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If ${\log _{10}}8 = 0.90$, find the value of
 (i) ${\log _{10}}4 = ?$
(ii) ${\log _{10}}\sqrt {32} = ?$
(iii) ${\log _{10}}0.125 = ?$

Answer Verified Verified
Hint: At first, we have to convert ${\log _{10}}$ in the base index of 2. And then we need to simplify the whole part. In this way, we will convert ${\log _{10}}\sqrt {32} $ in the base index of 2 and simplify the whole part. For the third question we have to use the formula of logarithm.

Complete step-by-step answer:
According to the question
     ${\log _{10}}8 = {\log _{10}}{2^3}$ [ As $8 = {2^3}$]
     $ = 3.\log {10^2}$
From the values given in the question, we have
$\therefore 3{\log _{10}}2 = 0.90{\text{ (as per question)}}$
Solving it,
     $ \Rightarrow {\log _{10}}2 = 0.3$…..(1)
(i) ${\log _{10}}4$
     $ = {\log _{10}}{2^2}$ [We know $4 = {2^2}$ so by putting the value in the next step]
     $ = 2.{\log _{10}}2{\text{ }}...{\text{(1)}}$ $\left[ \begin{gathered}
  {\text{We know}} \\
  {\text{lo}}{{\text{g}}_e}{a^b} = b{\log _e}a \\
\end{gathered} \right]$
So, ${\log _{10}}4 = 2.{\log _{10}}2{\text{ from (1)}}$
     $ = 2 \times \left( {0.3} \right) = 0.6{\text{ }}...{\text{(2)}}$
(2) ${\log _{10}}\sqrt {32} $
     $ = {\log _{10}}{\left( {32} \right)^{\dfrac{1}{2}}}{\text{ }}\left[ \begin{gathered}
  {\text{We know}} \\
  {\text{lo}}{{\text{g}}_e}{a^b} = b{\log _e}a \\
\end{gathered} \right]$
Applying the rule, we get
     $ = \dfrac{1}{2}{\log _{10}}32$
     $ = \dfrac{5}{2}{\log _{10}}2$
Thus, after simplification and value from eqn (2), we get
     $ = \dfrac{5}{2} \times 0.3 = 0.75 {\text{ }}\left[ {{\text{from (1)}}} \right]$
(3) ${\log _{10}}0.125$
     $ = {\log _{10}}\left( {\dfrac{{125}}{{1000}}} \right)$
(since$0.125 = \dfrac{{125}}{{1000}}$)
Solving it, we get
     ${\log _e}\left( {a/b} \right) = {\log _e}a - {\log _e}b$$ = {\log _{10}}\left( {\dfrac{1}{8}} \right){\text{ }}\left[ \begin{gathered}
  {\text{We know}} \\
  {\log _e}\left( {a/b} \right) = {\log _e}a - {\log _e}b \\
\end{gathered} \right]$
     $ = {\log _{10}}1 - {\log _{10}}8$
     $ = - 0.90$ (since ${\log _{10}}1 = 0$and ${\log _{10}}8 = 0.90$)
Thus solution is:
 $\begin{gathered}
  {\log _{10}}4 = 0.6 \\
  {\log _{10}}\sqrt {32} = 1.5 \\
  {\log _{10}}0.125 = - 0.90 \\
\end{gathered} $

Additional Information: Logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. In the simplest case, the logarithm counts the number of occurrences of the same factor in repeated multiplication; e.g., since \[1000{\text{ }} = {\text{ }}10\, \times \,10\, \times \,10{\text{ }} = {\text{ }}{10^3}\], the “\[logarithm{\text{ }}base{\text{ }}10\]" of \[1000\] , or \[lo{g_{10}}\left( {1000} \right){\text{ }} = {\text{ }}3\] \[lo{g_{10}}\left( {10} \right){\text{ }} = {\text{ }}3\].
Logarithms are also of various representation type,
Common logarithm: The common logarithm has \[base{\text{ }}10\], and is represented on the calculator as log(x).
Natural logarithm: The natural logarithm has \[base{\text{ }}e\], a famous irrational number, and is represented on the calculator by ln(x). The natural and common logarithm can be found throughout Algebra and Calculus.

Note: We have to be careful about the index because we all have to change the index of the given one and the one which we have to find. Do the calculative portion carefully to ensure that there would be no mistake instead of knowing the concept & formulas. Formulas used in the above questions are;
\[{\text{lo}}{{\text{g}}_e}{a^b} = b{\log _e}a\]
${\log _e}\left( {a/b} \right) = {\log _e}a - {\log _e}b$