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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.

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    $\begingroup$ 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. $\endgroup$
    – tdm
    May 17 at 15:16
  • $\begingroup$ @tdm so drop both monotonic transformation and just look at $3x_1+2x_2+x_3$? But wouldn't that yeild that $3=2=1$? $\endgroup$
    – user879295
    May 17 at 15:29
  • $\begingroup$ wouldn't it be. $\endgroup$
    – user879295
    May 17 at 15:40
  • $\begingroup$ no. please check how to solve a utility maximisation problem with perfect substitutes (i.e. when indifference curves are linear) for example $\endgroup$
    – tdm
    May 17 at 15:43
  • $\begingroup$ @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?. $\endgroup$
    – user879295
    May 17 at 15:49

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