If the interest is compounded continuously at 6% per annum, how much worth ₹ 1000 will be after ten years? How long will it take to double ₹ 1000?

[Given e^0.6 = 1.822]

Let the principal, rate and time be Rs P, r and t years.

Also, let the initial principal be P_o.

⇒

Integrating both sides, we have

⇒ ∫∫dt

⇒ log|P| = t + c ……(1)

Now, at t = 0, P = P_o

log| P_o | = 0 + c

⇒ c = log| P_o |……(2)

Putting the value of c in equation (1) we have,

log|P| = t + log|P_o|

⇒ log|P| – log|P_o| = t

⇒ (log |P| – log|P_o|) = t []

⇒ log ( = t ……(3)

Now, P_o = 1000, t = 10years, r = 6

∴ log ( = ×10

⇒ log ( = 0.6

⇒

⇒ P = ×1000

⇒ P = 1.822×1000 (Given: = 1.822)

⇒ P = 1822

Rs 1000 will be Rs 1822 after 10 years at 6% rate.