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October 27th, 2014, 01:21 PM   #1
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Logarithms

logx, logx/y, logx/y^2, logx/y^3 ...

Find the sum of the first 35 terms.
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October 27th, 2014, 01:41 PM   #2
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Originally Posted by a2910 View Post
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
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October 27th, 2014, 03:08 PM   #3
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Hello, a2910!

Another approach . . .


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Find the sum of the first 35 terms.





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October 27th, 2014, 03:14 PM   #4
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$\displaystyle \log\frac{x}{y}=\log{x}-\log{y}$
$\displaystyle \log\frac{x}{y^2}=\log{x}-2\log{y}$
$\displaystyle \log\frac{x}{y^3}=\log{x}-3\log{y}$
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.
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Last edited by Monox D. I-Fly; October 27th, 2014 at 03:22 PM.
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