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 e0.6 = 1.822]
Let the principal, rate and time be Rs P, r and t years.
Also, let the initial principal be Po.
⇒
Integrating both sides, we have
⇒ ∫∫dt
⇒ log|P| = t + c ……(1)
Now, at t = 0, P = Po
log| Po | = 0 + c
⇒ c = log| Po |……(2)
Putting the value of c in equation (1) we have,
log|P| = t + log|Po|
⇒ log|P| – log|Po| = t
⇒ (log |P| – log|Po|) = t [
]
⇒ log ( =
t ……(3)
Now, Po = 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.