Question

If \[{{\log }_{\text{e}}}\left( 4 \right)=1.3868\], then \[{{\log }_{\text{e}}}\left( 4.01 \right)=?\]
a)1.3968
b)1.3898
c)1.8393
d)None of these

Answer 1

Hint: Assume the function = $f\left( x \right)={{\log }_{\text{e}}}\left( x \right)$.Since the function to be calculated consists of a small change, convert the given logarithmic function in the form of \[{{\log }_{\text{e}}}\left( x+dx \right)\] where ‘x’ represents the original value and ‘dx’ represents the small change in the original value. In this question; x= 4 and dx=0.01

Complete step-by-step answer:
Then, differentiate the function with respect to x and substitute the values of ‘x’ and ‘dx’ to get the solution.
Let a function \[f\left( x \right)=y={{\log }_{\text{e}}}\left( x \right)......(1)\]
Differentiate both sides of equation (1) with respect to ‘x’, we get:
  \[f'\left( x \right)=\dfrac{dy}{dx}\]
Therefore, we can write: \[dy=f'\left( x \right)dx\]
So, from equation (1), we can say: \[dy=f'\left( {{\log }_{\text{e}}}\left( x \right) \right)dx\]
Since, \[f'\left( {{\log }_{\text{e}}}\left( x \right) \right)=\dfrac{1}{x}\]
So, we can write: \[dy=\dfrac{1}{x}dx......(2)\]

Now by increasing f(x) by an element ‘dx’, we get:
\[f\left( x+dx \right)={{\log }_{\text{e}}}\left( x+dx \right)\]
Comparing with equation (1), we can write:
\[y+dy={{\log }_{\text{e}}}\left( x+dx \right)\]
Therefore, \[dy={{\log }_{\text{e}}}\left( x+dx \right)-y\]
 \[\Rightarrow dy={{\log }_{\text{e}}}\left( x+dx \right)-{{\log }_{\text{e}}}\left( x \right)......(3)\]

Substitute the value of ‘dy’ from equation (2) in equation (4):
 \[\dfrac{1}{x}dx={{\log }_{\text{e}}}\left( x+dx \right)-{{\log }_{\text{e}}}\left( x \right)......(4)\]
Now, compare equation (4) with the function given in the question i.e. \[{{\log }_{\text{e}}}\left( 4.01 \right)\]
Consider x = 4 and dx = 0.01 and put the values in equation (4).
We get:
\[\left( \dfrac{1}{4}\times 0.01 \right)={{\log }_{\text{e}}}\left( 4.01 \right)-{{\log }_{\text{e}}}\left( 4 \right)\]
Therefore, \[{{\log }_{\text{e}}}\left( 4.01 \right)={{\log }_{\text{e}}}\left( 4 \right)+\left( \dfrac{1}{4}\times 0.01 \right)\]
\[\begin{align}
  & \Rightarrow {{\log }_{\text{e}}}\left( 4.01 \right)=1.3868-0.0025 \\
 & \Rightarrow {{\log }_{\text{e}}}\left( 4.01 \right)=1.3893 \\
\end{align}\]
So, the correct answer is “Option C”.

Note: As we know that, \[\begin{align}
  & \underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}=f'\left( x \right) \\
 & \Delta y=f'\left( x \right)\Delta x \\
 & dy=f'\left( x \right)dx \\
\end{align}\] \[\underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}=f'\left( x \right)\]
Therefore, \[\Delta y=f'\left( x \right)\Delta x\]
Also \[dy=f'\left( x \right)dx\]
Hence, we can use differentials to calculate small changes in the dependent variable (dy) of a function corresponding to small changes in the independent variable f(x).