Finding the domain of $f (x, y) = \ln (7 x^2 + 2 y + 1)$

Question

Find the domain of the function  $f(x,y) = \ln(7x^2 + 2y + 1)$.

I know that I should keep Domain Convention in mind when solving this question. I know that I can’t take the logarithm of a negative number or zero. It should be

$\frac{1 + 7x^2}{2}$

But I don't know how to determine whether it is $<,>$, or $\leq$, or $\geq$. I could really use the help please.

Answer 1

You are right that the argument of the logarithm has to be positive. Thus, the domain is all coordinate pairs $(x,y)$ satisfying,

$$7x^2+2y+1 > 0.$$

Rearranging so $y$ is on one side, the above inequality is equivalent to,

$$y > -\frac{7x^2+1}{2}.$$

Thus, the domain is,

$$\left\{ (x,y) \in \mathbb{R}^2: y > -\frac{7x^2+1}{2}\right\}.$$