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$x^x = 100$.

I have no clue on how to solve this. If you guys have, please show me your solution as well.

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    $\begingroup$ The solution is about $3.6$. I'm certain that it cannot be solved in closed form, without the use of a special function. $\endgroup$ Commented Jan 1, 2015 at 9:08
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    $\begingroup$ What kind of a solution are you looking for? Would it be enough to prove that there is only one positive real solution and calculate an approximate value for it? Where does $x$ take values? $\endgroup$ Commented Jan 1, 2015 at 9:09
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    $\begingroup$ Have a look at the following. math.stackexchange.com/questions/50316/xx-y-how-to-solve-for-x $\endgroup$
    – Radz
    Commented Jan 1, 2015 at 9:10
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    $\begingroup$ There is closed form; to solve it using graphics, take logarithms and plot $x\log(x)-\log(100)$. $\endgroup$ Commented Jan 1, 2015 at 9:15
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    $\begingroup$ See the Lambert W function. $\endgroup$
    – Lucian
    Commented Jan 1, 2015 at 9:16

3 Answers 3

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$$\begin{align} x^x&=100\\ x\log x&=\log 100\\ e^{\log x}\log x&=\log 100\\ \log x&=W(\log 100)\\ x&=e^{W(\log 100)}\\ x&\approx 3.597 \end{align}$$ Where $W(x)$ is the ProductLog function, defined as the inverse of $[f(x)=xe^x]$

Explanations: 1. Problem

2.Logs

3.$ e^{\log x}=x$

4.Definition of $W(x)$

5.Taking $e^x$

6.Numerical Solution

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  • $\begingroup$ He explains in the answer. Read the last line? $\endgroup$
    – H_B
    Commented Jan 1, 2015 at 9:17
  • $\begingroup$ Don't you mean $e^{\ln(x)} = x$? You wrote log instead of ln. $\endgroup$
    – DrZ214
    Commented Jan 31, 2017 at 9:35
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    $\begingroup$ @DrZ214 $\log$ is commonly used to denote the base e logarithm. You really only see log denoting the base 10 log in high school. $\endgroup$
    – Teoc
    Commented Jan 31, 2017 at 15:31
  • $\begingroup$ @Displayname I don't think I've ever seen it used that way before. I live in the states and got through calc 1/2/3 in undergrad, if it matters. $\endgroup$
    – DrZ214
    Commented Jan 31, 2017 at 22:45
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$$3^3=27\,\,\,\,\,\,\,\,\,\,\,\,\,\ 4^4=256$$ Therefore there are no integer solutions for this equation. The unique solution of this equation with $50$ decimal places can obtain using Mathematica as

$$3.5972850235404175054976522517822860691355430548866.$$

Also I plot the graph of $f(x)=x^x-100$ in $x\in[3,4].$

enter image description here

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x=log(100)/log(x)

x'=x/2+log(100)/2log(x)

After several iteration for x (that is x', x'' to xn) you'll come to a good value of x to so much decimal places.

What am doing up there is applying the Babylonian method (an iterative algorithm similar to Newton-Raphson method). I've reduced it to its empirical form. Self taught I should add.

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    $\begingroup$ Please format your answer for better clarity. Also, please explain that you're proposing a numerical approximation via an iterative method. Thanks! $\endgroup$ Commented Feb 2, 2019 at 5:17
  • $\begingroup$ This is a good answer that contributes to the question what other answers have not! You can make it even better by formatting your answer, and using mathjax to type your math expressions to improve readability. Cheers! $\endgroup$
    – YiFan Tey
    Commented Feb 2, 2019 at 5:48
  • $\begingroup$ I've no access to a computer (I'm using a feature phone) so formatting is not possible unfortunately. $\endgroup$
    – Adam Boit
    Commented Feb 2, 2019 at 9:07

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