October 27th, 2014, 01:21 PM  #1 
Newbie Joined: Oct 2014 From: Australia Posts: 1 Thanks: 0  Logarithms
logx, logx/y, logx/y^2, logx/y^3 ... Find the sum of the first 35 terms. 
October 27th, 2014, 01:41 PM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,596 Thanks: 620 Math Focus: Wibbly wobbly timeywimey stuff.  
October 27th, 2014, 03:08 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, a2910! Another approach . . . Quote:
 
October 27th, 2014, 03:14 PM  #4 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,354 Thanks: 116 
$\displaystyle \log\frac{x}{y}=\log{x}\log{y}$ $\displaystyle \log\frac{x}{y^2}=\log{x}2\log{y}$ $\displaystyle \log\frac{x}{y^3}=\log{x}3\log{y}$ Thus, the series become $\displaystyle \log{x},\log{x}\log{y},\log{x}2\log{y},\log{x}3\log{y},...$, an arithmatic series with $\displaystyle a=\log{x}$ and $\displaystyle d=\log{y}$. You can find the sum using arithmatic sequence formula and change it back to logarithm form after that. Last edited by Monox D. IFly; October 27th, 2014 at 03:22 PM. 

Tags 
geometric, logarithms, sequences, sum 
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