Suppose that, instead of the linear demand function of Example 4.5.4, one has the log-linear function $\ln Q = a − b \ln P.$
(a) Express Q as a function of P, and show that dQ/dP = −bQ/P.
(b) Express P as a function of Q, and find dP/dQ.
(c) Check that your answer satisfies the version dP/dQ = 1/(dQ/dP) of (7.3.4).

Question

Suppose that, instead of the linear demand function of Example 4.5.4, one has the log-linear function [latex] \ln Q = a − b \ln P.[/latex]
(a) Express Q as a function of P, and show that dQ/dP = −bQ/P.
(b) Express P as a function of Q, and find dP/dQ.
(c) Check that your answer satisfies the version dP/dQ = 1/(dQ/dP) of (7.3.4).

[latex]{\frac{\mathrm{d}x}{\mathrm{d}y}}={\frac{1}{\mathrm{d}y/\mathrm{d}x}}[/latex] (7.3.4)

Accepted Answer

(a) Taking exponentials gives [latex]Q = e^{a−b \ln P} = e^{a}(e^{\ln P})^{−b} = e^{a}P^{−b},[/latex] from which it follows that [latex]dQ/dP = −be^{a}P^{−b−1} = −bQ/P.[/latex]
(b) Solving [latex]Q = e^{a}P^{−b}[/latex] for P gives [latex]P = e^{a/b}Q^{−1/b},[/latex] so [latex]dP/dQ = (−1/b)e^{a/b}Q^{−1−1/b}.[/latex]
(c) From part (b) one has dP/dQ = (−1/b)P/Q = 1/(dQ/dP).

Question 7.3.4: Suppose that, instead of the linear demand function of Examp......

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