log₁₀10.1, it being given that log₁₀e = 0.4343

Let us assume that f(x) = log₁₀x

Also, let x = 10 so that x + Δx = 10.1

⇒ 10 + Δx = 10.1

∴ Δx = 0.1

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 = 0.1

⇒ Δf = (0.04343)(0.1)

∴ Δf = 0.004343

Now, we have f(10.1) = f(10) + Δf

⇒ f(10.1) = log₁₀10 + 0.004343

⇒ f(10.1) = 1 + 0.004343 [∵ log_aa = 1]

∴ f(10.1) = 1.004343

Thus, log₁₀10.1 ≈ 1.004343