Which of the following functions is not a Cobb-Douglas function?
$z(x,y) = 5x^{\tfrac{1}{2}}y^{\tfrac{1}{2}} - 3x^{\tfrac{1}{2}}y^{\tfrac{1}{2}}$.
$z(x,y) = \sqrt{2}x^{0.1}y^{0.9}$.
$z(x,y) = \sqrt[4]{16xy^3}$.
$z(x,y) = 5x^{-\tfrac{1}{2}}y^{\tfrac{3}{2}}$.
Correct: $z(x,y) = 5x^{-\tfrac{1}{2}}y^{\tfrac{3}{2}}$ can be rewritten to the form $k{x^\alpha}y^{1-\alpha}$, with $k>0$. However, $\alpha = -\tfrac{1}{2} < 0$.
Go on.
Wrong: $z(x,y)$ can be written in the right form:
$$z(x,y) = 5x^{\tfrac{1}{2}}y^{\tfrac{1}{2}} - 3x^{\tfrac{1}{2}}y^{\tfrac{1}{2}} = 2x^{\tfrac{1}{2}}y^{\tfrac{1}{2}}.$$
Hence, $z(x,y)$ is a Cobb-Douglas function with $k=2$ and $\alpha=\tfrac{1}{2}$.
See Cobb-Douglas functions.
Wrong: $z(x,y)$ is a Cobb-Douglas function with $k=\sqrt{2}$ and $\alpha=0.1$.
See Cobb-Douglas functions.
Wrong: $z(x,y)$ can be written in the right form:
$$z(x,y) = \sqrt[4]{16xy^3} = 2x^{\frac{1}{4}}y^{\frac{3}{4}}.$$
Hence, $z(x,y)$ is a Cobb-Douglas function with $k=2$ and $\alpha=\tfrac{1}{4}$.
See Power functions or Cobb-Douglas functions.