Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ y=f\left(x\right)={\log }_{10}x \\ \mathrm{Here}, \\ x=1000, \\ x+∆x=1005 \\ \Rightarrow ∆x=5 \\ \Rightarrow dx=∆x=5 \\ \mathrm{For}x=1000, \\ y={\log }_{10}1000={\log }_{10}{\left(10\right)}^3=3 \\ \mathrm{Now},y={\log }_{10}x=\frac{{\log }_{e}x}{{\log }_{e}10} \\ \therefore \frac{dy}{dx}=\frac{0.4343}{x} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=1000}=\frac{0.4343}{1000}=0.0004343 \\ ∆y=dy=\frac{dy}{dx}dx=0.0004343\times 5=0.0021715 \\ \therefore {\log }_{10}1005=y+∆y=3.0021715 \end{gathered}$$