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How can I know if one power is bigger than the other when the bases are different?

For example, considering $2^{10}$ and $10^{3}$ the former is the greater one, but how to prove this? Logarithms? I'll be working with big numbers, and though a more general solution is really appreciated, I will be comparing exactly powers of $2$ and $10$.

J. M. ain't a mathematician
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sidyll
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    Well, to prove that $2^{10}$ is larger than $10^3$ you can just compute them! $2^{10}=1024\gt 1000=10^3$. Nothing else to "prove", really. And, yes, you can use logarithms. Since $2^a\lt 10^b$ if and only if $a\lt b\log_{2}(10)$, if and only if (assuming both $a$ and $b$ are positive) $\frac{a}{b}\lt\log_2{10}$, you just need to compare $\frac{a}{b}$ with $\log_2{10}$. – Arturo Magidin Oct 12 '11 at 21:15
  • @ArturoMagidin I know :-) But which one is bigger: $2^{2000}$ or $10^{800}$? I'd like to know about cases like this one. Edit: You should post this edit in your comment as answer. – sidyll Oct 12 '11 at 21:18
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    @sidyll: $2^m=10^{m\log_{10}(2)}$. Thus $2^m>10^n$ precisely if $m\log_{10}(2) >n$. Now use $\log_{10}(2) \approx 0.30129996$. – André Nicolas Oct 12 '11 at 21:46
  • @Ross Millikan: Thanks, I will delete, change. – André Nicolas Oct 12 '11 at 22:46
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    @sidyll: $2^m=10^{m\log_{10}(2)}$. Thus $2^m \gt 10^n$ precisely if $m\log_{10}(2) \gt n$. Now use $\log_{10}(2) \approx 0.301029996$. – André Nicolas Oct 12 '11 at 22:49
  • @sidyll: In addition to the logs, note that $2^{10}\approx 10^3$ (actually, a bit bigger), so $2^{2000} = (2^{10})^{200} \approx (10^3)^{200} = 10^{600}$. The difference in exponents is sufficiently large to know that $10^{800}$ is larger. (Of course, if you have $2^a$ vs. $10^b$ and $a/3$ is only a little smaller than $b$, this kind of back-of-the-envelope calculation would not be good enough to tell which one is different; in that case, go with logarithms. – Arturo Magidin Oct 13 '11 at 03:10
  • Thanks once again for the explanation @ArturoMagidin – sidyll Oct 13 '11 at 14:16
  • @sidyll: There's a typo there: it should be "if $3a/10$ is only a little smaller than $b$..." – Arturo Magidin Oct 13 '11 at 15:58

2 Answers2

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$\mathrm{log}_2$ is the way to go.

$\mathrm{log}_2(2^{2000})=2000$

$\mathrm{log}_2(10^{800})=800\,\log_2(10)$

So which is bigger $20$ or $8\mathrm{log}_2(10)$?

Let's see $20$ is smaller than $8 \times 3=24$ and $\mathrm{log}_2(10) > \mathrm{log}_2(8)=\mathrm{log}_2(2^3)=3$. So it looks like: $$2^{2000} < 10^{800}$$ (no calculator required).

To compare: $3=\mathrm{log}_2(2^3)=\mathrm{log}_2(8)<\mathrm{log}_2(10)<\mathrm{log}_2(16)=\mathrm{log}_2(2^4)=4$

Quixotic
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Bill Cook
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  • Why not log (to the base 10) for example? – NoChance Oct 12 '11 at 22:02
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    If you're using a calculator (which has a log base 10 button), then go ahead and compute base 10. I used base 2 because it could be done by hand (base 10 is too course to do by hand). – Bill Cook Oct 12 '11 at 22:47
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Actually, you can solve problem above by much much easier method. No need for logarithm. Simply, give 2^2000 and 10^800 the same exponent. We know that (a^n)^m = a^(n*m). That is clear. Now, we can restate the different bases but with the same exponent so they'll look like this: (2^5)^400 and (10^2)^400 and they are absolutely the same as the before ones. 2^5 = 32 and 10^2 = 100 are easily calculated, even without calculator, so now we can say that 32^400 < 100^400. Nice and easy, without any trouble and logarithms :) Hope that helped

zelusp
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qwerty_
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