Dividing with exponential functions

[amaya244]

Homework Statement

When attempting to divide the following equality (a "tangency condition" in microeconomic consumer theory), I'm puzzled by the solution derived here, please explain the procedure for arriving at the solution. Many thanks!

Homework Equations

I differentiated the Lagrangian wrt x and y:

L = P₁x₁ + P₂x₂ + λ[U - x^αy^1-α]

First order conditions

L'(x) = P₁ - λαx^α-1y^1-α = 0

L'(y) = P₂ - λ(1-a)x^αy^-α = 0

The "tangency condition" requires that we divide L'(x) by L'(y): Doing so, and carrying the second half of each function to the other side, we get:

P₁/P₂ = λαx^α-1y^1-α / λ(1-a)x^αy^-α

The Attempt at a Solution

Here is where I get confused. How did the we arrive at the next step:

P₁/P₂ = α/(1-α) . x/y

Please explain you get x/y from the tangency condition.

 

[Muphrid]

Looks like they just simplified, but as written the factor should be y/x.

 

[jedishrfu]

I get P1/P2 = (a/1-a) * y/x

x^(a-1) / x^a --> x^(a-1-a) --> x^(-1)--> 1/x

and

y^(1-a) / y^(-a) --> y^(1-a+a) --> y^(1) --> y

 

[amaya244]

Looks like they just simplified, but as written the factor should be y/x.

You're right, it says y/x not x/y. My mistake. Does it all look okay to you then?

 

[amaya244]

I get P1/P2 = (a/1-a) * y/x

x^(a-1) / x^a --> x^(a-1-a) --> x^(-1)--> 1/x

and

y^(1-a) / y^(-a) --> y^(1-a+a) --> y^(1) --> y

Thanks.
Then, can we say 1/x times y is y/x?

 

[jedishrfu]

thats a yes