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 $\frac{c}{ab}$, 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, $\frac{c}{ab}=\frac{\left(\frac{5}{4}\right)}{\left(\frac{1}{2}\right)\left(\frac{5}{4}\right)}=2$.