what is missing in the calculation below? i have been stuck in this question for hours, pls help. when $x=-1$, the first one gives $2$ but the 2nd one does not meet the logarithm condition. I am putting the power $2$ out in the 2nd one for some cases that it is canceled out like $0.5\cdot\log_3(-3)^2 = 0.5\cdot2\cdot\log_3(-3)$.
Is it incorrect since $-1$ from $-3$ are not squared? is there a rule that $-1$ must be powered in similar cases like this?
Condition (left): $(x-2)^2 > 0$, i.e. $x\ne 2$
Condition (right): $x-2 > 0$, i.e. $x > 2$
right: $2\log_3(-3)$ is undefined...