In a bank, the principal increases continuously at the rate of 8% per year, that is, the rate of increase of the principal at any time is 8% of the principal at that time. The time in which a principal doubles itself is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{25}{2}\log_e 2\) years
Let P be the principal at time t (in years) and \(P_0\) the principal at t = 0. The given rate is a differential equation:
\( \displaystyle \frac{dP}{dt} = \frac{8}{100}P = \frac{2}{25}P \)
Separating the variables:
\( \displaystyle \frac{dP}{P} = \frac{2}{25}\,dt \)
\( \displaystyle \log_e P = \frac{2t}{25} + C \)
At t = 0, \(P = P_0\), so \(C = \log_e P_0\):
\( \displaystyle \log_e \frac{P}{P_0} = \frac{2t}{25} \)
The principal has doubled when \(\frac{P}{P_0} = 2\):
\( \displaystyle \log_e 2 = \frac{2t}{25} \)
\( \displaystyle t = \frac{25}{2}\log_e 2 \)
Why the other options are incorrect
Option (1) \(8\log_e 2\) years — Incorrect. This multiplies by the rate 8 instead of dividing by the rate \(\frac{8}{100}\).
Option (2) \(25\log_e 2\) years — Incorrect. This uses the rate \(\frac{1}{25} = 4\%\); at 8% the rate is \(\frac{2}{25}\), which halves the time.
Option (4) \(\frac{2}{25}\log_e 2\) years — Incorrect. This multiplies \(\log_e 2\) by the rate; the time is \(\log_e 2\) divided by the rate.