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Which is greater as $n$ gets larger, $f(n) = 2^{2^{2^n}}$ or $g(n) = 100^{100^n}$?

I tried differentiating the terms but it didn't really reveal anything. Can anyone come up with a solution?

Thanks in advance for any contributions.

martini
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MadChickenMan
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2 Answers2

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Hint: try comparing $$\ln \ln f(n) \quad \text{vs} \quad \ln \ln g(n).$$ That works in all similar cases.

Adayah
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  • I haven't come across the convention for taking the log of numbers raised to the power multiple times. Would you mind explaining it? – MadChickenMan May 06 '15 at 08:43
  • I think I get (2^n+1)(ln(2)) and (n+1)(ln(100))? I assume the answer is f(n), but is that proof enough? Sorry for the poor formatting, I'm on my phone. – MadChickenMan May 06 '15 at 08:52
  • @MadChickenMan: $\log a^b=b\log a$. It doesn't matter if $b$ happens to be an expression involving powers. Just remember that $r^{s^t}$ means $r^{(s^t)}$ and not $(r^s)^t$. So you have that $\log r^{s^t}= s^t\log r$. – MPW May 06 '15 at 09:46
  • Ok thanks. So is f(n) the answer? – MadChickenMan May 06 '15 at 11:22
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To anyone still left, is this sufficient proof?

$\lim_{n \to \infty} \frac{2^{2^{2^n}}}{100^{100^n}} \\=\lim_{n \to \infty} \frac{ \ln \ln 2^{2^{2^n}}}{\ln \ln 100^{100^n}} \\=\lim_{n \to \infty} \frac{2^n \ln 2 + \ln \ln 2}{n \ln 100 + \ln \ln 100} \\=\lim_{n \to \infty} \frac{2^n}{n} \\= +\infty$

MadChickenMan
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  • Not quite. It is not true in general that $$\lim_{n \to \infty} \frac{f(n)}{g(n)} = \lim_{n \to \infty} \frac{\ln \ln f(n)}{\ln \ln g(n)}$$ because logarithm softens the difference between orders of magnitude. But what you've written suffices to show that $\ln \ln f(n) > \ln \ln g(n)$ for large $n$, so $f(n) > g(n)$. – Adayah May 07 '15 at 11:39