Question #44f58

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Question #44f58

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Answer 1 · 2017-01-07T12:31:16

$log (\frac{7}{5})$ $log(7/5)$

Explanation:

$x^{7} - 1 = (x - 1) (1 + x + x^{2} + \dots + x^{6})$ $x^7-1=(x-1)(1+x+x^2+cdots+x^6)$
$x^{5} - 1 = (x - 1) (1 + x + x^{2} + \dots + x^{4})$ $x^5-1=(x-1)(1+x+x^2+cdots+x^4)$

so

$\frac{x^{7} - 1}{x^{5} - 1} = \frac{1 + x + x^{2} + \dots + x^{6}}{1 + x + x^{2} + \dots + x^{4}}$ $(x^7-1)/(x^5-1)=(1+x+x^2+cdots+x^6)/(1+x+x^2+cdots+x^4)$

then

$lim_(x->1)log((x^7-1)/(x^5-1))=log(lim_(x->1)(1+x+x^2+cdots+x^6)/(1+x+x^2+cdots+x^4))=log(7/5)$

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Question #44f58

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