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	$\text{ Natural solution of the equation }\sqrt{7^{2x^2-5x-6}}=(\sqrt{2}{)}^{3{\log }_{2}49}\text{ equals }$	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) $\sqrt{7^{2x^2-5x-6}}=(\sqrt{2}{)}^{3{\log }_{2}49}$
i.e.$7^{2x^2-5x-6}=2^{3{\log }_{2}49}=2^{{\log }_2(49{)}^3}$
$7^{2x^2-5x-6}=49^3$
$7^{2x^2-5x-6}=7^6$
$2x^2-5x-6=6$
i.e. $2x^2-5x-12=0$
So $x=4$ or $x=-\frac{3}{2}\Rightarrow$ absolute values are $x=4$ or $x=\frac{3}{2}$
(D)$\frac{{\log }_{2}27}{2+{\log }_{2}3}=\frac{3{\log }_{2}3}{2+{\log }_{2}3}=a\dots .(1)$
${\log }_{6}16=\frac{4}{1+{\log }_{2}3}\dots .(2)$
$\text{ from(1), }3{\log }_{2}3=2a+a{\log }_{2}3\Rightarrow {\log }_{2}3=\frac{2a}{3-a}$
$\text{ from(2), }{\log }_{6}16=\frac{4}{1+\frac{2a}{3-a}}\Rightarrow {\log }_{6}16=\frac{4(3-a)}{3+a}\Rightarrow K=4$