Question

$\log 0.132$ value.

Answer 1

Compute the required value:

$\log 0.132$ can be written as:

$$\begin{gathered} {\log }_{10}0.132={\log }_{10}[1.32\times {10}^{-1}] \\ \Rightarrow {\log }_{10}0.132={\log }_{10}1.32+{\log }_{10}{10}^{-1} \\ \Rightarrow {\log }_{10}0.132={\log }_{10}1.32+{\log }_{10}\frac{1}{10} \\ \Rightarrow {\log }_{10}0.132={\log }_{10}1.32+{\log }_{10}1-{\log }_{10}10 \\ \Rightarrow {\log }_{10}0.132={\log }_{10}1.32-1 \end{gathered}$$ $$\begin{gathered} \left[\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{=}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}\mathit{a}\mathbf{-}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}\mathit{b} \\ \mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}\mathbf{1}\mathbf{=}\mathbf{0} \\ \because \mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}\mathbf{10}\mathbf{=}\mathbf{1}\right] \end{gathered}$$

Using common log table, look the value for row $13$ and column $2$ which is $1206$.

Therefore, ${\log }_{10}1.32=0.1206$

$$\begin{gathered} \Rightarrow {\log }_{10}0.132=0.1206-1 \\ \Rightarrow {\log }_{10}0.132=-0.8794 \end{gathered}$$

Hence, the value of $\mathbf{log}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{132}$ is $\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{8794}$.