Question

Using differential, find the approximate value of the log_e 10.02, it being given that log_e10 = 2.3026 ?

Answer 1

\[\text { Consider the function } y = f\left( x \right) = \log_e x . \]

\[\text { Let }: \]

\[ x = 10 \]

\[ x + ∆ x = 10 . 02\]

\[\text { Then }, \]

\[ ∆ x = 0 . 02\]

\[\text { For }x = , \]

\[ y = \log_e 10 = 2 . 3026\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 02\]

\[\text { Now }, y = \log_e x\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 10} = \frac{1}{10}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 0 . 02 = 0 . 002\]

\[ \Rightarrow ∆ y = 0 . 002\]

\[ \therefore \log_e 10 . 02 = y + ∆ y = 2 . 3046\]