Posted by Wally on Sunday, February 19, 2012 at 5:04pm.
I will assume you meant
P(t) = 100/(1+48.2e^(-.52t) )
You did not define P(t), but I will assume it will be the
"efficiency" mastered after t months
so .....
.8 = 100/(1 + 48.2e^(.52t))
.8 + 38.56e^(.52t) = 100
e^(.52t) = (100-.8)/38.56 = 2.572614
taking ln of both sides, and knowing that ln e = 1
.52t = ln (2.572614)
t = 1.817 or appr 1.8 months
check:
let t=1.817
P(1.1817) = 100/(1 + 48.2e^(.52(1.817)
= 0.8
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