Question

Let 'a' denotes the logarithm of $0 . \overline{3}$ to the base $0 . \overline{1}$, 'b' denotes the logarithm of 243 to the base 81 and ' $c$ ' denotes the number whose logarithm to the base 0.64 is minus $\frac{1}{2}$. Then the value of $\frac{c}{a b}$, is

• A. lies between 4 and 5 .
• B. odd composite.
• C. odd prime.
• D. lies in $(1,3)$

Answer 1

Answer: (D)

$a =\log _{0 . T }(0 . \overline{3})=\log _{\frac{1}{9}}\left(\frac{1}{3}\right)=\frac{1}{2}$
$b=\log _{81} 243=\log _{3^4}(3)^5=\frac{5}{4}$
Let $\log _{0.64} N =\frac{-1}{2} \Rightarrow N =(0.64)^{\frac{-1}{2}}=\left(\frac{8}{10}\right)^{-1}=\frac{5}{4}$
$\therefore c =\frac{5}{4}$
Hence, $\frac{c}{a b}=\frac{\left(\frac{5}{4}\right)}{\left(\frac{1}{2}\right)\left(\frac{5}{4}\right)}=2$.