# Dividing with exponential functions

1. Jul 5, 2012

### amaya244

1. The problem statement, all variables and given/known data

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!

2. Relevant equations

I differentiated the Lagrangian wrt x and y:

L = P1x1 + P2x2 + λ[U - xαy1-α]

First order conditions

L'(x) = P1 - λαxα-1y1-α = 0

L'(y) = P2 - λ(1-a)xαy = 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 = λαxα-1y1-α / λ(1-a)xαy

3. 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.

2. Jul 5, 2012

### Muphrid

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

3. Jul 5, 2012

### Staff: Mentor

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

4. Jul 5, 2012

### amaya244

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

5. Jul 5, 2012

### amaya244

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

6. Jul 5, 2012

thats a yes