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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$\begingroup$$\mathrm{log}_2$ is the way to go.
$\mathrm{log}_2(2^{2000})=2000$
$\mathrm{log}_2(10^{800})=800\,\log_2(10)$
So which is bigger $20$ or $8\mathrm{log}_2(10)$?
Let's see $20$ is smaller than $8 \times 3=24$ and $\mathrm{log}_2(10) > \mathrm{log}_2(8)=\mathrm{log}_2(2^3)=3$. So it looks like: $$2^{2000} < 10^{800}$$ (no calculator required).
To compare: $3=\mathrm{log}_2(2^3)=\mathrm{log}_2(8)<\mathrm{log}_2(10)<\mathrm{log}_2(16)=\mathrm{log}_2(2^4)=4$
$\endgroup$ 2 $\begingroup$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
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