Q. 9 H5.0( 3 Votes )

# Using differentials, find the approximate values of the following:

log_{e}4.04, it being given that log_{10}4 = 0.6021 and log_{10}e = 0.4343

Answer :

log_{e}4.04, it being given that log_{10}4 = 0.6021 and log_{10}e = 0.4343

Let us assume that f(x) = log_{e}x

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_{e}4 + 0.01

⇒ f(4.04) = 1.3863689 + 0.01

∴ f(4.04) = 1.3963689

Thus, log_{e}4.04 ≈ 1.3963689

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