Given $x^n=y$ where $n=y$, I have no problem finding $x$ if $y$ is known. Problem is getting $y$ when only the value of $x$ is known. Is there a way? At the moment I'm working with $x$ and $y$ values less than $1$. Thanks.
Asked
Active
Viewed 42 times
0
-
In the elementary functions it's impossible. – Michael Rozenberg Jan 31 '19 at 18:00
-
$\ln(x^y)=y\ln(x)$ – hamam_Abdallah Jan 31 '19 at 18:01
2 Answers
1
So you are trying to solve for $y$ in the equation $x^y = y$.
The Lambert W function is useful for this.
$x^y=y \implies x^{-y} = y^{-1} \implies yx^{-y}=1 => ye^{-\ln(x)y} =1$
Multiply by $-\ln x$ on both sides:
$-y\ln xe^{-y\ln x}=-\ln x$
Now use the Lambert W function on both sides.
$-y\ln x=W(-\ln x) \implies y=-\frac {W(-\ln x)}{\ln x}$ For $x \lt 1$, $W(-\ln x)$ is well defined, so this is our solution.
For the love of maths
- 2,368
-
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot. – Adam Boit Jan 31 '19 at 18:46
-
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method? – Adam Boit Feb 01 '19 at 13:51
-
-
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math] – Adam Boit Feb 03 '19 at 11:58
-
Yes, this is trivial. However it does not help you find y for a given x like my solution. – Abhishek Vangipuram Feb 03 '19 at 19:52
-
-
This is what I meant to write in full. y=log(y)/log(x). From here I propose an iterative method of finding the value of y when x is given. y'=y/2+log (y)/2log(x) – Adam Boit Feb 05 '19 at 12:05
-
yes you could Euler's Method to approximate y using the easily known solution (1,1) or any other known solution, however Euler's Method is very inaccurate for far away x and with anything but a miniscule step. – Abhishek Vangipuram Feb 07 '19 at 00:07
