Question

Find the approximate value of  log⁡(11.01-log⁡(10.1)), Given log(10) = 2.30 and and log(8.69) = 2.16, all the log are in base ‘e’.

(a) 2.1654

(b) 2.1632

(c) 2.1645

(d) 2.1623

The question was posed to me in an online interview.

The above asked question is from Errors and Approximations topic in portion Partial Differentiation of Engineering Mathematics

Answer 1

The correct option is (d) 2.1623

To explain: Let, f(x,y) = log(x-log(y))

Now by differentiating,

\(\frac{∂f}{∂x}=\frac{1}{x-log⁡(y)}\) and \(\frac{∂f}{∂y}=\frac{1}{y(x-log⁡(y))}\)

Now, putting, x = 11, y = 10, δx=.01 and δy=.1

We get,

\(\frac{∂f}{∂x}∂f/∂x\)=1/8.69 and \(\frac{∂f}{∂y}∂f/∂y\)=1/86.9

Hence, df = 0.0023

Hence, f(x + δx, y + δy) = log⁡(11.01 – log⁡(10.1))= 2.16 + df = 2.1623.