Title is fairly self-explanatory. For example, for $y=27$, $x$ would be $3$. Specifically I was trying to find $x$ given $y = 10^{100}$, but I'd like to know how to solve it for any value of $y$.
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1$x=\exp W(\ln y)$ where $W$ is the Lambert $W$ function. – anon Jun 09 '15 at 23:07
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1In general, it requires a non-trivial function: http://math.stackexchange.com/questions/54210/is-xx-y-solvable-for-x – Brenton Jun 09 '15 at 23:07
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Re: your question of which areas of math this falls into, it falls into any of the categories of numerical methods, special functions and good old calculus. – anon Jun 09 '15 at 23:10
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@Timothy smith, it's seems to determine nature of x and y ? do you meant in integer ? or real numbers ? – zeraoulia rafik Jun 09 '15 at 23:10
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This answer answers a very similar question, and presents an algorithm to compute Lambert W. – robjohn Jun 09 '15 at 23:40
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This question asks *how* to solve this equation. The answers in the cited original do not show how to get the formula given in the accepted answer, nor do they show how to compute Lambert W without a CAS to do so. – robjohn Jun 10 '15 at 06:02
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$$\begin{align} x^x&=y\\ x\log x&=\log y\\ e^{\log x}\log x&=\log y\\ \log x&=W(\log y)\\ x&=e^{W(\log y)}\\ \end{align}$$ Where $W(x)$ is the ProductLog function, defined as the inverse of $[f(x)=xe^x]$
For your number: $$\begin{align} x^x&=10^{100}\\ x\log x&=\log 10^{100}\\ e^{\log x}\log x&=\log 10^{100}\\ \log x&=W(\log 10^{100})\\ x&=e^{W(100\log 10)}\\ x&\approx 56.9 \end{align}$$
Teoc
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Well, $x \ln x = \ln y$, so you can use binary search on $f(x) = x \ln x$ to get a pretty good estimate of the value. But you're not going to get a "closed form" formula for $x$ in this case.
It may be easier to use $\log_{10}$. For instance, in this case, you'd get $$ x \log_{10}(x) = 100 $$ so $x$ is between $0$ and $100$. Trying $x = 10$ gives a too-small value (namely 10), so $x = 10^{1.5}$ might be better: you get $10^{1.5} \cdot 1.5 \approx 31.62 \cdot 1.5 \approx 45$. So the right $x$ is somewhere between 30 and 100.
John Hughes
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Thanks, I'll research it. I edited the comment because I didn't see all the answers the first time. – Timothy Smith Jun 09 '15 at 23:08
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I'm afraid this math is beyond me but it was nice to satisfy my curiosity. (I approximated x to four decimal places given y = 10^100.) – Timothy Smith Jun 09 '15 at 23:14
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Hint :look Lambert W function. solutions represented as:
$$x=\frac{\ln(y)}{W(\ln y)}$$
zeraoulia rafik
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