Question

Using differential, find the approximate value of the  log₁₀ 10.1, it being given that log₁₀e = 0.4343 ?

Answer 1

\[\text { Consider the function y } = f\left( x \right) = \log_{10} x . \]

\[\text { Let }: \]

\[ x = 10 \]

\[x + ∆ x = 10 . 1\]

\[\text { Then }, \]

\[ ∆ x = 0 . 1\]

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

\[ y = \log_{10} 10 = 1\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 1\]

\[\text { Now,} y = \log_{10} x = \frac{\log_e x}{\log_e 10}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2 . 3025x}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 10} = 0 . 04343\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = 0 . 04343 \times 0 . 1 = 0 . 004343\]

\[ \Rightarrow ∆ y = 0 . 004343\]

\[ \therefore \log_{10} 10 . 1 = y + ∆ y = 1 . 004343\]