How to get a variable in a closed form given an equation

Question

I need to express a variable in a closed form given an equation, $k^4 = 2^k.$ Taking the logarithm from both sides I get $k^{\frac{1}{k}} = 2^{\frac{1}{4}}$ which is not helping much. I tried other things but without getting a satisfactory solution. Do you have any idea how to get a closed form for $k$ ? Thanks.

Answer 1

$$k^4=2^k$$ $$k^4=e^{\ln(2)k}$$ $$k^4e^{-\ln(2)k}=1$$

We see, your equation is an algebraic equation in dependence of two algebraically independent monomials ($k$, $e^{\ln(2)k}$). Therefore, the elementary function $k\to k^4e^{-\ln(2)k}$ doesn't have a partial inverse that is an elementary function. Therefore, the equation cannot be rearranged for $k$ by applying only elementary functions/operations we can read from the equation.

Your equation is an exponential polynomial equation, and some simple cases of such kind of equations can be solved by Lambert W. Your equation is solvable by Lambert W:

$$k^4e^{-\ln(2)k}=1$$ $$\left(k^4e^{-\ln(2)k}\right)^\frac{1}{4}=1^\frac{1}{4}$$ $$\left(k^4e^{-\ln(2)k}\right)^\frac{1}{4}=1$$ $$-ke^{-\frac{\ln(2)}{4}k}=1;\ -kie^{-\frac{\ln(2)}{4}k}=1;\ ke^{-\frac{\ln(2)}{4}k}=1;\ kie^{-\frac{\ln(2)}{4}k}=1$$ $\forall n\in\mathbb{Z}$: $$-\frac{1}{4}\ln(2)k=W_n\left(-\frac{1}{4}\ln(2)\right),\ W_n\left(\frac{1}{4}\ln(2)i\right),\ W_n\left(\frac{1}{4}\ln(2)\right),\ W_n\left(-\frac{1}{4}\ln(2)i\right)$$ $$k=-\frac{4}{\ln(2)}W_n\left(-\frac{1}{4}\ln(2)\right),\ -\frac{4}{\ln(2)}W_n\left(\frac{1}{4}\ln(2)i\right),\ -\frac{4}{\ln(2)}W_n\left(\frac{1}{4}\ln(2)\right),\ -\frac{4}{\ln(2)}W_n\left(-\frac{1}{4}\ln(2)i\right)$$