Question

The value of \[\log 0.008\] is

Answer 1

Hint: By using the basic logarithm formula, solve the expression. Consider 0.008, convert it into fraction form and apply the log formula, then simplify it.

Complete step-by-step answer:
A logarithm is the opposite of power. In other words, if we take a logarithm of a number, we undo an exponentiation.

Here we have been asked to find the value of \[\log 0.008\].

Let us consider \[\log 0.008\] as \[{{\log }_{10}}0.008\].

log base 10 is also known as the common logarithm. The common logarithm of x is the power to which the number 10 must be raised to obtain the value of x. The common logarithm of 10 is 1.

Thus, \[{{\log }_{10}}0.008\].

0.008 can be converted from decimal form to fraction,

\[0.008=\dfrac{8}{1000}=\dfrac{8}{{{10}^{3}}}=8\times {{10}^{-3}}\]

Thus the expression becomes,

\[{{\log }_{10}}0.008={{\log }_{10}}8\times {{10}^{-3}}\]

This expression is of the form \[\log ab\].

We know that, \[\log \left( ab \right)=\log a+\log b\].

\[\therefore {{\log }_{10}}8\times {{10}^{-3}}=\log 8+\log {{10}^{-3}}\] \[\left( {{2}^{3}}=8 \right)\].

\[{{\log }_{10}}8\times {{10}^{-3}}={{\log }_{10}}{{2}^{3}}+{{\log }_{10}}{{10}^{-3}}\]

We know the formula, \[\log {{a}^{b}}=b\log a\].

\[\therefore {{\log }_{10}}{{2}^{3}}+{{\log }_{10}}{{10}^{-3}}=3{{\log }_{10}}2+\left( -3 \right){{\log }_{10}}10\]

We said that, \[{{\log }_{10}}10=1\] and the value of \[{{\log }_{10}}2=0.301\].

Thus substitute these values,

\[3{{\log }_{10}}2+\left( -3 \right){{\log }_{10}}10=3\times 0.301-3\times 1\]

      \[\begin{align}

  & =3\times 0.301-3 \\

 & =3\left( 0.301-1 \right) \\

 & =-2.097 \\

\end{align}\]

Thus we got the value of \[\log 0.008\] = -2.097.


Note: We can also solve this by,

\[\log 0.008=\log \left( \dfrac{8}{1000} \right)\]

                   \[\begin{align}

  & =\log \left( \dfrac{{{2}^{3}}}{{{10}^{3}}} \right)=\log {{\left( \dfrac{2}{10} \right)}^{3}} \\

 & =\log {{\left( \dfrac{1}{5} \right)}^{3}} \\

 & =\log {{\left( 5 \right)}^{-3}} \\

\end{align}\]

Thus by, \[\log {{a}^{b}}=b\log a\]

\[\log {{\left( 5 \right)}^{-3}}=-3\log 5\]

The value of \[\log 5=0.699\].

\[\therefore -3\log 5=-3\times 0.699=-2.097\]