Question

If ${\log }_{10}3=0.4771212$, without using log tables , find ${\log }_{10}9,{\log }_{10}\sqrt{3},\log \left(\frac{1}{9}\right)$ and ${\log }_{10}\left(0.3\right)$.

Answer 1

Given:${\log }_{10}3=0.4771212$

$\left(i\right){\log }_{10}9={\log }_{10}{3}^2=2{\log }_{10}3=2\times 0.4771212=0.9542424‬$

$\left(ii\right){\log }_{10}\sqrt{3}={\log }_{10}{3}^{\frac{1}{2}}=\frac{1}{2}{\log }_{10}3=\frac{1}{2}\times 0.4771212=0.2385606‬$

$\left(iii\right){\log }_{10}\frac{1}{9}={\log }_{10}{3}^{-2}=-2{\log }_{10}3=-2\times 0.4771212=-0.9542424$

$=-1+1-0.9542424=\overline{1}+0.0457576‬=\overline{1}.0457576$

$\left(iv\right){\log }_{10}0.3={\log }_{10}\frac{3}{10}={\log }_{10}3-{\log }_{10}10$

$={\log }_{10}3-1=0.4771212-1$

$=-0.5228788=-1+1-0.5228788=\overline{1}+0.4771212‬=\overline{1}.4771212‬$