Question

Column I		Column II
A	Anti logarithm of $(0.\overline{6})$ to the base 27 has the value equal to	P	5
B	Characteristic of the logarithm of 2008 to the base 2 is	Q	7
C	The value of b satisfying the equation, ${\log }_{e}2\cdot {\log }_{b}625={\log }_{10}16\cdot {\log }_{e}10\text{ is }$	R	9
D	Number of naughts after decimal before a significant figure comes in the number ${\left(\frac{5}{6}\right)}^{100}$, is	S	10

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

Answer 1

Answer: (A)

(A) Antilog ${\log }_{27}(2/3)=x\Rightarrow (2/3)={\log }_{27}x\Rightarrow x=(27{)}^{2/3}=9$ Ans
(B) We have to find char of ${\log }_{2}2008$
we know , ${\log }_{2}1024=10$ and ${\log }_{2}2048=11$
$\therefore 10<{\log }_{2}2008<11\therefore$ it has char $=10$ Ans.
(C) $\frac{\log 2}{\log e}\times \frac{\log 625}{\log b}=\frac{4\log 2}{\log 10}\times \frac{\log 10}{\log e}\Rightarrow \log 5=\log b\Rightarrow b=5$ Ans.
(D)$y={\left(\frac{5}{6}\right)}^{100}\Rightarrow {\log }_{10}y=100\left({\log }_{10}5-{\log }_{10}6\right)=100\left(1-{\log }_{10}2-{\log }_{10}3-{\log }_{10}2\right)$
$\Rightarrow -7.91\Rightarrow -8+0.09$
since char is -8 . Hence number of zeros after decimal $=7$