Question

If ${\log }_{ab}a=\frac{1}{3}$, then the value of ${\log }_{ab}\left(\frac{\sqrt{a}}{\sqrt[3]{3b}}\right)$ is equal to

• A. $-\frac{1}{3}$
• B. $-\frac{1}{9}$
• C. $-\frac{1}{18}$
• D. $-\frac{1}{24}$

Answer 1

Answer: (C)

$-\frac{1}{18}$

${\log }_{ab}a=\frac{1}{3}\Rightarrow a={\left(ab\right)}^{\frac{1}{3}}{a}^{\frac{2}{3}}=b^{\frac{1}{3}}...\left(i\right)$

${\log }_{ab}\left(\frac{\sqrt{a}}{\sqrt[3]{3b}}\right)={\log }_{ab}\left(\frac{a^{\frac{1}{2}}}{b^{\frac{1}{3}}}\right)$

$\text{Substituting the value of b from eq.(i)}{\log }_{ab}\left(\frac{\sqrt{a}}{\sqrt[3]{3b}}\right)={\log }_{ab}\left(\frac{a^{\frac{1}{2}}}{a^{\frac{2}{3}}}\right)={\log }_{ab}(a^{\frac{1}{2}-\frac{2}{3}})={\log }_{ab}(a^{\frac{3-4}{6}})={\log }_{ab}{a}^{\frac{-1}{6}}=x{a}^{\frac{-1}{6}}=({ab)}^x$

$\text{Substituting value of a we get}$

$a^{\frac{-1}{6}}={\left(ab\right)}^{\frac{1}{3}\times \frac{-1}{6}}={\left(ab\right)}^{\frac{-1}{18}}\therefore x=\frac{-1}{18}{\log }_{ab}\left(\frac{\sqrt{a}}{\sqrt[3]{3b}}\right)=\frac{-1}{18}$