Why does Wolfram Alpha say that $\underbrace{\ln(y(y^{2}-4))}_\text{EQ1} \neq \underbrace{\ln(y) + \ln(y-2) + \ln(y+2)}_\text{EQ2}$ (seen here)?
The Produce Rule of Logarithms states that: $$\log_b{(M*N)}=\log_b{M} + \log_b{N}$$
Therefore, $$ \begin{align} \ln(y) + \ln(y-2) + \ln(y+2)& = \\ \ln(y) + \ln((y-2)(y+2)) & = \\ \ln(y) + \ln(y^2-4) & = \ln(y(y^2 - 4)) = \mathbf{EQ1} \ \checkmark \end{align} $$
However, Wolfram Alpha is saying that EQ1 "is not always equal to" EQ2. The graph Wolfram provides feels problematic, because it is showing two dramatically different curves when in reality both sizes are the same curve. What am I missing here?
For example: $\ln(y(y^2-4))$ is defined for $y=-1$, but $\ln(y) + \ln(y-2) + \ln(y+2)$ is not defined for $y=-1$.
The rule $\log_b{(M*N)}=\log_b{M} + \log_b{N}$ is only valid if both $M$ and $N$ are positive !