CAT 2018 Set-2 | Question: 100

Question

$\frac{1}{\log_{2}100} – \frac{1}{\log_{4}100} + \frac{1}{\log_{5}100} – \frac{1}{\log_{10}100} + \frac{1}{\log_{20}100} – \frac{1}{\log_{25}100} + \frac{1}{\log_{50}100}=?$

• A. $1/2$
• B. $0$
• C. $10$
• D. $-4$

Comments

option A

Answer 1

Let, $ \text{S} = \frac{1}{ \log_{2}100} – \frac{1}{ \log_{4}100} +\frac{1}{\log_{5}100} – \frac{1}{\log_{10}100} + \frac{1}{\log_{20}100} – \frac{1}{\log_{25}100} + \frac{1}{\log_{50}100} $

$ \Rightarrow \text{S} = \log_{100}2 – \log_{100}4 + \log_{100}5 – \log_{100}10 + \log_{100}20 – \log_{100}25 + \log_{100}50  \quad \left[ \because \log_{b}a = \frac{1}{\log_{a}b} \right] $

$ \Rightarrow \text{S} = \log_{100}2 – \log_{100}4 + \log_{100}5 – \log_{100}10 + \log_{100}(5 \cdot 4) – \log_{100}(5 \cdot 5) + \log_{100}(5 \cdot 10) $

$ \Rightarrow \text{S} = \require{cancel}\log_{100}2 – {\color{Red}{\cancel{\log_{100}4}}} + \log_{100}5 – {\color{Blue}{\cancel{\log_{100}10}}} + {\color{Magenta}{\cancel{\log_{100}5}}} +{\color{Red}{\cancel{\log_{100}4}}} – {\color{Magenta}{\cancel{\log_{100}5}}} – {\color{Magenta}{\cancel{\log_{100}5}}} + {\color{Magenta}{\cancel{\log_{100}5}}} + {\color{Blue}{\cancel{\log_{100}10}}} \quad \left[ \because \log_{x}(ab) = \log_{x}a + \log_{x}b \right] $

$ \Rightarrow \text{S} = \log_{100}2 + \log_{100}5 $

$ \Rightarrow \text{S} = \log_{100}(2 \cdot 5) $

$ \Rightarrow \text{S} = \log_{100}10 $

$ \Rightarrow \text{S} = \log_{10^{2}} 10 $

$ \Rightarrow \text{S} = \frac{1}{2} \; \log_{10}10 \quad \left[ \because \log_{b^{x}}a = \frac{1}{x} \; \log_{b}a \right] $

$ \Rightarrow \boxed{\text{S} = \frac{1}{2}} \quad [ \because \log_{a}a = 1 ] $

Correct Answer $ : \text{A}$