Question

Using differentials, find the approximate values of the following:
(x) $lo{g}_{10}10.1$, it being given that $lo{g}_{10}e=0.4343$

Answer 1

Let $y=lo{g}_{10}x\Rightarrow \frac{dy}{dx}=\frac{1}{x}lo{g}_{10}e$
And x =10 and $\Delta x=0.1$
Then,
$\Delta y=lo{g}_{10}(x+\Delta x)-lo{g}_{10}x$
$\Rightarrow \Delta y=lo{g}_{10}(10+0.1)-lo{g}_{10}10$
$\Rightarrow \Delta y=lo{g}_{10}10.1-1$
$\Rightarrow lo{g}_{10}10.1=1+\Delta y$ ...(1)
Now, approximate change in value of y is given by,
$\because \Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{x}lo{g}_{10}e\times \Delta x$
$\Rightarrow \Delta y\approx \frac{1}{10}\times \left(0.4343\right)\times 0.1\approx 0.004343$
From equation (1)
$lo{g}_{10}10.1\approx 1+0.004343$
$\therefore lo{g}_{10}10.1\approx 1.004343$