Question

Using differentials, find the approximate value of: log_e 10.02, given that log_e 10 = 2.3026

Answer 1

Consider y = log_e x

Here x = 10 and Δ x = 0.02

By differentiating w.r.t. x

So we get

On further calculation

Δ y = 0.02/10 = 0.002

We know that

Δ y = f(x + Δ x) – f(x)

By substituting the values

0.002 = log_e (10 + 0.02) – log_e 10

It can be written as

0.002 = log_e 10.02 – 2.3026

We get

log_e 10.02 = 2.3046