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