Question

Column I		Column II
A	If $N={\log }_{\frac{1}{5}}7\cdot {\log }_{(2-\sqrt{3})}5\cdot {\log }_{\frac{1}{9}}(2+\sqrt{3})\cdot {\log }_{\sqrt{\frac{1}{7}}}3$ then $N$ is coprime with	P	2
B	If $M=\frac{6}{{\log }_{3}1500}+\frac{6}{{\log }_{5}1500}+\frac{12}{{\log }_{10}1500}$ then $M$ is less than or equal to	Q	3
C	If $P={\log }_{\sqrt[5]{3.5}}(1+2+3÷6)$ then $P$ is twin prime with	R	5
D	If ${\left(\sqrt[6]{x^9{y}^{-8}}\right)}^{-3}=x^a\cdot y^b$ then $(2a+4b)$ is greater or equal to	S	6
		T	7

• A. (A) P, P, R, S, T; (B) S, T; (C) Q, T; (D) P, Q, R, S, T
• B. (A) P, Q, R, S, T; (B) S, T; (C) Q, T; (D) P, Q, R, S, T
• C. (A) P, R, R, S, T; (B) S, T; (C) Q, T; (D) P, Q, R, S, T
• D. (A) P, S, R, S, T; (B) S, T; (C) Q, T; (D) P, Q, R, S, T

Answer 1

Answer: (B)

(A)$N={\log }_{\frac{1}{5}}7\cdot {\log }_{(2-\sqrt{3})}5\cdot {\log }_{\frac{1}{9}}(2+\sqrt{3})\cdot {\log }_{\sqrt{\frac{1}{7}}}3$
$=\left(-{\log }_{5}7\right)\left({\log }_{(2-\sqrt{3})}5\right)\left(\frac{1}{2}{\log }_3(2-\sqrt{3})\right)\left(-2{\log }_{7}3\right)=1$
and 1 is coprime with all natural numbers.
(B)$M=6{\log }_{1500}3+6{\log }_{1500}5+12{\log }_{1500}10$
$=6\left({\log }_{1500}3+{\log }_{1500}5+{\log }_{1500}100\right)=6{\log }_{1500}1500=6$
(C) $P={\log }_{\sqrt[5]{3.5}}\left(1+2+\frac{3}{6}\right)={\log }_{(3.5{)}^{1/5}}\left(3+\frac{1}{2}\right)=5{\log }_{\frac{7}{2}}\frac{7}{2}=5$ 5 is twin prime with 3 or 7 .
(D) ${\left(\sqrt[6]{x^9{y}^{-8}}\right)}^{-3}={\left({\left(x^9{y}^{-8}\right)}^{1/6}\right)}^{-3}={\left(x^9{y}^{-8}\right)}^{-1/2}=x^{-\frac{9}{2}}\cdot y^4=x^a\cdot y^b$
$\therefore 2a+4b=-9+16=7$