Let us assume that f(x) = log₁₀x

Also, let x = 1000 so that x + Δx = 1005

⇒ 1000 + Δx = 1005

∴ Δx = 5

On differentiating f(x) with respect to x, we get

We know

When x = 1000, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 5

⇒ Δf = (0.0004343)(5)

∴ Δf = 0.0021715

Now, we have f(1005) = f(1000) + Δf

⇒ f(1005) = log₁₀1000 + 0.0021715

⇒ f(1005) = log₁₀10³ + 0.0021715

⇒ f(1005) = 3 × log₁₀10 + 0.0021715

⇒ f(1005) = 3 + 0.0021715 [∵ log_aa = 1]

∴ f(1005) = 3.0021715

Thus, log₁₀1005 = 3.0021715