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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:
 $\log(x),\;\log\left(\frac{x}{y\right),\;\log\left( \frac{x}{y^2}\right),\;\log \left(\frac{x}{y^3}\right),\,\cdots$ Find the sum of the first 35 terms.

$\log(\frac{x}{1})\,+\,\log \left(\frac{x}{y^1}\right)\,+\,\log \left(\frac{x}{y^2}\right)\,+\,\log \left(\frac{x}{y^3}\right)\,+\,\cdots\,+\,\frac{x} {y^{34}}$

$\;\;\;\;=\;\log\left(\frac{x}{1}\cdot\,\frac{x}{y^ 1}\,\cdot\, \frac{x}{y^2}\,\cdot\, \frac{x}{y^3}\, \cdot\, \cdot\, \cdot\, \frac{x}{y^{34}}\right) \;=\;\log\left(\frac{x^{35}}{y^{595}}\right) \;\;\text{ . . . etc.}$

 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.

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