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If $n=(2017) !$, then what is $[2018-1]$
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots .+\frac{1}{\log _{2017} n}$
equal to?
Options:
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots .+\frac{1}{\log _{2017} n}$
equal to?
Solution:
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Verified Answer
The correct answer is:
1
$n=(2017) !$
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots .+\frac{1}{\log _{2017} n}$
$=\frac{1}{\frac{\log _{n}}{\log _{2}}}+\frac{1}{\frac{\log _{n}}{\log _{3}}}+\frac{1}{\frac{\log _{n}}{\log _{4}}}+\ldots .+\frac{1}{\frac{\log _{n}}{\log _{2017}}}$
$\left(\because \log _{\mathrm{a}} \mathrm{b}=\frac{\log _{\mathrm{b}}}{\log _{\mathrm{a}}}\right)$
$=\frac{\log _{2}}{\log _{n}}+\frac{\log _{3}}{\log _{n}}+\frac{\log _{4}}{\log _{n}}+\ldots .+\frac{\log _{2017}}{\log _{n}}$
$=\frac{\log _{2}+\log _{3}+\log _{4}+\ldots .+\log _{2017}}{\log _{n}}$
$=\frac{\log (2.3 .4 \ldots .2017)}{\log _{\mathrm{n}}}$
$(\because \log a+\log b+\log c+\ldots=\log a . b . c \ldots \ldots)$
$=\frac{\log (2017 !)}{\log _{\mathrm{n}}}=\frac{\log _{\mathrm{n}}}{\log _{\mathrm{n}}}=1$
$\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots .+\frac{1}{\log _{2017} n}$
$=\frac{1}{\frac{\log _{n}}{\log _{2}}}+\frac{1}{\frac{\log _{n}}{\log _{3}}}+\frac{1}{\frac{\log _{n}}{\log _{4}}}+\ldots .+\frac{1}{\frac{\log _{n}}{\log _{2017}}}$
$\left(\because \log _{\mathrm{a}} \mathrm{b}=\frac{\log _{\mathrm{b}}}{\log _{\mathrm{a}}}\right)$
$=\frac{\log _{2}}{\log _{n}}+\frac{\log _{3}}{\log _{n}}+\frac{\log _{4}}{\log _{n}}+\ldots .+\frac{\log _{2017}}{\log _{n}}$
$=\frac{\log _{2}+\log _{3}+\log _{4}+\ldots .+\log _{2017}}{\log _{n}}$
$=\frac{\log (2.3 .4 \ldots .2017)}{\log _{\mathrm{n}}}$
$(\because \log a+\log b+\log c+\ldots=\log a . b . c \ldots \ldots)$
$=\frac{\log (2017 !)}{\log _{\mathrm{n}}}=\frac{\log _{\mathrm{n}}}{\log _{\mathrm{n}}}=1$
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