Question

Using differentials, find the approximate values of the following:
(ix) $lo{g}_{e}10.02$, it being given that $lo{g}_{e}10=2.3026$

Answer 1

Let $y=lo{g}_{e}x\Rightarrow \frac{dy}{dx}=\frac{1}{x}$
And x =10 and $\Delta x=0.02$
Then,
$\Delta y=lo{g}_e(x+\Delta x)-lo{g}_{e}x$
$\Rightarrow \Delta y=lo{g}_e(10+0.02)-lo{g}_{e}10$
$\Rightarrow \Delta y=lo{g}_{e}10.02-2.3026$
$\Rightarrow lo{g}_{e}10.02=2.3026+\Delta y$ ...(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{x}\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{10}\times \left(0.02\right)\approx 0.002$
From equation (1)
$lo{g}_{e}10.02\approx 2.3026+0.002$
$\therefore lo{g}_{e}10.02\approx 2.3046$