Question

Find the approximate value of (1.04)^3.01.

(a) 1.14

(b) 1.13

(c) 1.11

(d) 1.12

This question was posed to me by my school teacher while I was bunking the class.

This intriguing question comes from Errors and Approximations in division Partial Differentiation of Engineering Mathematics

Answer 1

To approximate the value of (1.04)3.01(1.04)^{3.01}(1.04)3.01, we can use a technique based on linear approximations or the binomial expansion for small changes.

We know that:

(1+x)n≈1+n⋅xfor small values of x.(1 + x)^n \approx 1 + n \cdot x \quad \text{for small values of } x.(1+x)n≈1+n⋅xfor small values of x.

In this case, we can write:

(1.04)3.01=(1+0.04)3.01.(1.04)^{3.01} = (1 + 0.04)^{3.01}.(1.04)3.01=(1+0.04)3.01.

Here, $x=0.04$ and $n=3.01$. Using the linear approximation:

(1.04)3.01≈1+3.01⋅0.04.(1.04)^{3.01} \approx 1 + 3.01 \cdot 0.04.(1.04)3.01≈1+3.01⋅0.04.

Now, calculate the value:

$$3.01\cdot 0.04=\mathrm{0.1204.}$$

Thus:

(1.04)3.01≈1+0.1204=1.1204.(1.04)^{3.01} \approx 1 + 0.1204 = 1.1204.(1.04)3.01≈1+0.1204=1.1204.

Rounding to two decimal places, we get approximately:

$$\mathrm{1.12.}$$

Thus, the correct answer is:

(d) 1.12.