Answer: 23 years
=======================================================
Work Shown:
P = 1000 is the amount deposited
We want this value to double to A = 2000 which is the amount in the account at time t (in years).
r = 0.03 represents the interest rate in decimal form.
The value of t is unknown but we can solve for it like so
[tex]A = Pe^{rt}\\\\2000 = 1000e^{0.03t}\\\\2 = e^{0.03t}\\\\\ln(2) = \ln\left(e^{0.03t}\right)\\\\\ln(2) = 0.03t\ln\left(e\right)\\\\\ln(2) = 0.03t*1\\\\\ln(2) = 0.03t\\\\[/tex]
[tex]0.03t = \ln(2)\\\\t = \frac{\ln(2)}{0.03}\\\\t \approx 23.1049060186649\\\\t \approx 23\\\\[/tex]
It will take about 23 years for the amount to double.
As a check,
[tex]A = 1000e^{0.03t}\\\\A \approx 1000e^{0.03*23.104906}\\\\A \approx 1,999.9999988806\\\\A \approx 2000\\\\[/tex]
which helps show that after roughly 23 years, we'll have about 2000 dollars in the account.
------------
Side note: Use of the rule of 72 leads to 72/3 = 24. The '3' is from the 3% interest rate. So the rule of 72 says it will take about 24 years for the amount to double. This isn't too far off from the 23 answer we got.
