log_e4.04, it being given that log₁₀4 = 0.6021 and log₁₀e = 0.4343

Let us assume that f(x) = log_ex

Also, let x = 4 so that x + Δx = 4.04

⇒ 4 + Δx = 4.04

∴ Δx = 0.04

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

We know

When x = 4, 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 = 0.04

⇒ Δf = (0.25)(0.04)

∴ Δf = 0.01

Now, we have f(4.04) = f(4) + Δf

⇒ f(4.04) = log_e4 + 0.01

⇒ f(4.04) = 1.3863689 + 0.01

∴ f(4.04) = 1.3963689

Thus, log_e4.04 ≈ 1.3963689