What is the solution of given equation?

Question

Is the solution correct for the given equation?

$\log (x) + \frac {2\cdot u}{x}=0$

On differentiation, $\frac 1x - \frac {2\cdot u}{x^2}=0$

$x=2\cdot u$

Answer 1

This is a transcendental equation that has no analytical solution. Anyway, you can gain some insight by considering

$$u=-\frac{x\log x}2.$$

This function is initially positive, reaches the maximum $(1/e,e/2)$, then crosses the axis at $(1,0)$ and decreases monotonically.

So the number of roots is

$$\begin{cases} u < 0 &\to 1,&x>1\\ u = 0 &\to 2,&x=0\text{ and }x=1\\ 0 < u < 1/e &\to 2,&0<x<1/e\text{ and } 1/e<x<1\\ u= 1/e &\to 1\text{ (double)},&x=e/2\\ u > 1/e &\to 0. \end{cases}$$