Logarithmic elasticities

[elfy]

Homework Statement

Find an expression (in terms of x and y) for the elasticity of y wrt x:
ln(y) = 10 + 0.9x - 0.6x^2

Homework Equations

ln(y) = 10 + 0.9x - 0.6x^2

The Attempt at a Solution

I first tried the normal way of doing elasticities:

ElxY = (x/y)*y'(x) which I got to:

[x / 10 + 0.9x - 0.6x^2] * (0.9 - 1.2x) --> 0.9x - 1.2x^2 / (10 + 0.9x -0.6x^2)

Furthermore, the log stated on the left-hand side is somewhat worrying me so I also tried the logarithmic rule in relation to elasticities:
ElxY = dlnY/dlnX but I got a really weird answer when I started to differentiate wrt x.

Could someone just point me in the right direction so that I could have an attempt at the problem? I would very much appreciate any help I can get!

EDIT: If that one is too complex to explain I also have a simpler version which is similar:
lny=ax + bx
ElxY = (x/ax + bx) * a + bx ---> ax / ax = 1.
However, the same problem happens here with regards to the lny, which I assume makes my answer incorrect.

 

[mighty2000]

First, let's review the definition of elasticity. Elasticity is a measure of the responsiveness of one variable (y) to changes in another variable (x). In other words, it tells us how much y changes in response to a change in x.

In this case, we are looking for the elasticity of y with respect to x, which is denoted as ElxY. This can be calculated using the following formula:

ElxY = (x/y) * dy/dx

where dy/dx is the derivative of y with respect to x.

Now, let's apply this formula to the given function:

ln(y) = 10 + 0.9x - 0.6x^2

To find dy/dx, we need to take the derivative of both sides of the equation with respect to x. Remember, the derivative of ln(y) is 1/y, and the derivative of a constant (10 in this case) is 0. So, we get:

1/y * dy/dx = 0.9 - 1.2x

Now, we can rearrange this equation to solve for dy/dx:

dy/dx = (0.9 - 1.2x) * y

Substituting this expression for dy/dx into our original formula for ElxY, we get:

ElxY = (x/y) * (0.9 - 1.2x) * y

Simplifying, we get:

ElxY = 0.9 - 1.2x

So, the elasticity of y with respect to x for the given function is 0.9 - 1.2x. This means that for every 1 unit increase in x, y will decrease by 1.2 units, and for every 1 unit decrease in x, y will increase by 0.9 units.

I hope this helps! Let me know if you have any further questions.