Show in an algebraic way, not using calculator if possible that: $$2^{123}<5^{53}$$
- First I want to tell that I'm trying to make this post a reference post.
There are duplicates out there, but none really offers a non-calculator proof, at most they rely on a fine approximation of $\log_{10}(2)$, and moreover they are all getting deleted one by one due to no effort showing from the original poster.
Since I found a solution, I'd like to expose it here, of course other solutions are welcome as well.
- Secondly, the problem is not as easy as it looks and I'll show why:
The standard idea would be to compare $2^7\approx 5^3$ ($128$ vs $125$) but this inequality is the wrong order $5^3 < 2^7$.
By using it we can prove that $\ 2^{123}>16\cdot 5^{51}\ $ while we want in fact $\ 2^{123}<25\cdot 5^{51}$
There is not much room between $16$ and $25$ so the inequality is quite tight.
Another choice would be to consider other approximations like $\ 2^{16}<5^7\ $ or $\ 2^{30}<5^{13}$
The latter one is already quite tedious to calculate manually, yet it is not sufficiently tight:
You get $2^{123}=2^{30\times 4+3}<8\times 5^{13\times 4}<8\times 5^{52}$ but still $8>5$ and cannot conclude.
Note:
Some other posters have shown that using $\log_{10}(2)\approx 0.3010 < 0.3011$ was adequate
to prove the claim since $176\times 0.3011 < 53$.