The equation has infinitely many solutions when x = 0, so you're going to have to partition the domain anyway. For negative values of x specifically, you might try toying with the inverse of x^{u}...
Even if x is negative, the ln x's still cancel out.
so, u = 1 for all x such that x != 0.
Thank you for the suggestion chiro.
One final quick question; using a Maclaurin or Taylor series would it be possible to evaluate ln x when x = 0? Sorry if this question sounds foolish, I haven't learned much about them yet.
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling We Value Civility
• Positive and compassionate attitudes
• Patience while debating We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving