Question 1205986
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{{{A = P*e^(r*t)}}}


{{{20000 = 6000*e^(0.05*t)}}}


{{{20000/6000 = e^(0.05*t)}}}


{{{20/6 = e^(0.05*t)}}}


{{{10/3 = e^(0.05*t)}}}


{{{0.05t = ln(10/3)}}}


{{{0.05t = 1.203973}}} approximately


{{{t = 1.203973/0.05}}} 


{{{t = 24.07946}}} approximately


It takes <font color=red>about 24.079 years</font> depending how you round the final answer.


Check:
{{{A = P*e^(r*t)}}}


{{{A = 6000*e^(0.05*24.07946)}}}


{{{A = 20000.00391348}}} approximately
This rounds to 20,000 when rounding to the nearest cent.
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