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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?

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    Related, yes. Inspired by, quite possibly. Duplicate, no.2012-02-06

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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 $x<1294.9$

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    @DougSpoonwood Divide both sides by $99^{n+1}$2012-02-07
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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 $e$.)

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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.2012-02-07