how can i solve x/ln(n)=100

danielhaish

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Hi i am worikng on some encryption program and i need to solve this problem thereis any chance thus is possible using w Limburg function and how
 
Hi i am worikng on some encryption program and i need to solve this problem thereis any chance thus is possible using w Limburg function and how
The expression in the subject line

x/lenx=n

does not make any sense to me. Please "correct" it and re-post.
 
I would use the function:

[MATH]f(x)=x-n\ln(x)=0[/MATH]
And use the Newton-Raphson method to recursively find the root to a desired accuracy.
 
hi i olso got something like that but cant solve it
nt-e^t=0
I would use the function:

[MATH]f(x)=x-n\ln(x)=0[/MATH]
And use the Newton-Raphson method to recursively find the root to a desired accuracy.
in the end i simplified it to 100t-e^t while t=len(x)
and i use regula falsi method tenx
 
To solve xln(x)=n\displaystyle \frac{x}{ln(x)}= n for x, write it as ln(x)x=1n\displaystyle \frac{ln(x)}{x}= \frac{1}{n} and let y= 1/x. The equation becomes yln(1/y)=yln(y)=1n\displaystyle y ln(1/y)= -y ln(y)= \frac{1}{n} or yln(y)=1n\displaystyle y ln(y)= -\frac{1}{n}. NOW you can use Lambert's W function that I think you intended in your first post: y=W(1n)\displaystyle y= W\left(-\frac{1}{n}\right) and then x=1W(1n)\displaystyle x= \frac{1}{W\left(-\frac{1}{n}\right)}.
 
To solve xln(x)=n\displaystyle \frac{x}{ln(x)}= n for x, write it as ln(x)x=1n\displaystyle \frac{ln(x)}{x}= \frac{1}{n} and let y= 1/x. The equation becomes yln(1/y)=yln(y)=1n\displaystyle y ln(1/y)= -y ln(y)= \frac{1}{n} or yln(y)=1n\displaystyle y ln(y)= -\frac{1}{n}. NOW you can use Lambert's W function that I think you intended in your first post: y=W(1n)\displaystyle y= W\left(-\frac{1}{n}\right) and then x=1W(1n)\displaystyle x= \frac{1}{W\left(-\frac{1}{n}\right)}.
tenx it make my program alot more simple becuase now i just use function that allready exsits
 
To solve xln(x)=n\displaystyle \frac{x}{ln(x)}= n for x, write it as ln(x)x=1n\displaystyle \frac{ln(x)}{x}= \frac{1}{n} and let y= 1/x. The equation becomes yln(1/y)=yln(y)=1n\displaystyle y ln(1/y)= -y ln(y)= \frac{1}{n} or yln(y)=1n\displaystyle y ln(y)= -\frac{1}{n}. NOW you can use Lambert's W function that I think you intended in your first post: y=W(1n)\displaystyle y= W\left(-\frac{1}{n}\right) and then x=1W(1n)\displaystyle x= \frac{1}{W\left(-\frac{1}{n}\right)}.
hi i tried possed n=10 and then take the x and poss it in education and it doesn't return 10 it return 4
 
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