High School Advanced Functions: Clarifying log rules in a log equation - $\log(x^2) = 2$, Solve for x. I got in an argument with my teacher for the possible solutions of x. From some sources i found that because x is squared, negative values should be possible; however, my teacher insists that:
$$
\log(x^2) = 2\log x
$$
and that they have to be interchangeable, no matter what was given originally. Therefore the negative solution (-10) would be extraneous.
Please help me understand which is correct! Thanks.
 A: Your teacher is wrong here. As you are noting
$$\log((-10)^2)=\log(100)=2$$
hence $-10$ is a solution to that equation and is just as valid as $10$. There is no escaping this fact.
Frankly, the conclusion that should be reached is that
$$\log(x^2)=2\log(x)$$
doesn't hold for negative values - in fact, it barely makes sense, given that the logarithm is (in elementary settings) only defined for positive $x$. A more accurate statement of this would be
$$\log(x^2)=2\log(|x|)$$
which clearly holds since $x^2=|x|^2$ - which admits $-10$ as a solution.
Your teacher's argument is little different from saying
$$|x|=5$$
has $x=5$ as it's only solution, due to the identity $|x|=x$ (which only holds for positive $x$).

If we're being really sure about things, we could wander into complex-number-land and talk about the natural logarithm, where, owing to Euler's identity that $e^{i\pi}=-1$, we could reasonably define $\ln(-1)=\pi i$. Then, if we were solving the equation
$$\ln(x^2)=0$$
we would, naturally, have solutions $x=\pm 1$. We could write out
$$\ln(x^2)=2\ln(x)$$
but, if we plug $-1$ into this, we get
$$\ln(1)=2\ln(-1)=2\pi i$$
which, at first, looks like an issue, since $\ln(1)=0$, but if $e^{i\pi}=-1$, then squaring that gives $e^{2i\pi}=1$ - so, actually, saying $\ln(1)=2\pi i$ is, in some sense, not that crazy, and can be considered an equally valid solution. Essentially, the problem arises because the logarithm is inherently multi-valued - if $e^x=y$, then so does $e^{x+2\pi i}=y$. A more accurate version and general of the identity would be:
$$2ni\pi= \ln(x^2)-2\ln(x)$$
for some integer $n$ and this suffices to work over any choice of the value of $\ln(x)$ and works over the entire complex plane, whereas other identities likely only work in the positive real numbers. If we are working in only the positive reals numbers, the difference $\ln(x^2)-\ln(x)$ would have no imaginary component, so this would reduce to $0=\ln(x^2)-2\ln(x)$, but we need to be more careful in larger domains.
A: Note that $\log x^2=2\log x$ holds only if $x>0$.
For the equation: $\log x^2=2$, It is easy to see that $x^2=100$, and hence $x=\pm 10.$
A: As always in algebra, the bit inside brackets is calculated first.  So if you substitute $x=-10$ into the equation you get
$$LHS=\log((-10)^2)=\log(100)=2=RHS\ .$$
Therefore $x=-10$ is a solution of the equation.
A: Let $\xi \in \mathbb R$ be such that:
$$\xi = \ln |x| \text{ for some $x\in\mathbb R$ }\Rightarrow e^{\xi}=|x|\Rightarrow e^{m\xi }=|x|^m \Rightarrow m\xi=\ln |x|^m \Rightarrow \boldsymbol{m\ln |x| = \ln |x|^m}.$$
 This works not only to natural logarithm but to every base of a logarithm. 
In particular $-\log(x^2)= -\log(|x|^2)=2\log |x|.$ Therefore you are right and your teacher is wrong.
