There are two mistakes in your arguments.
First, the derivatives of $\log_{10}x$ and $\ln x$ are not the same. Since $\log_{b}x=\ln x/\ln b$, we have
$$\frac{\mathrm d}{\mathrm dx}\log_{b}x=\frac{\mathrm d}{\mathrm dx}\frac{\ln x}{\ln b}=\frac1{\ln b}\frac{\mathrm d}{\mathrm dx}\ln x=\frac1{\ln b}\frac1x\;.$$
Second, it's not true that different functions must have different derivatives. Functions that differ only by an additive constant have the same derivative. To use an example close to yours,
$$\frac{\mathrm d}{\mathrm dx}\ln(cx)=c\frac1{cx}=\frac1x\;.$$
This might seem mysterious in that form, but is becomes clearer if you instead write
$$\frac{\mathrm d}{\mathrm dx}\ln(cx)=\frac{\mathrm d}{\mathrm dx}\left(\ln x+\ln c\right)=\frac{\mathrm d}{\mathrm dx}\ln x=\frac1x\;,$$
which shows that this is due to the fact that $\ln (cx)$ and $\ln x$ only differ by an additive constant.
$\mathrm{log}_2$ 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: $3=\mathrm{log}_2(2^3)=\mathrm{log}_2(8)<\mathrm{log}_2(10)<\mathrm{log}_2(16)=\mathrm{log}_2(2^4)=4$
Best Answer
Play around with the values. Estimate them. See what happens. For example:
Let's see... $\log_4 (1/15)$.... hmmm.... well, $\log_4(1/16) = -2$... and $\log_4(1/4)=-1$.... so, $\log_4(1/15)$ is between $-2$ and $-1$.... okay, that's pretty good.
How about this one? $\log_3(1/2)$... hmmm..... $\log_3(1/3)=-1$,... and, hey I see it... $\log_3(1)=0$, so $\log_3(1/2)$ is between $-1$ and $0$, so $$\log_4(1/15) < -1 < \log_3(1/2) $$