X^2*log(2)x=256^2 (2) refers to base 2Is ther any rule to solve

  • Context: Undergrad 
  • Thread starter Thread starter kind
  • Start date Start date
  • Tags Tags
    Base
Click For Summary
SUMMARY

The equation x^2*log(2)x=256^2 can be solved using the Lambert W function, which is the inverse of the function y=x*e^(x). This method involves manipulating the equation to isolate log(x) and applying the W function to derive the solution for x. The final expression for x is given by x=±√(2*log(2)*(256)^2/W[2*log(2)*(256)^2]). The Lambert W function is essential for solving this type of equation as it is not an elementary function.

PREREQUISITES
  • Understanding of logarithmic functions
  • Familiarity with the Lambert W function
  • Basic algebraic manipulation skills
  • Knowledge of exponential functions
NEXT STEPS
  • Study the properties and applications of the Lambert W function
  • Learn how to manipulate logarithmic equations
  • Explore other special functions such as Bessel and Hankel functions
  • Practice solving equations involving exponential and logarithmic terms
USEFUL FOR

Mathematicians, students studying advanced algebra, and anyone interested in solving complex equations involving logarithmic and exponential functions.

kind
Messages
1
Reaction score
0
x^2*log(2)x=256^2

(2) refers to base 2
Is ther any rule to solve for x.
 
Mathematics news on Phys.org


No, there isn't any general rule to get an exact solution to that equation.
 


x can be solved for here, but it involves a function that you've probably not heard of, called the "Product Log function", or the "Lambert-W function". It's defined as the inverse function of the function y=x*e^(x). In other words, if x=y*e^(y), then y=W(x).

This is the only way to solve for x in your function, by using Lambert's W Function.

EDIT: Or, if c=1, then x=0.
 


kind said:
x^2*log(2)x=256^2

(2) refers to base 2
Is ther any rule to solve for x.

Here's how to do it via the Lambert W function:

\frac{x^2}{\log(2)}\log(x)=(256)^2

\log(x)=\frac{\log(2)(256)^2}{x^2}

x=\exp\left\{\frac{\log(2)(256)^2}{x^2}\right\}

Now square both sides and divide by x^2:

1=\frac{1}{x^2} \exp\left\{\frac{2\log(2)(256)^2}{x^2}\right\}

2\log(2)(256)^2=\frac{2\log(2)(256)^2}{x^2}\exp\left\{\frac{2\log(2)(256)^2}{x^2}\right\}

and that's where we take the W-function of both sides:

\frac{2\log(2)(256)^2}{x^2}=W\left[2\log(2)(256)^2\right]

x=\pm \sqrt{\frac{2\log(2)(256)^2}{W\left[2\log(2)(256)^2\right]}}

Nothin' wrong with the W function is it guys?
 
Last edited:


It is not an elementary function.

To just about any equation you might define a function by which the solution can be represented.

So what?

That Lambert functions, Hankel functions, Bessel functions and whatnot are useful is not the issue here.
 


arildno said:
It is not an elementary function.

To just about any equation you might define a function by which the solution can be represented.

So what?

That Lambert functions, Hankel functions, Bessel functions and whatnot are useful is not the issue here.

Come on Arildno. You made a boo-boo. Just fess-up. He asked how to solve it without qualification and what I showed is conceptually no different than using sines and cosines to solve an equation.

Stop special function discrimination: Equal rights for special functions.
 


jackmell said:
Stop special function discrimination: Equal rights for special functions.


Nope. :smile:
 

Similar threads

  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 16 ·
Replies
16
Views
1K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K