Solving natural log equations

Question

The equation $\ln(|y+1|)= x-2$ where you solve for $y$, I am just unsure of how the absolute value plays into this. I am assuming that I would convert to exponential form to get $|y+1|=e^{x-2}$ and then I am stuck. Sorry if this is an easy question and I'm just missing something obvious. :/

Answer 1

Notice, $$\ln|y+1|=x-2$$ $$|y+1|=e^{x-2}$$ Since, $e^{x-2}>0 \ \forall\ x\in R$ $$ y+1=\pm e^{x-2}$$

$$\bbox[5px, border:2px solid #C0A000]{y=\pm e^{x-2}-1}$$