Find the approximate value of log10 1005, given that log10 e = 0.4343.

Let:

$y=f(x)=\log _{10} x$

Here,

$x=1000$

$x+\Delta x=1005$

$\Rightarrow \Delta x=5$

$\Rightarrow d x=\Delta x=5$

For $x=1000$

$y=\log _{10} 1000=\log _{10}(10)^{3}=3$

Now, $y=\log _{10} x=\frac{\log _{e} x}{\log _{e} 10}$

$\therefore \frac{d y}{d x}=\frac{0.4343}{x}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=1000}=\frac{0.4343}{1000}=0.0004343$

$\Delta y=d y=\frac{d y}{d x} d x=0.0004343 \times 5=0.0021715$

$\therefore \log _{10} 1005=y+\Delta y=3.0021715$