Comparing $3^{1431}$ with $2^{2010}$ without logarithms

Question

I want to compare : $3^{1431}$ and $2^{2010}$

I tried logarithms, $\mathrm{log}_2$ is the way to go.

$\mathrm{log}_2(2^{2010})=2010$

$\mathrm{log}_2(3^{1431})=1431\,\log_2(3)=2268.08$

since $2268.08$ > $2010$

Then we have, $3^{1431}$ > $2^{2010}$.

but I am trying to figure out another way to prove this without logarithms to explain it to middle school students.

Answer 1

Since $$3^7=2187\gt 1024= 2^{10}$$ we have $$3^{1431}\gt 3^{1407}=(3^7)^{201}\gt (2^{10})^{201}=2^{2010}$$