Question

Column I		Column II
A	Let $p={\log }_5\left({\log }_{5}3\right)$. If $3^{C+5^{-p}}=45$, then the value of $C$ is	P	1
B	If $x$ and $y$ satisfy simultaneously the equations,$(2x{)}^{\log 2}=(3y{)}^{\log 3}\text{ and }{3}^{\log x}=2^{\log y}\text{, }$then the value of $\left(x^{-1}+y^{-1}\right)$, is	Q	2
C	The value of $x$ satisfying the equation $3^{{\log }_{\sqrt{3}}\sqrt{x}}+9^{{\log }_{3}x}+27^{{\log }_{3}2x^2}=3$ is coprime with	R	3
		S	5

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

Answer 1

Answer: (D)

(A)$\text{ Consider }{5}^{-p}=5^{-{\log }_5\left({\log }_{5}3\right)}=\frac{1}{5^{{\log }_5\left({\log }_{5}3\right)}}=\frac{1}{{\log }_{5}3}\left(a^{{\log }_{a}N}=N\right)$
$\therefore 5-p=\frac{1}{{\log }_{5}3}={\log }_{3}5$
Now, $3^{C+{\log }_{3}5}=45$
$\Rightarrow 3^C\cdot 3^{{\log }_{3}5}=45\Rightarrow 3^C\cdot 5=45\Rightarrow 3^C=9=3^2\Rightarrow C=2\text{. Ans. }$
(B) We have $(2x{)}^{\log 2}=(3y{)}^{\log 3}$
taking log on both side, we get
$\Rightarrow (\log 2)(\log 2+\log x)=\log 3(\log 3+\log y)$
$\Rightarrow {\log }^{2}2-{\log }^{2}3=\log 3\cdot \log y-\log 2\cdot \log x$ ....(1)
$\text{ Again, }{3}^{\log x}=2^{\log y}$
taking log on both the sides, we get
$\Rightarrow \log x\cdot \log 3=\log y\cdot \log 2\Rightarrow \frac{\log x}{\log 2}=\frac{\log y}{\log 3}=k\text{ (let) }$
$\Rightarrow \log x=k\cdot \log 2\text{ and }\log y=k\cdot \log 3(\text{ put in (1)) }$
$\therefore \text{ we will get }k=-1\Rightarrow x=1/2,y=1/3$
Hence $\left(x^{-1}+y^{-1}\right)=(2+3)=5$. Ans
(C)$3^{{\log }_{3}x}+3^{2{\log }_{3}x}+3^{3{\log }_{3}x}=3$
$\Rightarrow x+x^2+x^3=3\Rightarrow (x-1)+\left(x^2-1\right)+\left(x^3-1\right)=0$
$\Rightarrow (x-1)\left(1+x+1+x^2+x+1\right)=0$
$\Rightarrow x=1\left(\because x^2+2x+3\ne 0\right)$