log_e10.02, it being given that log_e10 = 2.3026

Let us assume that f(x) = log_ex

Also, let x = 10 so that x + Δx = 10.02

⇒ 10 + Δx = 10.02

∴ Δx = 0.02

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

We know

When x = 10, 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 = 0.02

⇒ Δf = (0.1)(0.02)

∴ Δf = 0.002

Now, we have f(10.02) = f(10) + Δf

⇒ f(10.02) = log_e10 + 0.002

⇒ f(10.02) = 2.3026 + 0.002

∴ f(10.02) = 2.3046

Thus, log_e10.02 ≈ 2.3046