove the inequality to the nearest integer (457) using inequalities based on the exponential, but this will likely just be a taylor expansion at some point "removing" the logarithm which is implicitly there. It is very likely that any other solution will just be hiding the logarithms in some way, since to even express the cut off point we need to use logarithms or equivalents. – Eric Naslund Feb 06 '12 at 07:25
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    related/duplicate for $n=99$? – draks ... Feb 06 '12 at 07:46
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    Related, yes. Inspired by, quite possibly. Duplicate, no. – Gerry Myerson Feb 06 '12 at 10:16

2 Answers2

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$99^{n+1}>(99+1)^n={n\choose0}99^n+{n\choose1}99^{n-1}+{n\choose2}99^{n-2}$

$1>\frac{1}{99}+\frac{n\choose1}{99^2}+\frac{n\choose2}{99^3}+...$

Upper bounds for n can be derived from the first couple of terms. For example, $1>\frac{1}{99}+\frac{n}{99^2}+\frac{n^2-n}{2.99^3}$ is a quadratic which solves as $x<1294.9$

Angela Pretorius
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We can rewrite $99^{n+1} < 100^n$ as $99(99)^n < (99 + 1)^{n}$ and so $99 < (1 + \frac{1}{99})^n$. Rewrite the right-hand side as $((1 + \frac{1}{99})^{99})^{n/99}$. The quantity inside is a bit less than $e$ (actually, the convergence is sort of slow!). Taking a log (but just one!) we find that $e^{4.6} > 100$. So when $n \geq 456$, the inequality holds. (Or almost holds, since I am underestimating $e$.)

Sam Nead
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  • I admit I did use the plural form of "logarithm", but I meant to ask how the problem can get solved without using a single logarithm. – Doug Spoonwood Feb 07 '12 at 02:24