# Calculation of simple elasticity problem

I'm reading Advanced Macroeconomics by Romer and having trouble to understand very simple calculation provided in just one sentence.

Define real money market equilibrium condition as $$(\frac{M}{P})^s=L(Y,i)=(\frac{M}{P})^d$$
Then the price leve is $$P=\frac{M}{L(Y,i)}$$
If the interest elasticity of money demand is $$-0.2$$, we need almost $$32$$ times rise in interest rate to increase price level double.

I can't understand where "$$32$$ times" comes from.
From the definition of elasticity, and using the fact that
if we want to make price level double then $$L(Y,i)$$ should be reduced by half,
$$\frac{\partial L}{\partial i}\frac{i}{L}=\frac{\partial L}{L}\frac{i}{\partial i}=-\frac{1}{2}\frac{i}{\partial i}=-0.2$$ then $$\frac{\partial i}{i}=2.3$$
Can anyone tell me how to solve this problem?

• Hi! Where did the $-1/2$ come from in your equation $$\frac{\partial L}{\partial i}\frac{i}{L}=\frac{\partial L}{\partial i}\frac{i}{\partial i}=-\frac{1}{2}\frac{i}{\partial i}=-0.2$$? Also, it seems like in this case we would have $\frac{\partial i}{i} = 2.5$? It feels like you are only sharing half of your thoughts with us. Commented Nov 15, 2023 at 7:17
• @Giskard I edited my post to prove why $\frac{\partial L}{L}=-\frac{1}{2}$ Commented Nov 15, 2023 at 7:28
• Can you please tell us a bit about your background, so that someone can answer in an understandable way? Are you familiar with differentials and calculus? Commented Nov 15, 2023 at 8:28

If $$L(Y,i)$$ has a constant elasticity of $$-0.2$$ then you can write it as: $$L(Y,i) = g(Y)i^{-0.2}.$$ Then for the period 0 and period 1 price levels you have: $$P_0 = \frac{M}{g(Y)i_0^{-0.2}} \text{ and } P_1 = \frac{M}{g(Y)i_1^{-0.2}}.$$ We have that $$P_1 = 2 P_0$$ so: $$2 \frac{M}{g(y) i_0^{-0.2}} = \frac{M}{g(Y)i_1^{-0.2}}.$$ Simplifying gives: $$2 i_1^{-0.2} = i_0^{-0.2}.$$ This gives: \begin{align*} \frac{i_1}{i_0} = 2^5 = 32 \end{align*}
• Hi @tdm. For my own benefit, could you include any additional steps in going from: $$2i_1^{-0.2} = i_0^{-0.2}$$ to $$\frac{i_1}{i_0} = 2^5$$. Would be appreciated. Commented Nov 17, 2023 at 14:39
• @ EB3112 It is enough to understand that $0.2=\frac{1}{5}$. Commented Nov 18, 2023 at 5:24