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 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
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Quote:
 Originally Posted by a2910 logx, logx/y, logx/y^2, logx/y^3 ... Find the sum of the first 35 terms.
Hint: $\displaystyle log \left ( \frac{x}{y^n} \right ) = log(x) - log \left ( y^n \right ) = log(x) - n~log(y)$

What does that do to your series?

-Dan October 27th, 2014, 03:08 PM   #3
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Hello, a2910!

Another approach . . .

Quote:
 Find the sum of the first 35 terms. October 27th, 2014, 03:14 PM #4 Senior Member   Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry $\displaystyle \log\frac{x}{y}=\log{x}-\log{y}$ $\displaystyle \log\frac{x}{y^2}=\log{x}-2\log{y}$ $\displaystyle \log\frac{x}{y^3}=\log{x}-3\log{y}$ 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. Thanks from topsquark Last edited by Monox D. I-Fly; October 27th, 2014 at 03:22 PM. Tags geometric, logarithms, sequences, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ian McPherson Algebra 2 January 15th, 2013 03:04 PM vic Algebra 1 January 4th, 2013 02:15 PM empiricus Algebra 8 July 9th, 2010 08:31 AM stekemrt Abstract Algebra 11 November 21st, 2007 10:37 AM Ian McPherson Calculus 2 December 31st, 1969 04:00 PM

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