find the approximate values of 10.02, given that 10 = 2.3026

Question

Philoid · Accepted Answer

Let

Let x =10 and Δx = 0.02.

As

⇒

We, know

⇒

∴

⇒

∴ Δy = 0.002

Also,

Δy = f(x+Δx)-f(x)

∴ 0.002 = log_e(10+0.02) – log_e(10)

⇒ 0.002 = log_e(10.02) – 2.3026

⇒ log_e(10.02) = 2.3046.