Comparing $2^{317}$ and $81^{50}$ by hand

Question

How to compare these two numbers without calculator:

$2^{317}$ and $81^{50}$

(Pen & paper test)

I thought about using logarithms and doing Taylor approximation, but these numbers are close to one another and I'd need a lot of Taylor expansion summands which defeats the purpose as raising 3 to some power of 15 and operating those fractions is not something you'd do by hand.

I've seen similar questions but there the powers were "nice" and are possible to simplify / reduce, yet there I don't see such an opportunity.

Answer 1

We have $$ 2^{317} > 81^{50} \iff 2^{17} > \left(\frac98\right)^{100} \iff \log_e2 > \frac{100}{17}\log_e\frac98, $$ and $$ \log_e2 = \log_e\frac{1 + 1/3}{1 - 1/3} > 2\left(\frac13 + \frac1{3\cdot3^3} + \frac1{5\cdot3^5}\right) = \frac23\left(1 + \frac1{27} + \frac1{405}\right) = \frac{842}{1215}, $$ and \begin{gather*} \frac{100}{17}\log_e\frac98 = \frac{100}{17}\log_e\frac{1 + 1/17}{1 - 1/17} = \frac{200}{17^2}\left(1 + \frac1{3\cdot17^2} + \frac1{5\cdot17^4} + \cdots\right) \\ < \frac{200}{17^2}\left(1 + \frac1{3\cdot17^2}\left(1 + \frac1{17} + \frac1{17^2} + \cdots\right)\right) = \frac{200}{17^2}\left(1 + \frac1{3\cdot16\cdot17}\right) \\ = \frac{200\cdot817}{3\cdot16\cdot17^3} = \frac{25\cdot817}{6\cdot4913} = \frac{25\cdot817}{29478} < \frac{25\cdot817}{29475} = \frac{817}{1179}, \end{gather*} and $$ 842\cdot1179 = 992718 > 992655 = 1215\cdot817, $$ therefore $$ \log_e2 > \frac{842}{1215} > \frac{817}{1179} > \frac{100}{17}\log_e\frac98, $$ therefore $$ 2^{317} > 81^{50}. $$

Answer 2

I'm not sure if this is something you consider as 'by hand', but the only thing you really need to calculate to a couple decimals is $2^{1.585}$.

We want to know the ratio $r = \frac{2^{317}}{3^{200}}$. Taking logarithms on both sides $$ \ln r = \ln\left(\frac{2^{317}}{3^{200}}\right) = 317 \ln 2 - 200 \ln 3.$$

Note that $317/200 = 1.585$ and we have that $2^{1.585}$ is slightly bigger than $3$. So in fact we have a lower bound for $\ln r$, hence the ratio by $$ \ln r = 317 \ln 2 - 200 \ln 3 > 317 \ln 2 - 200 \ln 2^{1.585} = 317 \ln 2 - 317 \ln 2 = 0, $$

so $r > e^0 = 1$, hence $2^{317}$ is slightly bigger.

Answer 3

I think it is not exactly easy to use approximations in this case, as both numbers are extremely close. Instead, just calculate each number with square and multiply: $$ 2^{317}=2\cdot 2^{316}=2\cdot 16^{79} = 32\cdot 16^{78} = 32\cdot 256^{39} = 8192 \cdot 65536^{19} = 536870912\cdot 4294967296^9 = 2305843009213693952\cdot 4294967296^8 = 2305843009213693952\cdot 18446744073709551616^4 = 2305843009213693952\cdot 340282366920938463463374607431768211456^2 = 2305843009213693952\cdot 115792089237316195423570985008687907853269984665640564039457584007913129639936 = 266998379490113760299377713271194014325338065294581596243380200977777465722580068752870260867072$$ Similarly $$ 81^{50} = 6561^{25} = 6561 \cdot43046721^{12} = 6561 \cdot 1853020188851841^6 = 6561 \cdot 3433683820292512484657849089281^3 = 22528399544939174411840147874772641 \cdot 11790184577738583171520872861412518665678211592275841109096961 = 265613988875874769338781322035779626829233452653394495974574961739092490901302182994384699044001$$

By writing the results under each other

$\begin{align*} 266998379490113760299377713271194014325338065294581596243380200977777465722580068752870260867072& \\ 265613988875874769338781322035779626829233452653394495974574961739092490901302182994384699044001& \end{align*}$

we see that the results are of same length and the third digit differs, making $2^{317}$ larger than $81^{50}$

Answer 4

This is not an answer, but too long for a comment-box

An option to work with the logarithms is possibly, to use the Mercator-series for $\log(1+x)$ which is well known and "handy" and can be used in this way: $$\small \log(1+x/2)-\log(1-x/2) \small = x + 1/12x^3 + 1/80x^5 + 1/448x^7 + 1/2304x^9 + 1/11264x^{11} + 1/53248x^{13} + O(x^{15}) $$ and assigning $x=1$ for the approximation-series of $\log(3)$: $$ \log(3) = 1+ 1/12 + 1/80 + 1/448 + 1/2304 + 1/11264 + ... \tag 1 $$ The explicite terms of this approximation-series can be found by hand to some leading terms.

Similarly from $\small \log(2)=\log(1+1/3)-\log(1-1/3)$ one can get $$ \log(2) = 2/3 + 2/81 + 2/1215 + 2/15309 + 2/177147 + 2/1948617 + 2/20726199 + ... \tag 2$$ Now we can write the subtraction $$ \Lambda_n = 317 \cdot P(\log(2),n) - 200 \cdot P(\log(3),n) \tag 3$$ for $n=1$ to some index, where $P(\log(2),n)$ means the partial series of first $n$ terms of the logarithmic approximation-series for the argument.

This is just to record the idea, I didn't do this by hand towards the solution. Using Pari/GP the termwise differences for $n=1 .. 16$ are anyway

                                          34/3
                                       -716/81
                                    -4807/2430
                                 -49603/122472
                               -471787/5668704
                            -4347523/249422976
                         -39533467/10611813888
                       -357424243/440798423040
                      -189606491/1057916215296
                  -29035761763/723614691262464
               -261425730427/28792247715495936
            -2353247072083/1135237195639554048
          -21180885641707/44422325046765158400
       -190634618747203/1727139997818229358592
     -1715738160612187/66782746582304868532224
  -15441749813059123/2569984316753525285584896
 ...

It remains to show, that the partial summation of this terms from 34/3 minus 716/81 and so on stays positive.
I don't know whether this is doable with any reasonable effort; it seems to be difficult anyway, when one looks at the sequence of partial sum in decimal representation:

     11.33333333
     2.493827160
    0.5156378601
    0.1106228362
   0.02739622924
  0.009965906383
  0.006240486071
  0.005429629625
  0.005250403247
  0.005210277249
  0.005201197523
  0.005199124611
  0.005198647803
  0.005198537427
  0.005198511736
  0.005198505728
  ...

Answer 5

It is worth noting as in one of the comments that the standard estimates in base $10$ of $\log(2)\sim 0.3010$ and $\log(3)\sim 0.4771$ do not yield the correct answer.

To prove that $2^{317}>81^{50}=3^{200},$ the following is doable by hand: $$2^{317}>3^{200}$$ $$\Leftrightarrow 2^{1.585}>3$$ $$\Leftrightarrow 2^{0.585}>1.5$$ $$\Leftrightarrow 2^{1.17}>2.25$$ $$\Leftrightarrow 2^{0.17}>1.125$$ $$\Leftrightarrow \left(2^{0.17}\right)^6>1.125^6$$ $$\Leftarrow 2^{1.02}>1.42383^2~ (\because 1.125^3=1.423828125)$$ $$\Leftrightarrow 2^{1.02}>2.027291869$$ $$\Leftarrow 2^{0.02}>\frac{2.027292}2=1.013646$$ $$\Leftrightarrow 2^2>1.013646^{100}$$ $$\Leftarrow 2^2>1.0137^{100}$$ $$\Leftrightarrow 2>1.0137^{50}$$ $$\Leftrightarrow 2>1.02758769^{25}$$ $$\Leftarrow 2>1.028^{25}$$ $$\Leftarrow 2>1.1481^5,$$ where one uses $1.028^2=1.056784$ and $1.028^3=1.086373952$ and hence $1.028^5$ is bounded by $1.05679\times 1.08638=1.14807552.$ The arrows of implications continue as follows $$\Leftarrow 2> (1.15)(1.1481^4)$$ $$\Leftrightarrow \frac {40}{23}>1.1481^4$$ $$\Leftrightarrow \frac {40}{23}>1.31813361^2$$ $$\Leftarrow \frac {40}{23} > 1.3182^2,$$ which is true, since $$1.7391\cdots > 1.73765124.$$

Answer 6

I (think) I found a solution that does not involve doing natural logarithms / logarithms at all and relies on basic algebra + approximation.

First step: multiply/divide ${2}^{317}$ by 8 and expand $81={3}^{4}$ we will have:

$$\frac{{2}^{320}}{8}\space?\space{3}^{200}$$

Next we can take the 40th root from both sides to arrive at this:

$$\frac{2^8}{3^5}\space?\space\sqrt[40]{8}$$

Or in other words:

$${(1+\frac{13}{243})}^{40}\space?\space8$$

So here we will reduce the powers very carefully by expanding the binomial as a square one step at a time and then replacing some of the numbers with those which are close to the power of 3. Here's how I do this:

$${(1+\frac{13}{243})}^{40} = {(1+\frac{2\cdot13}{3^5}+\frac{169}{3^{10}})}^{20} > {(1+\frac{2\cdot13}{3^5}+\frac{162}{3^{10}})}^{20} = {(1+\frac{2\cdot13}{3^5}+\frac{2\cdot3^4}{3^{10}})}^{20}$$ $$={(1+\frac{2\cdot13}{3^5}+\frac{2}{3^6})}^{20} = {(1+\frac{2\cdot3\cdot13+2}{3^6})}^{20}= {(1+\frac{80}{3^6})}^{20}$$

And we've successfully reduced the power by half while still retaining the manageable numbers under the calculation. So, next step:

$${(1+\frac{80}{3^6})}^{20}={(1+\frac{2\cdot80}{3^6}+\frac{6400}{3^{12}})}^{10}>{(1+\frac{2\cdot80}{3^6}+\frac{6399}{3^{12}})}^{10}={(1+\frac{2\cdot80}{3^6}+\frac{79\cdot3^4}{3^{12}})}^{10}$$ $$={(1+\frac{2\cdot80}{3^6}+\frac{79}{3^8})}^{10}={(1+\frac{160\cdot3^2+79}{3^8})}^{10}={(1+\frac{1519}{3^8})}^{10}>{(1+\frac{1518}{3^8})}^{10}={(1+\frac{3\cdot506}{3^8})}^{10} ={(1+\frac{506}{3^7})}^{10}$$

So we did halve it again and the numbers are still doable. Next step involves calculating $506^2$ but I assume one time is fine and it's a three-digit number close to 500, so I did it as $506^2=(500+6)^2=500^2+12\cdot500+36$.

Another tricky part in this step is finding out that $256036>255879$ i.e. choosing that "magic" number. Here's the algorithm I employ here (basically this is how I found these numbers for all steps):

1. Divide the original number by 3 with a remainder, check what that remainder is
2. Divide it again by 3 and continue to do so until we're down to the number less than 3. Note down the remainders on each step.
3. Compare the remainders. The smallest one will signify which power of 3 we want to choose. And of course the division result on that step is our second factor ($13$ in the case below)
4. For the division by 3 I use the "fast division" that's excellently described here.

Anyways,back to our inequality:

$${(1+\frac{506}{3^7})}^{10}={(1+\frac{2\cdot506}{3^7}+\frac{256036}{3^{14}})}^{5}>{(1+\frac{2\cdot506}{3^7}+\frac{255879}{3^{14}})}^{5}={(1+\frac{2\cdot506}{3^7}+\frac{13\cdot3^9}{3^{14}})}^{5}$$ $$={(1+\frac{2\cdot506}{3^7}+\frac{13}{3^{5}})}^{5}={(1+\frac{2\cdot506+13\cdot3^2}{3^7})}^{5}={(1+\frac{1129}{3^7})}^{5}={(\frac{3316}{3^7})}^{5}$$

And we've completed the power halving. Next step is - we want to estimate $\sqrt[5]{8}$ and to do so we can notice that ${(\frac{3}{2})}^5\approx8$ but still pretty close which makes it a good guess. We can "notice" this since $8-(\frac{3}{2})^5=8-\frac{243}{32}=\frac{13}{256}$ (close to $0$).

Here's what Newton's method gives. Suppose we have $x_{0}=\frac{3}{2}$ (we chose that guess earlier, see above). Then:

$$x_{1} = \frac{4}{5}\cdot x_0 + \frac{8}{5}\cdot\frac{1}{{x_0}^{4}}=\frac{4}{5}\cdot(\frac{3}{2}+2\cdot{(\frac{2}{3})}^4)=\frac{614}{405}=\frac{2\cdot307}{5\cdot3^4}>\sqrt[5]{8}$$

Critical bit here is that $x_{1}>\sqrt[5]{8}$. We can indeed claim it because we're approximating $f(x)=x^5-8$ and geometrically we are setting the tangent line to the $f(x)$ at the point $x_0=\frac{3}{2}$ which will then intersect the $x$-axis at $x_{1}$: but $f(x)$ is convex down for all $x>0$ therefore our tangent line at $x_0$ can never cross the graph of $f(x)$ to the right of our $x_0$. That means that it will cross the $x$-axis below the graph of $f(x)$ meaning that $x_1$ will be to the right of the true root which is $\sqrt[5]{8}$ and hence $x_{1}>\sqrt[5]{8}$

All that's left is to compare our $x_1$ with the fraction we've built before:

$$\frac{3316}{3^7}=\frac{2^2\cdot829}{3^7}\space?\space\frac{2\cdot307}{5\cdot3^4}$$

Removing common terms:

$$\frac{2\cdot829}{27}\space?\space\frac{307}{5}$$ $$10\cdot829\space?\space27\cdot307$$ $$8290>8289$$

(By the way,a difference of 1 is as close as it can possibly be which I find hilarious) So we managed to build this:

$${(1+\frac{13}{243})}^{40}>{(1+\frac{80}{3^6})}^{20}>{(1+\frac{506}{3^7})}^{10}>{(\frac{3316}{3^7})}^{5}>(\frac{614}{405})^{5}>8$$

Hence $\frac{2^8}{3^5}>\sqrt[40]{8}$ and finally

$${2}^{317}>{3}^{200}={81}^{50}$$

Answer 7

Alternative solution:

We have $$2^{317} > 81^{50} ~ \iff ~ 2^{317\cdot 3/50} > 81^3 ~ \iff ~ 2^{1/50}2^{19} > 81^3$$ $$~ \Leftarrow ~ \left(1 + \frac{1}{50}\ln 2\right)2^{19} > 81^3 ~ \Leftarrow ~ \left(1 + \frac{1}{50}\cdot \frac{11}{16}\right)2^{19} > 81^3$$ $$~ \iff ~ 811 \cdot 2^{16} > 81^3 \cdot 100 ~ \iff ~ (10 + 1/81)\cdot 2^{16} > 81^2 \cdot 100$$ $$ ~ \iff ~ \frac{1}{81}\cdot 2^{16} > 81^2 \cdot 100 - 10 \cdot 2^{16}$$ $$ ~ \iff ~ \frac{1}{81}\cdot 65536 > 6561 \cdot 100 - 10 \cdot 65536 = 740$$ $$~ \iff ~ 65536 > 81 \cdot 740$$ which is true; here, we have used $\mathrm{e}^{y} \ge 1 + y$ for all $y\in \mathbb{R}$, and $\ln 2 > \frac{11}{16}$ which follows from (I learned it from @Jack D'Aurizio) $$0 < \int_0^1 \frac{x^2(1 - x)^2}{1 + x}\,\mathrm{d} x = \int_0^1 \left(x^3 - 3x^2 + 4x - 4 + \frac{4}{1 + x}\right)\mathrm{d} x = 4\ln 2 - \frac{11}{4}.$$