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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 $176\times 0.3011 < 53$.

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2 Answers 2

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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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  • $\begingroup$ Why do have defined $a$ and $R$? What is the reason? Why $a=\ln(1.024)$? $\endgroup$
    – Sebastiano
    Nov 11, 2021 at 21:15
  • $\begingroup$ @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 $\endgroup$
    – zwim
    Nov 14, 2021 at 22:56
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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 $$\frac{5^{53}}{2^{123}} \;>\;1.024\,=\,\frac{2^7}{5^3}\,.$$

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  • $\begingroup$ Many compliments. :-) $\endgroup$
    – Sebastiano
    Nov 11, 2021 at 21:14

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