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, ${{\log }_{\text{e}}}\left( 4.01 \right)={{\log }_{\text{e}}}\left( 4 \right)+\left( \dfrac{1}{4}\times 0.01 \right)$

& \Rightarrow {{\log }_{\text{e}}}\left( 4.01 \right)=1.3868-0.0025 \\\ & \Rightarrow {{\log }_{\text{e}}}\left( 4.01 \right)=1.3893 \\\ \end{aligned}$$ **So, the correct answer is “Option C”.** **Note** : As we know that, $$\begin{aligned} & \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{aligned}$$ $$\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).