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Can anyone help me figure out how to go from the first expression to the second?

$$ \begin{equation} \ln D=u+\delta(e-p)+\gamma y-\sigma r \end{equation} $$

$$ \begin{equation} \pi \ \ln (D/Y)= \pi[u+\delta(e-p)+(\gamma -1) y-\sigma r] \end{equation} $$ where $y=\ln(Y)$

$u,\delta,\gamma,\sigma=$ Constants

$e=$ Current exchange rate

$p=$ Current price level

$r=$ Domestic interest rate

$D=$ Demand

GoodDeeds
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Alim Teacher
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2 Answers2

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First of you don't need that $\pi$ messing thins even more(*).

So, you have: $$\ln D=u+\delta(e-p)+\gamma y-\sigma r$$ Since $\ln Y=y$,

we can apply the rule of the logarithm: $$\ln(a/b)=lna-lnb$$ for $$b=Y$$ and $$a=u+\delta(e-p)+\gamma y-\sigma r$$

Then your equation simply follows.

(*)Note that: $c\ln (a/b)=c(\ln a-\ln b )$ So for $c=\pi$ you got the desired expression.

GoodDeeds
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MathematicianByMistake
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Starting with the second equation, divide both sides by $\pi$, then distribute the $y$ on the RHS. On the LHS, use the fact that $\ln \left(\dfrac{D}{Y}\right) = \ln D - \ln Y$. Since $y = \ln Y$, rewrite the LHS as $\ln D - y$. Then add $y$ to both sides and you end up with the first equation.

To go from the first equation to the second, just work backwards:

  1. Subtract $y$ from both sides in the first equation.
  2. On the LHS, rewrite $y$ as $\ln Y$.
  3. On the LHS, rewrite $\ln D - \ln Y$ as $\ln \left(\dfrac{D}{Y}\right)$.
  4. On the RHS, rewrite $\gamma y -y$ as $(\gamma -1)y$. (i.e., factor the $y$)
  5. Multiply both sides by $\pi$. Done!