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

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 $$176\times 0.3011 < 53$$.

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 $$2^{123}<5^{53}$$

• Why do have defined $a$ and $R$? What is the reason? Why $a=\ln(1.024)$? Nov 11, 2021 at 21:15
• @Sebastiano the reason we define $a$ is to have something like $\ln(1+x)$ with small $x$ that has a good approximation for the log around $1$. we define $R$ simply because comparing log of big quantities is easier than comparing them directly. the main idea is to reduce the coefficients before $\ln(2)$ and $\ln(5)$ in $R$ so that we end up again with logarithms of quantities close to $1$ where we have good approximations and inequalities for the log. I used also this technique here for instance math.stackexchange.com/a/4299385/399263
– zwim
Nov 14, 2021 at 22:56

No logarithms around in this solution, and a calculator need not be at hand.

Let's start by sharpening the claimed bound: consider $$1 \;<\;\frac{5^3}{2^7}\cdot\frac{5^{53}}{2^{123}} \;<\;\frac{5^{53}}{2^{123}}$$ then focus on the left "$$<$$" whilst the right one follows from $$\,5^3=125 <128=2^7$$.

The cube $$\,63^3=250\,047\,$$ is just a few bits away from $$\,250\,000$$, thus $$1.26^3 \;=\;\left(\frac{63}{50}\right)^3 \;=\;\frac{250047}{125000} \;\gtrsim\;\frac{250000}{125000} \;=\;2$$ so $$1.26$$ is slightly above $$\sqrt[3\,]{2}\,$$. That yields $$\frac{5^{56}}{2^{130}} \;=\;\frac{5^{54}}{2^{108}\cdot 2^{18}}\cdot\frac{5^2}{2^4} \;>\;\left(\frac{1.25}{1.26}\right)^{54}\cdot\left(\frac{5}{4}\right)^2$$ and the claimed bound is established if $$\left(\frac{1.25}{1.26}\right)^{54} =\left(1- \frac1{126}\right)^{54} >\big(\frac{4}{5}\big)^2\,$$ is shown. By truncating the Binomial expansion after four terms and upon simplification (without calculator $$\ddot\smile$$) we obtain \begin{align}\left(1- \frac1{126}\right)^{54} & \;>\;1-\frac{54}{126} \;\;+\;\frac{54\cdot 53}{2}\cdot\frac1{126^2} \,-\,\frac{54\cdot 53\cdot 52}{2\cdot 3}\cdot\frac1{126^3} \\[2ex] & \;=\quad\frac8{14} \quad +\;\frac{27\cdot 53}{126^2} \left(1-\frac{52}{3\cdot 126}\right) \\[2ex] & \;=\quad\frac8{14} \quad +\;\frac{53\cdot 163}{7\cdot 126^2} \\[2ex] & \;>\quad\frac8{14} \quad +\;\frac{49\cdot 162}{7\cdot 7^2\cdot 18^2} \;=\;\frac9{14} \end{align} The truncation is okay as the Binomial expansion has alternating signs and each plus-minus-pair is positive. Because of $$\,\frac9{14} - \frac{16}{25} = \frac1{14\cdot 25}\,$$ we are done.

In summary it has been shown that $$\frac{5^{53}}{2^{123}} \;>\;1.024\,=\,\frac{2^7}{5^3}\,.$$

• Many compliments. :-) Nov 11, 2021 at 21:14