# Unsolveable Demand/Utility Problem?

A consume has a preference relation on $$\mathbb{R}^4_+$$ with a utility function defined as

$$U(x_1,x_2,x_3)=(\ln(3x_1+2x_2+x_3))^3$$

Find the demand at prices $$p=(1,1,1)$$ and wage $$4$$.

Attempt

I thought to either use Lagrange or condition

$$\frac{MU_{1}}{p_1}=\frac{MU_{2}}{p_2}=\frac{MU_{2}}{p_2}$$

but just considering the inner function i.e $$U(x_1,x_2,x_3)=\ln(3x_1+2x_2+x_3)$$ since they represent the same preferences. I get the 3 partial derivatives

$$\frac{\partial U}{\partial x_1}=\frac{3}{3x_1+2x_2+x_3}$$ $$\frac{\partial U}{\partial x_2}=\frac{2}{3x_1+2x_2+x_3}$$ $$\frac{\partial U}{\partial x_3}=\frac{1}{3x_1+2x_2+x_3}$$

Now, i would isolate the variables in $$\frac{MU_{1}}{p_1}=\frac{MU_{2}}{p_2}=\frac{MU_{2}}{p_2}$$ but a solution to that system does not exist since they are just $$MU_1=MU_2=MU_3$$ which is of course not true (for any variables). Am I missunderstanding the approach?

*I would then use the expressions in the budget constrain and solve for the variables ones again.

• You can drop the $\ln$ function as this is a monotonic transformation; Then you will see that this is a utility function with perfect substitutes.
– tdm
May 17 at 15:16
• @tdm so drop both monotonic transformation and just look at $3x_1+2x_2+x_3$? But wouldn't that yeild that $3=2=1$? May 17 at 15:29
• wouldn't it be. May 17 at 15:40
• no. please check how to solve a utility maximisation problem with perfect substitutes (i.e. when indifference curves are linear) for example
– tdm
May 17 at 15:43
• @tdm So that would be spending it all on good $x_1$ since the marginal utility of it is the highest after dividing by the price. So that the demand is $4/1=4$, is this correctly understood for 3 goods?. May 17 at 15:49