Question

If $lo{g}_{e}4=1.3868$, then $lo{g}_{e}4.01=$

• A. 1.3968
• B. 1.3898
• C. 1.3893
• D. none of these

Answer 1

The correct option is C 1.3893

Let $f\left(x\right)=y={\log }_{e}x$
Then, $y+\Delta y={\log }_e(x+\Delta x)$
$\therefore \Delta y={\log }_e(x+\Delta x)-lo{g}_{e}x$ .....(i)
Since,$\Delta y=f^{\prime }\left(x\right).\Delta x$
$\Rightarrow {\log }_e(x+\Delta x)-lo{g}_{e}x=f^{\prime }\left(x\right).\Delta x$
$\Rightarrow {\log }_e(x+\Delta x)-lo{g}_{e}x=\frac{1}{x}.\Delta x$
Put $x=4$, $\Delta x=0.01$
${\log }_{e}4.01-lo{g}_{e}4=\frac{1}{4}\times \left(0.01\right)=1.3868+0.0025=1.3893$