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 $=7$