Question

The logarithm of 27 $\sqrt[4]{49}\sqrt[3]{39}$ to the base 3 is :

• A. $8\frac{1}{2}$
• B. $4\frac{1}{6}$
• C. 5
• D. 3
• E. none of these

Answer 1

Answer: (B)

$4\frac{1}{6}$

$27\sqrt[4]{49}\sqrt[3]{39}=3^3\cdot 3^{\frac{1}{2}}\cdot 3^{\frac{2}{3}}=3^{\frac{25}{6}}{\log }_3\left(27\sqrt[4]{49}\sqrt[3]{39}\right)={\log }_3\left(3^{\frac{25}{6}}\right)=\frac{25}{6}{\log }_{3}3=4\frac{1}{6}$