The approximate value of $(1.0002)^{3000}$, is
Answer & explanation
Correct answer: option 3
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,
$(1.0002)^{3000}=y+\Delta y=1+0.6=1.6$