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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 $10$.

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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
@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
@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
up vote 6 down vote accepted

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



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

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Why not log (to the base 10) for example? – NoChance Oct 12 '11 at 22:02
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
Thanks for the hint. – NoChance Oct 12 '11 at 23:03

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, what we can do with our numbers is, put them on 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 32^800 < 100^800. Nice and easy, without any trouble and logarithms :) Hope that helped

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Awesome, thanks for posting this – sidyll Sep 25 '15 at 1:39

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