Question

Find the logarithms of
$\frac{1}{256}$ to base $2\surd 2$, and $\cdot \dot{3}$ to base $9$.

Answer 1

Logarithm of $\frac{1}{256}$ to base $2\sqrt{2}$ is

${\log }_{2\sqrt{2}}\frac{1}{256}={\log }_{2^{\frac{3}{2}}}{2}^{-8}=\frac{2}{3}(-8)=-\frac{16}{3}$

Logarithm of $0.3$ to base $9$ is

${\log }_{9}0.3={\log }_{3^2}0.3=\frac{1}{2}\left({\log }_{3}0.3\right)=\frac{1}{2}\left({\log }_{3}3\left(0.1\right)\right)=\frac{1}{2}({\log }_{3}3+{\log }_{3}0.1)=\frac{1}{2}(1-{\log }_{3}10)$