Question

Column I		Column II
A	$ 3 \log 4+2 \log 5-\frac{1}{3} \log 64-\frac{1}{2} \log 16 $where base of the logarithm is 10 , equals	P	1
B	$ 2^{\log 3-\log 5} \cdot 3^{\log 5-\log 2} \cdot 5^{\log 2-\log 3} $ where base of the logarithm is 10 , equals	Q	2
C	$\text { Natural solution of the equation } \sqrt{7^{2 x^2-5 x-6}}=(\sqrt{2})^{3 \log _2 49} \text { equals }$	R	3
D	If $\log _{12} 27=a$ and $\log _6 16=K\left(\frac{3-a}{3+a}\right)$, then $K$ equals	S	4
		T	5

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

Answer 1

Answer: (D)

(A)$3 \log 4+2 \log 5-\frac{1}{3} \log 64-\frac{1}{2} \log 16$
$=\log 4^3+\log 5^2-\log (64)^{1 / 3}-\log (16)^{1 / 2}$
$=\log \frac{4^3 \cdot 5^2}{(64)^{1 / 3}(16)^{1 / 2}}=\log \frac{4^3 \cdot 5^2}{4 \cdot 4}=\log 4 \cdot 5^2=\log 100=2 \text { Ans. }$
(B) $2^{\log 3-\log 5} \cdot 3^{\log 5-\log 2} \cdot 5^{\log 2-\log 3}$
$=\frac{2^{\log 3}}{2^{\log 5}} \cdot \frac{3^{\log 5}}{3^{\log 2}} \cdot \frac{5^{\log 2}}{5^{\log 3}}=1 \text { Ans. }$
(C) $\sqrt{7^{2 x^2-5 x-6}}=(\sqrt{2})^{3 \log _2 49}$
i.e.$7^{2 x^2-5 x-6}=2^{3 \log _2 49}=2^{\log _2(49)^3}$
$7^{2 x^2-5 x-6}=49^3 $
$7^{2 x^2-5 x-6}=7^6 $
$2 x^2-5 x-6=6$
i.e. $ 2 x^2-5 x-12=0$
So $ x=4$ or $x=-\frac{3}{2} \Rightarrow $ absolute values are $x=4$ or $x=\frac{3}{2}$
(D)$\frac{\log _2 27}{2+\log _2 3}=\frac{3 \log _2 3}{2+\log _2 3}= a \ldots .(1) $
$\log _6 16=\frac{4}{1+\log _2 3} \ldots .(2)$
$\text { from(1), } 3 \log _2 3=2 a+a \log _2 3 \Rightarrow \log _2 3=\frac{2 a}{3-a}$
$\text { from(2), } \log _6 16=\frac{4}{1+\frac{2 a}{3-a}} \Rightarrow \log _6 16=\frac{4(3-a)}{3+a} \Rightarrow K =4$