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April 27th, 2014, 11:51 PM  #1 
Member Joined: Oct 2009 Posts: 50 Thanks: 0  how to derive log approximation?
Hello! This is probably an easy question. I'm trying to find out how to derive the log approximation of an exchange rate percentage change. So the actual change is: ( S[t+1]  St )/St How do you derive from this that an approximation is log(St+1)  log(St)? Thank you! 
May 2nd, 2014, 08:52 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  
May 2nd, 2014, 09:30 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,654 Thanks: 2632 Math Focus: Mainly analysis and algebra 
I think the OPs equation is not as you worked on, but $$\frac{s_{t+1}  s_t}{s_t} = \frac{s_{t+1}}{s_t}1$$ Although, $$\log{\frac{s_{t+1}  s_t}{s_t}} = \log{\left(s_{t+1}  s_t\right)}  \log{s_t}$$ I don't know what the question is either. Last edited by v8archie; May 2nd, 2014 at 09:33 PM. 
May 3rd, 2014, 11:03 PM  #4  
Member Joined: Oct 2009 Posts: 50 Thanks: 0  Quote:
The thing is, my textbook does this "derivation"/aproximation (the factis,I don't know what their doing...): They assume: $${rr*} = \frac{S_{t+1}S_{t}}{S_t}$$ And conclude from this that (This step I don't get): $$rr* = \log{S_{t+1}}  \log{S_t}$$ I understand your derivation, but its different from my textbook, since they don't take the log on both sides. They're effectively saying: $$\frac{S_{t+1}S_{t}}{S_t} = \log{S_{t+1}}  \log{S_t}$$ And I don't understand why that would be true (or approximately true, since I don't know if its an aproximation or not.) I hope my question is clearer now. Thanks for the help, either way! Last edited by questioner1; May 3rd, 2014 at 11:10 PM.  

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