Question

If $\log 2=0.3010$ and $\log 3=0.4771$, find the value of $\log 1.2$.

Answer 1

Hint: We first use the logarithmic formulas of ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$, \[{{\log }_{x}}{{a}^{b}}=b{{\log }_{x}}a\] to find the factorisation of 12 with the log function. The base of the logarithm in any case will be taken as 10. Then we find the relation between $\log 12$ and $\log 1.2$ to use the formula of ${{\log }_{x}}\dfrac{a}{b}={{\log }_{x}}a-{{\log }_{x}}b$. We put the given values to find the solution.

Complete step by step answer:
We are going to use the logarithmic formula of adding and power reduction to solve the problem.
We know ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$ and \[{{\log }_{x}}{{a}^{b}}=b{{\log }_{x}}a\].
We know $12={{2}^{2}}\times 3$. Taking logarithm both sides we get $\log \left( 12 \right)=\log \left( {{2}^{2}}\times 3 \right)$.
We apply the theorems and find
${{\log }_{10}}\left( 12 \right)={{\log }_{10}}{{2}^{2}}+{{\log }_{10}}3=2{{\log }_{10}}2+{{\log }_{10}}3$
Now we put the values of $\log 2=0.3010$ and $\log 3=0.4771$.
${{\log }_{10}}\left( 12 \right)=2\left( 0.3010 \right)+\left( 0.4771 \right)=1.0791$.
We have to find the value of $\log 1.2$. We know ${{\log }_{x}}\dfrac{a}{b}={{\log }_{x}}a-{{\log }_{x}}b$.
We also have the identity of \[{{\log }_{x}}x=1\].
$\begin{align}
  & {{\log }_{10}}1.2={{\log }_{10}}\dfrac{12}{10} \\
 & \Rightarrow {{\log }_{10}}12-{{\log }_{10}}10 \\
 & \Rightarrow 1.0791-1 \\
 & \Rightarrow 0.0791 \\
\end{align}$

Therefore, the value of $\log 1.2$ is $0.0791$.

Note: We need to be careful about the base here. If otherwise mentioned we will always take the base as 10. The special case of taking ‘e’ as a base of logarithm is called ‘ln’. The base in general can be anything. Depending on the number we are taking, the base of the logarithm changes.