We solve for t:
Take logs of both sides:
Use a rule of the log of a product:
Subtract from both sides:
Use a rule of the log of a power:
Divide both sides by
In your problem,
P = 3000
A = 6000
r = 6% = .06
n = 1 (per annum)
The 11th year the amount would be
[Interest-payers always round down to the lower penny,
never up to the higher penny]
The 12th year it would be
So the 12th year would be the first time it would
have been at least $6000.
Edwin