Find the number of vectors $(x, x_1, x_2)$ , where $x, x_1, x_2 > 0$ , for which the following holds

Question

$$\left | \log(xx_{1}) \right |+\left | \log(xx_{2}) \right |+\left | \log\frac{x}{x_{1}} \right |+\left | \log\frac{x}{x_{2}} \right |= \left | \log x_{1}+\log x_{2} \right |$$

Please kindly help with this, I can't figure out how to proceed. Thanks in advance!

Answer 1

At least $(1,1,1)$ is a solution.

The pesky bits here are the absolute value function $\lvert . \rvert$ applications, they can be removed if one does the case distinctions on the arguments. $$ \lvert x \rvert = \begin{cases} x & x\ge 0 \\ -x & x < 0 \end{cases} $$

There are $5$ arguments, so one has to deal with $2^5 = 32$ cases at worst.

Then log computation rules can be used to simplify.

Answer 2

Write $x:=e^t$ with $t\in{\mathbb R}$, and similarly for the other variables. We then obtain the equation $$|t+t_1|+|t+t_2|+|t-t_1|+|t-t_2|=|t_1+t_2|\ .\tag{1}$$ The left hand side of $(1)$ can be written as $$\eqalign{|t_1-t|+|t-(-t_1)|+|t_2-t|+|t-(-t_2)|&\geq|t_1-(-t_1)|+|t_2-(-t_2)|\cr&=2|t_1|+2|t_2|\ ,\cr}$$ and the right hand side of $(1)$ is $\leq|t_1|+|t_2|$. It follows that for any solution $(t,t_1,t_2)$ of $(1)$ we have $$2|t_1|+2|t_2|\leq|t_1|+|t_2|\ .$$ This immediately implies $t_1=t_2=0$, and $(1)$ then gives $t=0$.

It follows that $(1,1,1)$ is the only solution to the given equation.