Question

Find the approximate value of log₁₀ 1005, given that log₁₀ e = 0.4343 ?

Answer 1

\[\text { Let }: \]

\[ y = f\left( x \right) = \log_{10} x\]

\[\text { Here }, \]

\[x = 1000, \]

\[x + ∆ x = 1005\]

\[ \Rightarrow ∆ x = 5\]

\[ \Rightarrow dx = ∆ x = 5\]

\[\text{ For } x = 1000, \]

\[y = \log_{10} 1000 = \log_{10} \left( 10 \right)^3 = 3\]

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

\[ \therefore \frac{dy}{dx} = \frac{0 . 4343}{x}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 1000} = \frac{0 . 4343}{1000} = 0 . 0004343\]

\[ ∆ y = dy = \frac{dy}{dx}dx = 0 . 0004343 \times 5 = 0 . 0021715\]

\[ \therefore \log_{10} 1005 = y + ∆ y = 3 . 0021715\]