The approximate value of $(1.0002)^{3000}$, is

1.6

Let $y=x^{3000}, x=1$ and $x+\Delta x=1.0002$

Then, $\Delta x=1.0002-1=0.0002$

Also, y = 1 when x = 1

Now,

$y=x^{3000}$

$\Rightarrow \frac{d y}{d x}=3000 x^{2999} \Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=3000$

∴ $\Delta y=\frac{d y}{d x} \Delta x$

$\Rightarrow \Delta y=3000 \times 0.0002=\frac{6}{10}=0.6$

Hence,

$(1.0002)^{3000}=y+\Delta y=1+0.6=1.6$