Without using a calculator how to solve $x^x = 100$?

Question

Without using a calculator how to solve $x^x = 100$ ? A way of finding an approximation to 2 decimals would be good neough.

I know about the Lmabert W function but one cannot compute it mentally. This is why I believe my question not to be a duplicate.

Sorry, but how would you do that with using a calculator? Then, you do the same with mental arithmetic. — , Dec 20 '17 at 14:46
See here: How to solve $x^x=100$? or Solving for $x$: $x^x=N$ or How do you solve this equations where the unknown is to the power of the unknown? — projectilemotion, Dec 20 '17 at 14:46
You need something called the "Lambert W function" for a closed form, but it numerical methods can be applied to get estimates. — Thomas Andrews, Dec 20 '17 at 14:49
@ThomasAndrews I do not think it is a duplicate, because this question explicitely prohibits using calculators. — M. Winter, Dec 20 '17 at 14:50
See https://en.wikipedia.org/wiki/Lambert_W_function#Example_2 — Barry Cipra, Dec 20 '17 at 14:51
@M.Winter, I agree the prohibition on calculators gives the question some novelty, but in that case it'd help for the OP to specify what it means to "solve" the equation. — Barry Cipra, Dec 20 '17 at 14:54
I meant finding an approximate answer at 2 decimals. For the lambert W function, we cannot compute mentally taht is why it is a different from the one you pointed out. — astudentofmaths, Dec 20 '17 at 15:06
@Arthurim Maybe you could ask how to approximate Lambert $W$ without a calculator instead? Wikipedia has quite some information, e.g. here. — M. Winter, Dec 20 '17 at 15:19

score 2 · Accepted Answer · answered Dec 20 '17 at 15:11

2

If you know that $\log2\approx0.3$ and $\log3\approx0.48$ (where "$\log$" is log base $10$), then you can show that

$$\begin{align} \log(3.6^{3.6}) &=3.6\log(3.6)=7.2(\log2+\log3)-3.6\\ &\approx7.2\cdot0.78-3.6\\ &=0.001((75-3)(75+3))-3.6\\&=0.001(75^2-9)-3.6\\ &=0.001(5625-9)-3.6\\ &=5.616-3.6=2.016\\ &\approx2 \end{align}$$

Note the "trick" $(a5)^2=(a(a+1))25$ for squaring $75$ (i.e., $56=7\cdot8$).

Remark: $\log2\approx0.3$ comes from $2^{10}=1024\approx10^3$. The estimate for $\log3$ can be obtained with a bit a rounding from $3^4\approx80$.

answered Dec 20 '17 at 15:11

Barry Cipra

79,832

How do you get the $3.6\log(3.6)=7.2(\log2+\log3)-3.6$ ? – astudentofmaths Dec 20 '17 at 15:20
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@Arthurim, $\log(3.6)=\log36-\log10=2\log6-1=2(\log2+\log3)-1$ – Barry Cipra Dec 20 '17 at 15:24
Just, how do you get to first estimate 3.6? We know that x is between 3 and 5, but unless you knoiw the answer you have to go by trial and error right? – astudentofmaths Dec 20 '17 at 15:34
@Arthurim, I started with the obvious bounds $3\lt x\lt4$, as in gimusi's answer, then knew that $\log3.6$ would be easy to estimate from easy estimates of $\log2$ and $\log3$. What's missing, of course, is any estimate of how close $3.6$ is to the actual answer. That would require a good deal more work. – Barry Cipra Dec 20 '17 at 21:12

score 1 · Answer 2 · answered Dec 20 '17 at 14:45

1

You can try:

$3^3=27$

$4^4=256$

thus $3<x<4$.

Then you can try with $3.5^{3.5}$ and so on.

answered Dec 20 '17 at 14:45

user

154,566

2

But already calculating $3.5^{3.5}$ in your head (or even on paper) is something only a few people can do today :P – M. Winter Dec 20 '17 at 14:49
Yes ok but at least you can get a good approximation! – user Dec 20 '17 at 14:52
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You can compute $(7/2)^7$ and compare to $100^2$, @M.Winter So is $7^7>1,280,000$? – Thomas Andrews Dec 20 '17 at 14:53

M. Winter · Answer 3 · 2017-12-20T15:10:14.783

You can try to approximate the final solution using rational numbers $p/q$. For this you try to find $p$ and $q$ so that

$$\left(\frac pq\right)^{p/q}\approx100\qquad\Leftrightarrow\qquad p^p\approx 100^q\cdot q^p.$$

With some time and patience, the latter form can be computed on paper. What makes an approximation good? Probably when you have that the quotient of $p^p$ and $100^q\cdot q^p$ is near one.

An algorithm

A good starting point (as seen in @gimusi's answer) would be $p=7$ and $q=2$. You would find that $823543=7^7 < 100^2\cdot 2^7=1280000$. Now you can do the following:

If $p^p < 100^q\cdot q^p$ (as in this example), then replace $p\mapsto 3p$ and $q\mapsto2q$.
If $p^p > 100^q\cdot q^p$, then only replace $q\mapsto2q$.

This is basically binary bisection. So according to above, the next approximation to check would be $p=21$ and $q=8$. And you do this again and again until the result is good enough for you. The exact result is irrational, so you cannot hope to find it this way.

score 0 · Answer 4 · answered Dec 20 '17 at 15:07

You could potentially use a sequence of approximations. Suppose you want to approximate this with rational numbers, so let $x=\frac{a}{b}$.

$$ (\frac{a}{b})^{\frac{a}{b}}=100 $$

Rearranging this gives $(100b)^b=a^a$. Now let $b=1,2,3\ldots$ and find a suitable $a$ that bounds the answer above or below. For example if $b=1$:

$$ 27=3^3<100=(100\cdot1)^1<4^4=256, $$

$b=2$: $$ 3125=5^5< 40,000=(100\cdot 2)^2 <6^6=46656, $$

and $b=3$:

$$ 16,777,216=8^8< 27,000,000=(100\cdot 3)^3 <9^9=387,420,489, $$

one can obtain an upper and lower sequence of rational approximations:

$$ 3,\frac{5}{2},\frac{8}{3}\ldots,\\ 4,\frac{6}{2},\frac{9}{3}\ldots\\ $$

although this approach doesn't appear to converge terribly quickly.

Without using a calculator how to solve $x^x = 100$?

4 Answers4