If the demand function is q = m - 2(ln p)over some range of values of p, then at all such values of p, the absolute value of the price elasticity of demand
A)increases as p increases.
B)decreases as p increases.
C)is constant as p changes.
D)increases with p at small values and decreases with p at large values.
E)decreases with p at large values and increases with p at small values.

Question

Quizplus · Accepted Answer

The price elasticity of demand can be calculated as:
$$\epsilon = \frac{d\ln(q)}{d\ln(p)} = \frac{d\ln(F_1S_1)}{d\ln(p)} = \frac{d\ln(F_1)}{d\ln(p)} + \frac{d\ln(S_1)}{d\ln(p)}$$
Using the chain rule and the provided demand function, we get:
$$\frac{d\ln(F_1)}{d\ln(p)} = \frac{1}{p}\frac{dF_1}{dp} = \frac{-F_1}{mS_1- F_1}$$
$$\frac{d\ln(S_1)}{d\ln(p)} = \frac{1}{p}\frac{dS_1}{dp} = \frac{mF_1}{mS_1- F_1}$$
Therefore, the absolute value of the elasticity is:
$$\lvert\epsilon\rvert = \frac{\lvert F_1\rvert}{mS_1-F_1} + \frac{m\lvert F_1\rvert}{mS_1-F_1} = \frac{m+1}{\bigg\lvert\frac{mS_1}{F_1}-1\bigg\rvert}$$
Since $S_1$ and $F_1$ are positive, $\frac{mS_1}{F_1}$ can take on any positive value. As $p$ increases, $\frac{1}{p}$ decreases, which makes both $\frac{d\ln(F_1)}{d\ln(p)}$ and $\frac{d\ln(S_1)}{d\ln(p)}$ more negative. As a result, $\lvert\epsilon\rvert$ increases as $p$ increases. Therefore, the correct answer is A.

If the Demand Function Is Q = M - 2(ln