# Transformation of Cobb Douglas Utility

So, I need to determine whether or not $u=x^{0.5}y^{0.5}$ exhibits the same preferences as $u'= log(x) + log (y)$. Any tricks I can use here? I've done the logarithmic transformation so that I now have $(0.5)log (x) + (0.5)log (y)$. My guess is, if I multiply these by $2$ I'll get $log (x) + log (y)$ and thereby proving that u and u' refer to the same preferences, but I'm not sure if this is correct?

Hint: how does a positive monotone transformation of the utility function affect the consumer's maximization problem?

Edit,

Suppose the optimal bundle for some agent with some utility function $u$ is given by $x^{*} = (x_{1}^{*},x_{2}^{*})$. Suppose then that I take the transformation of $u$ by some strictly monotone function $f$. I claim that $x^{*} = (x_{1}^{*},x_{2}^{*})$ is the optimal bundle for the new utility function $v = f \circ u$.

Suppose not, suppose that there is another bundle $\hat{x} = (\hat{x}_{1},\hat{x}_{2})$ such that $v(\hat{x}) > v(x^{*})$. But by the definition of $v$, we have $$\tag{In}\label{In} f(u(\hat{x})) > f(u(x^{*}))$$ Then, since $f$ is strictly monotone, it is invertible and so from \ref{In} we must have $u(\hat{x}) > u(x^{*})$, which contradicts optimality of $x^{*}$.

The $\ln$ function is one such strictly monotone function on domain $\Re_{+}$.

• I guess the idea behind monotonic transformation is to make it easier to solve but the outcome of the maximization should be the same.
– jjj
Nov 19, 2017 at 17:49
• So I should solve them both and see if I get the same result?
– jjj
Nov 19, 2017 at 17:50
• So could I take both the utility functions i have here, differentiate w.r.t. to x1 for example and then set this equal to zero for both of them and solving each and see if they give the same result?
– jjj
Nov 19, 2017 at 17:53
• Can you show that for any bundle $z = (x,y)$, $u(z) \geq u(z')$ for CB utility, then $v(z) \geq v(z')$ for ln utility? You should not need to differentiate.
– user11305
Nov 19, 2017 at 18:17

To see if the preferences between $u$ and $u'$ are the same, simply look at the $MRS$ of both of them to see if they are equivalent.

i.e.

$$MRS_{u}=\frac{MU_x}{MU_y}=\frac{\left(0.5x^{-0.5}y^{0.5}\right)}{\left(0.5x^{0.5}y^{-0.5}\right)}=\frac{y}{x}$$

$$MRS_{u'}=\frac{\frac{1}{x}}{\frac{1}{y}}=\frac{y}{x}$$

$$\Rightarrow MRS_u=MRS_{u'}$$

$\therefore$ $u$ and $u'$ have the same preferences.