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

• kind
In summary, the equation x^2*log(2)x=256^2 can be solved using the Lambert-W function, which is defined as the inverse function of y=x*e^(x). This function is the only way to solve for x in this equation. Alternatively, if c=1, then x=0. However, it is not an elementary function and is on par with other useful functions such as sines and cosines. Therefore, it should not be discriminated against and given equal rights as other special functions.
kind
x^2*log(2)x=256^2

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

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.

## 1. What is the value of x in the equation X^2*log(2)x=256^2 (2)?

The value of x in this equation can be found by using logarithmic rules and algebraic manipulation. First, take the logarithm of both sides of the equation to get log(2)x = (256^2)/x. Then, use the power rule of logarithms to simplify the right side, giving log(2)x = 65536/x. Finally, cross-multiply and solve for x to get the value of x as approximately 4.209.

## 2. Can this equation be solved using any specific rules or methods?

Yes, this equation can be solved by using logarithmic rules and algebraic manipulation. It is also helpful to have a basic understanding of exponent rules and logarithmic properties.

## 3. Is the number 2 referring to the base in this equation?

Yes, the number 2 in this equation refers to the base of the logarithm. The base indicates what number is being raised to a certain power in the logarithm.

## 4. Are there any special considerations for using base 2 in this equation?

Using base 2 in this equation allows for easier calculations since the logarithm of 2 is equal to 1. This means that the equation can be simplified to become X^2 = 256^2, making it easier to solve for x.

## 5. Can this equation be solved for values other than base 2?

Yes, this equation can be solved for any base as long as the logarithm of that base is a known value. However, using a base other than 2 may result in more complex calculations and a more difficult solution process.

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