Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ \log y = \log a + b \log x $$

Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:

$$ \eta = b = \frac{ d \log y}{d \log x} $$

Now, I understand that $ d \log y = \frac 1y dy $ and $ d \log x = \frac 1x \ dx $. So I understand why we can write $ \eta = \frac {d \log y}{d \log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ \eta $?

Here's a snapshot from the book:

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  • 1
    $\begingroup$ About half of your questions have been answered, consider accepting some. $\endgroup$
    – Giskard
    Jan 13, 2019 at 13:24
  • 1
    $\begingroup$ Thank you for pointing that out. I'll do so right away. $\endgroup$
    – WorldGov
    Jan 13, 2019 at 13:44

2 Answers 2


Because $a$ is a parameter, and so $$ \eta = \frac{ d \log y}{d \log x} = \frac{ d \log a + d \ b \log x}{d \log x} = 0 + b. $$


Differentiating both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$: $$\frac{1}{y}\frac{dy}{dx}=b\frac{1}{x}$$ Rearranging: $$\frac{dy/y}{dx/x}=\eta=b$$


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