CAT 2018 Question Paper Shift 2 Solved Paper

© examsiri.com

Question : 70 of 100

Marks: +1, -0

1/{\log_3 100}-

1/{\log_4 100}+

1/{\log_5 100}

-

1/{\log_{10} 100}+

1/{\log_{20} 100}-

1/{\log_{25} 100}+

1/{\log_{50} 100}=

?

0
- 4
10
$1/2$

Solution:

We know that

\;{1}/{\log _{b} a}=\log _{a} b,

therefore,

\;{1}/{\log _{2} 100}-\;{1}/{\log _{4} 100}+\;{1}/{\log _{s} 100}-\;{1}/{\log _{10} 100}+\;{1}/{\log _{20} 100}-\;{1}/{\log _{25} 100}+\;{1}/{\log _{50} 100}

=\log _{100} 2-\log _{100} 4+\log _{100} 5-\log _{100} 10+\log _{100} 20-\log _{100} 25+\log _{100} 50

=\log _{100}(\;{2}/{4} × \;{5}/{10} × \;{20}/{25} × 50)

=\log _{160} 10

Using the relation

\log _{a^{m}} b=\;{1}/{m} \log _{a} b

\log _{100} 10=\log _{10^{2}} 10=\;{1}/{2} \log _{10} 10=\;{1}/{2}

© examsiri.com

Go to Question:

Prev Question

Next Question