Prove $2^{123}<5^{53}$ without calculator.

arithmeticexponentiationinequalitylogarithms

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 $176\times 0.3011 < 53$.

Best Answer

Let's call $\begin{cases}R=\ln\left(\dfrac{2^{123}}{5^{53}}\right)=123\ln(2)-53\ln(5)\\\\a=\ln(1.024)\end{cases}$


We expand the logarithm of $2$ now:

$10\ln(2) = \ln(1024) = \ln(1.024)+3\ln(10) = a + 3\ln(2)+3\ln(5)$

$$a = 7\ln(2)-3\ln(5)$$


And we use it to reduce $R$:

$R=119\ln(2)+4\ln(2)-51\ln(5)-2\ln(5)=17a+4\ln(2)-2\ln(5)=17a-\ln(\frac{25}{16})$


We will now use the inequality $$\ln(1+x)\le x\le\frac 12\ln\left(\dfrac{1+x}{1-x}\right)$$

Note that $\frac{1+x}{1-x}=y\iff x=\frac{y-1}{y+1}$

So we have $\begin{cases}17a\le \frac{17\times 24}{1000}=\frac{51}{125}\\\\ \ln(\frac{25}{16})\ge 2\times\frac{25/16-1}{25/16+1}=\frac{18}{41}\end{cases}$

This allows us to conclude that $R\le \dfrac{51}{125}-\dfrac{18}{41}=-\dfrac{159}{5125}<0$

Numerical verification:

$R \approx -0.0431 < -\frac{159}{5125}\approx -0.03102$

Since $R<0\iff e^R<1$ we conclude that $2^{123}<5^{53}$

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