Question

Find approximate value of ${\log _e}\left( {4.04} \right)$ if ${\log _{10}}4 = 0.6021$ and \[{\log _{10}}e = 0.4343\].

Answer 1

Hint: In this particular question use the concept that according to the logarithmic property ${\log _a}b = \dfrac{{{{\log }_{10}}b}}{{{{\log }_{10}}a}}$, ${\log _{10}}\left( {\dfrac{m}{n}} \right) = {\log _{10}}m - {\log _{10}}n$, ${\log _{10}}\left( {mn} \right) = {\log _{10}}m + {\log _{10}}n$ and ${\log _{10}}100 = 2$ so use these properties to reach the solution of the question.

Complete step by step answer:
Given logarithmic equation
${\log _e}\left( {4.04} \right)$.................... (1)
We have to find the approximate value of this equation.
Now as we all know that according to the logarithmic property ${\log _a}b = \dfrac{{{{\log }_{10}}b}}{{{{\log }_{10}}a}}$ so use this property in the above equation we have,
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{{{\log }_{10}}\left( {4.04} \right)}}{{{{\log }_{10}}e}}$
Now the above equation is also written as,
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{{{\log }_{10}}\left( {\dfrac{{404}}{{100}}} \right)}}{{{{\log }_{10}}e}}$
Now again according to the logarithmic property ${\log _{10}}\left( {\dfrac{m}{n}} \right) = {\log _{10}}m - {\log _{10}}n$, So use this property in the above equation we have,
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{{{\log }_{10}}\left( {404} \right) - {{\log }_{10}}\left( {100} \right)}}{{{{\log }_{10}}e}}$
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{{{\log }_{10}}\left( {4 \times 101} \right) - {{\log }_{10}}\left( {100} \right)}}{{{{\log }_{10}}e}}$
Now again according to the logarithmic property ${\log _{10}}\left( {mn} \right) = {\log _{10}}m + {\log _{10}}n$, So use this property in the above equation we have,
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{{{\log }_{10}}\left( 4 \right) + {{\log }_{10}}\left( {101} \right) - {{\log }_{10}}\left( {100} \right)}}{{{{\log }_{10}}e}}$
Now as we know that ${\log _{10}}100 = 2$, so we can say that ${\log _{10}}101 \simeq 2$
So use these values in the above equation we have,
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{{{\log }_{10}}\left( 4 \right) + 2 - 2}}{{{{\log }_{10}}e}} = \dfrac{{{{\log }_{10}}\left( 4 \right)}}{{{{\log }_{10}}e}}$
Now it is given that ${\log _{10}}4 = 0.6021$ and \[{\log _{10}}e = 0.4343\] so we have,
$ \Rightarrow {\log _e}\left( {4.04} \right) = \dfrac{{0.6021}}{{0.4343}}$
$ \Rightarrow {\log _e}\left( {4.04} \right) = 1.386$
So this is the required approximate value of ${\log _e}\left( {4.04} \right)$.
Approximate value is because we use the approximate value of ${\log _{10}}101 \simeq 2$.

So, 1.386 is the required approximate value of ${\log _e}\left( {4.04} \right)$.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic logarithmic properties which is all stated above then use the approximate value of ${\log _{10}}101$ which is approximately equal to 2, then simplify the given equation as above we will get the required answer.