$(x*(\log_{2}(x))^2)/2 = x^{3/2}$ how to solve it?

Question

Is there a manual solution for this equation? Or I should use Wolfram: result from Wolfram.

Answer 1

There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:

The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $\mathbb{C}$ and there are two possible functions $W_0$ with $\textrm{Re}(W_0(x))\geq -1$ and $W_{-1}$ with $\textrm{Re}(W_{-1}) < -1$.

Then: \begin{eqnarray} \frac{1}{2}x\ \left(\log_2(x)\right)^2 &=& x^{3/2} \\ \frac{1}{2}x^{-1/2}\ \left(\log_2(x)\right)^2 &=& 1 \\ \frac{1}{\sqrt{2}}x^{-1/4}\log_2(x) &=& \pm 1 \\ x^{-1/4}\frac{\ln(x)}{\ln(2)} &=& \pm \sqrt{2} \\ x^{-1/4}\ln(x) &=& \pm \sqrt{2}\ln(2) \\ e^{(-1/4)\ln(x)}\ln(x) &=& \pm \sqrt{2}\ln(2) \\ e^{(-1/4)\ln(x)}\left(-\frac{1}{4}\ln(x)\right) &=& \pm \frac{1}{4}\sqrt{2}\ln(2) \\ \end{eqnarray} Then with Lambert function: \begin{eqnarray} -\frac{1}{4}\ln(x) &=& W\left(\pm\frac{1}{4}\sqrt{2}\ln(2)\right) \\ x &=& \exp \left\lbrace-4W\left(\pm\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace \\ \end{eqnarray} Then the solutions are: \begin{eqnarray} x_1 &=& \exp \left\lbrace-4W_0\left(\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace = 0.44836908960\ldots \\ x_2 &=& \exp \left\lbrace-4W_0\left(-\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace=4 \\ x_3 &=& \exp \left\lbrace-4W_{-1}\left(\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace=-153792.65205358\ldots-i101297.96245405\ldots \\ x_4 &=& \exp \left\lbrace-4W_{-1}\left(-\frac{1}{4}\sqrt{2}\ln(2) \right)\right\rbrace=6380.45994697086\ldots \\ \end{eqnarray}

The function is numeric too, but it is very elegant.

There is an article in wikipedia for this Lambert W function.

Answer 2

This equation has three real solutions: $$x\approx 0.4483690898$$ $$x=4$$ $$x\approx 6380.459941$$ This can be obtained by the Newton Raphson method.

Answer 3

Welcome to the world of Lambert function !

Using Matti P.'s suggestion $x=2^y$, the equation becomes $$2^{y-1} y^2=2^{\frac{3y}{2}}$$ that is to say $$y^2=2^{\frac{y}{2}+1}=2 \times2^{\frac{y}{2}}$$ Now, let $2^{\frac{y}{2}}=t$, that is to say $y=\frac{2 t}{\log (2)}$ to make the equation $$e^{2 t} \left(\frac{2 t^2}{\log ^2(2)}-e^t\right)=0\implies e^t=\frac{2 t^2}{\log ^2(2)}$$ and then the three roots $$t_1=-2 W_0\left(\frac{\log (2)}{2 \sqrt{2}}\right)\qquad t_2=-2 W_0\left(-\frac{\log (2)}{2 \sqrt{2}}\right)\qquad t_3=-2 W_{-1}\left(-\frac{\log (2)}{2 \sqrt{2}}\right)$$

It have been faster to use $x=2^{e^t}$ for the same result.