How to solve this mathematical equation for x

Question

I am trying to solve this equation for $x$: $$ y = 1 - e^\left( \frac{{ x ^ {0.22} } ({e^{-cx^{0.22}}-1)}}{c} \right) $$ I tried by transferring 1 to left hand side and then taking ln on both sides but that doesn't gives me the solution for x

Can anyone help ?

Answer 1

It was important to be precise about the parenthesis because the slightest change is likely to make the equation solvable or not solvable analytically.

For example the quation $(1)$ is solvable thanks to a special function
$$\boxed{y = 1 - e^ \frac{({ x ^ {0.22} } {e^{-cx^{0.22}}})-1}{c} }\tag 1$$ $\frac{({ x ^ {0.22} } {e^{-cx^{0.22}}})-1}{c}=\ln|1-y|$

$ x ^ {0.22} e^{-cx^{0.22}}=1+c\ln|1-y|$

$ (-cx ^ {0.22}) e^{(-cx^{0.22})}=-c-c^2\ln|1-y|$

With $\begin{cases}X=(-cx ^ {0.22}) \\ Y=-c-c^2\ln|1-y| \end{cases}$ the equation is transformed into $\quad Xe^X=Y$

The solution is $\quad X=W(Y)$ .

$W$ is the Lambert W function : https://mathworld.wolfram.com/LambertW-Function.html

$-cx ^ {0.22}=W\left(-c-c^2\ln|1-y| \right)$

$$x=\left(-\frac{1}{c}W\left(-c-c^2\ln|1-y| \right)\right)^{1/0.22}$$

This is not the same for equation $(2)$ :

$$\boxed{y = 1 - e^ \frac{{ x ^ {0.22} } ({e^{-cx^{0.22}}-1)}}{c} }\tag 2$$

$\frac{{ x ^ {0.22} } ({e^{-cx^{0.22}}-1)}}{c}=\ln|1-y|$

With $\begin{cases}X=(-cx ^ {0.22}) \\ Y=-c^2\ln|1-y| \end{cases}$ the equation is transformed into $\quad Xe^X-X=Y$

This is not the Lambert equation $Xe^X=Y$. The Lambert W function is no longer convenient.

A different special function would be necessary. As far as I know, up to now, no convenient special function has been standardized for solving this kind of equation.

Do not expect an explicit solution for equation $(2)$. One have to solve it approximately with numerical calculus. They are a lot of numerical methods, for example Newton-Raphson, etc. But the parameters $c$ and $y$ have to be known on numerical form.

Answer 2

Let $t = \frac{x^{0.22}e^{-cx^{0.22}}}{c}$.

Then, $-cx^{0.22}e^{-cx^{0.22}} = -c^2t$.

$-cx^{0.22} = W(-c^2t)$, where $W(x)$ is the Lambert's function, which is defined as an inverse to the function $f(w) = we^w$.

Now, we have

$x^{0.22} = -\frac{1}{c}W(-c^2t).$

$x^{\frac{11}{50}} = -\frac{1}{c}W(-c^2t).$

$x = \left(-\frac{1}{c}W(-c^2t)\right)^{\frac{50}{11}}.$

Let us define $t$:

$y = 1-e^t \Leftrightarrow t = \log{(1-y)} \Rightarrow$

$x = \left(-\frac{1}{c}W(-c^2\log(1-y))\right)^{\frac{50}{11}}.$