Question 1153619
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Plug in P(t) = 900 and solve for t
P(t) = 2000/(1 + 9*e^(-0.6*t))
900 = 2000/(1 + 9*e^(-0.6*t))
900(1 + 9*e^(-0.6*t)) = 2000
900*1 + 900*9*e^(-0.6*t) = 2000
900 + 8100*e^(-0.6*t) = 2000
8100*e^(-0.6*t) = 2000-900
8100*e^(-0.6*t) = 1100
e^(-0.6*t) = (1100)/(8100)
e^(-0.6*t) = 0.135802469135802
Ln( e^(-0.6*t) ) = Ln( 0.135802469135802 )
-0.6*t*Ln( e ) = Ln( 0.135802469135802 )
-0.6*t*1= -1.99655388187407
-0.6*t= -1.99655388187407
t= (-1.99655388187407)/(-0.6)
t= 3.32758980312345



If we plug t= 3.32758980312345 into the function, we get
P(t) = 2000/(1+9*e^(-0.6*t))
P(3.32758980312345) = 2000/(1+9*e^(-0.6*3.32758980312345))
P(3.32758980312345) = 2000/(1+9*e^(-1.99655388187407))
P(3.32758980312345) = 2000/(1+9*0.135802469135802)
P(3.32758980312345) = 2000/(1+1.22222222222222)
P(3.32758980312345) = 2000/(2.22222222222222)
P(3.32758980312345) = 900.000000000001
which is really really close to 900. The error is due to rounding


When rounding 3.32758980312345 to the nearest tenth, you should get 3.3


So the answer is <font color=red>approximately 3.3 years</font> 
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