Artificial intelligent assistant

Range of $\log(16-4x^2-4y^2-z^2)$ I am trying to find the range of $f(x,y,z) = ln(16-4x^2-4y^2-z^2)$ I know that the natural log function of one variable has range of all real numbers, so would this function be the same ? I tried to find it on Wolfram Alpha, however it said indeterminable.

$$\text{The argument of the log must be positive:}$$ $$\ 16-4x^2-4y^2-z^2>0$$ $$\ 4x^2+4y^2+z^2<16$$ $$\ \frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{16}<1$$ $$\ (\frac{x}{2})^2+(\frac{y}{2})^2+(\frac{z}{4})^2<1$$ $$\text {so the domain is a region enclosed in an ellipsoid.}$$ $$\ \\\ \text{Now you calculate what happens to the range 'moving in the domain':}$$ $$\text{note that }$$ $$\ log(16-4x^2-4y^2-z^2)=log[16-(4x^2+4y^2+z^2)]=log[16-p(x,y,z)]$$ $$\ \text{and }\ p(x,y,z)≥0 \forall(x,y,z).$$ $$\text{In the origin }p(x,y,z)=0 \text{ so you have the maximum value in the range: }$$ $$\ f(x,y,z)_{max}=log16$$ $$\text{Approaching the surface of the ellipsoid instead, the argument of the log tends to }\ 0^+$$ $$\text{so }\ f(x,y,z)\to -\infty$$ $$\text{The total range should be therefore: }\ (-\infty,log16]$$

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