Question

The approximate value of ${\log }_e(4.01)$ is $\dots A\dots$, if ${\log }_{e}4=1.3863$. Here, $A$ refers to

• A. $1.3688$
• B. $0.3888$
• C. $1.3888$
• D. None of these

Answer 1

Answer: (C)

Let $y=f(x)={\log }_{e}x,x=4$
and $x+\Delta x=4.01$ Then,
$\Delta x=0.01$
Now, for $x=4$ we have,
$y=f(4)={\log }_{e}4=1.3863$
let $dx=\Delta x=0.01$
Now, $y={\log }_{e}x$
$\Rightarrow \frac{dy}{dx}=\frac{1}{x}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=4}=\frac{1}{4}$
$\therefore \Delta y=\frac{dy}{dx}\Delta x$
$\Rightarrow \Delta y=\frac{1}{4}\times 0.01$
$\Rightarrow \Delta y=0.0025(\because dy\approx \Delta y)$
$\therefore {\log }_e(4.01)=y+\Delta y$
$=1.3863+0.0025$
$=1.3888$