"Impossible" logarithm question?

Debatable- some would argue x = -1, y = 2 is a solution, but that requires a little fiddling with the second equation for it to be defined.

The question you have to ask is are 2log(x) and log(x^2) equivalent for -ve x.

log(x) is undefined for x<=0, so x=-1 cannot possibly be a solution.

log(x) has no real value for x<= 0, but that doesn't mean it's undefined. In this case the solutions have to be real numbers but that doesn't prevent you working through the complex plane to get a result.

the standard definition of the logarithm applies exactly the same for numbers < 0, there are simply no real solutions to log(-x).
It's fairly simple to show this without ever needing to consider something undefined- Start with the Euler Identity

$-1 = e^{\pi i}$

$-x = xe^{\pi i}$

$log(-x) = log(xe^{\pi i})$

$2log(-x) = 2log(xe^{\pi i})$

$2log(-x) = log(x^2 e^{2 \pi i})$

$e^{2 \pi i} = 1$, therefore $2log(-x) = log(x^2)$

You have to define what you mean by log before you use it. We can do this easily in the real case, in the complex case there are infinite valid definitions and the usual one rules out negative reals.

I'm working with the same definition everbody uses- log(x) is the number you have to raise the base to to get x. As far as I can see, this poses no restriction to negative x.

Indeed, it seems a strange reason to cut out valid solutions midway through evaluating an expression simply because you have decided you don't want to.

That is not the definition of logs for the complex plane. It only works on the positive really line because exponents are 1-1. In the complex plane such a definition makes no sense because there are infinite different numbers that would work. For example what is log(1) (base e)? Is it 0, or 2*pi*i, or 4*pi*i?

These are all valid solutions to the equation x = ln(1). We're not trying to define a one-one function here, this is just solving an equation.

Just because the square root function only gives out positive values, it doesn't mean -1 isn't a solution to the equation x^2 = 1

I was showing why the definition of log you used did not work (it would lead to a multi-valued function which is not allowed here). You still haven't said what you mean by log, and until you define it you cannot use it.

Point not taken there. Just exactly why is are "multi-valued functons not allowed here"?

This seems an obvious parallel with the square root function- just because we neglect -ve values when we want a one-one function, it doesn't mean we have to impose restrictions when solving equations.

I think this has become somewhat beyond the point. All the question is asking is to find values of x and y for which these relationships are true. If you plug in these values, you can evaluate both suides to the same real number, hence those x and y values are solutions.

Anything beyond this is not necessary for the purposes of this question.