<p>Taking logs of the equation $MV=PY$ gives: $ln(M) + ln(V) = ln(P) + ln(Y)$. </p>

<p>Next, differentiating with respect to time yields $\frac{dln(M)}{dt} + \frac{dln(V)}{dt} = \frac{dln(P)}{dt} + \frac{dln(Y)}{dt}$. </p>

<p>Or, using $\frac{dln(X)}{dt} = \frac{\dot{X}}{X}$, with $\dot{X} = \frac{dX}{dt}$, the derivative of a variable $X$ with respect to time, you can immediately derive $\frac{\dot{M}}{M} + \frac{\dot{V}}{V} = \frac{\dot{P}}{P} + \frac{\dot{Y}}{Y}$, which equals $\%\Delta M + \%\Delta V = \%\Delta P + \%\Delta Y$.</p>

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2016-02-09T08:54:48.347Z
2016-02-09T08:54:48.347Z
Wecon
<p>Equation of exchange states:<br>
$ MV=PY $<br>
where,<br>
       $ M $ is Quantity of Money<br>
       $ V $ is Velocity of Money<br>
       $ P $ is Price Level<br>
       $ Y $ is Real GDP<br>
Could someone please share how the above common form of equation is converted to the growth rate form:<br>
$ \%\Delta M + \%\Delta V = \%\Delta P + \%\Delta Y $  </p>

<p>Edit: Sorry the latex doesn't seem to be working. Unable to figure that out.</p>

<macroeconomics><economic-growth><quantity-theory-of-money>
How is the growth rate form of the exchange equation derived?
Loop
# How is the growth rate form of the exchange equation derived?

(Asked by: username_0 on 2016-02-09)

Equation of exchange states:  
$ MV=PY $  
where,  
$ M $ is Quantity of Money  
$ V $ is Velocity of Money  
$ P $ is Price Level  
$ Y $ is Real GDP  
Could someone please share how the above common form of equation is converted to the growth rate form:  
$ \%\Delta M + \%\Delta V = \%\Delta P + \%\Delta Y $

Edit: Sorry the latex doesn't seem to be working. Unable to figure that out.

**Tags:** macroeconomics, economic-growth, quantity-theory-of-money

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## 1 answers

### Answer by username_1 on 2016-02-09. Score: 2 | Accepted answer

Taking logs of the equation $MV=PY$ gives: $ln(M) + ln(V) = ln(P) + ln(Y)$.

Next, differentiating with respect to time yields $\frac{dln(M)}{dt} + \frac{dln(V)}{dt} = \frac{dln(P)}{dt} + \frac{dln(Y)}{dt}$.

Or, using $\frac{dln(X)}{dt} = \frac{\dot{X}}{X}$, with $\dot{X} = \frac{dX}{dt}$, the derivative of a variable $X$ with respect to time, you can immediately derive $\frac{\dot{M}}{M} + \frac{\dot{V}}{V} = \frac{\dot{P}}{P} + \frac{\dot{Y}}{Y}$, which equals $\%\Delta M + \%\Delta V = \%\Delta P + \%\Delta Y$.


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