Prompt:
Find the $x$- and $y$-intercepts of the graph of $x = (y+2)\ln(3y+4)$.
Given solution:
For $x = 0$, solve $(y+2)\ln(3y+4) = 0$ for $y$: \begin{align} y + 2 &= 0 & \ln(3y + 4) &= 0 \\ y &= -2 & 3y + 4 &= e^0 \\ \rlap{\text{not a solution}}\qquad && 3y + 4 &= 1 \\ && 3y &= -3 \\ && y &= -1 \end{align} So, only $(0, -1)$.
For $y = 0$, evaluate $x = (0 + 2) \ln(3(0) + 4) = 2\ln 4 = 4\ln 2$, so only $(4\ln2, 0)$.
My question:
Why does $y \neq -2$? Can someone help me in simple terms.
The domain of the natural logarithm function only consist of positive real numbers, but with $y = -2$, the input to the log is $3y + 4 = -2 < 0$.
You can see that the graph only contains points with $y$-coordinate satisfying $3y + 4 > 0$, i.e., $y > -\tfrac43$.