# Finding the elasticity of a function with respect to a variable from logarithm

I tried to ask this question on Math Stackexchange, but got no answer. I'll try here.

I am a bit confused as to how to see elasticity of a function with respect to a variable from logarithm. Lets say we have the following function:

$$y^* =\beta^{\frac{1}{1-a}}(\frac{s}{n+\delta})^{\frac{a}{1-a}} \leftrightarrow$$ $$ln y^* = \frac{1}{1-a} * ln (\beta) + \frac{a}{1-a}*ln(\frac{s}{n+\delta})$$

How does one see from here what the elasticity of y with respect to $$n + \delta$$ is?

Kind regards

Edit:

Elasticity is defined as "In economics, elasticity measures the percentage change of one economic variable in response to a change in another" via Wikipedia. To find the elasticity of y with respect to $$n+\delta$$, you can use the following formula:

$$\frac{\partial y}{\partial (n+\delta)} * \frac{n+\delta}{y} = -\frac{a}{1-a}$$

But my books suggests I should be able to spot this from the $$lny*$$ equation. Anyone got any idea how?

Elasticity of any differentiable function $$f(x)$$ wrt $$x$$ by definition is:

$$\epsilon = \frac{df(x)}{dx}\frac{x}{f(x)}$$

For a particular type of function given by:

$$f(x) = Ax^e$$

$$\epsilon = \frac{df(x)}{dx}\frac{x}{f(x)} = e$$

Moreover, for this sort of function if you take a log of both sides you will find that:

$$\ln f(x) = \ln A + e \ln x$$

where the coefficient before logged variable gives you the elasticity.

In your case, if you take log of the following:

$$y^* =\beta^{\frac{1}{1-a}}(\frac{s}{n+\delta})^{\frac{a}{1-a}}$$

you will get:

$$\ln y^* = \frac{1}{1-a} \ln (\beta) + \frac{a}{1-a}\ln s - \frac{a}{1-a} \ln (n+\delta)$$

where $$\frac{1}{1-a}$$, $$\frac{a}{1-a}$$ and $$- \frac{a}{1-a}$$ are all elasticities of the original function $$y$$ wrt $$\beta$$, $$s$$ and $$n+\delta$$ respectively.

You can verify that by explicitly calculating:

$$\epsilon_{\beta y} = \frac{\partial y}{\partial \beta}\frac{\beta}{y} = \frac{1}{1-a}$$

$$\epsilon_{s y} = \frac{\partial y}{\partial s}\frac{s}{y} = \frac{a}{1-a}$$

$$\epsilon_{(n-\delta); y} = \frac{\partial y}{\partial (n-\delta)}\frac{(n-\delta)}{y} = -\frac{a}{1-a}$$

• Thank you, I appreciate it. Jan 7 at 20:35