Answer
(a) $t=-\frac{1}{k}ln(\frac{M-P(t)}{C})$
(b) $23.3$months
(c) see graph.
Work Step by Step
(a) Express the learning time t as a function of the performance level P.
Rewrite the function as $e^{-kt}=\frac{M-P(t)}{C}, -kt=ln(\frac{M-P(t)}{C})$
so we have $t=-\frac{1}{k}ln(\frac{M-P(t)}{C})$
(b) Given $P=12,k=0.024,M=20,C=14$, we have $t=-\frac{1}{0.024}ln(\frac{20-12}{14})=23.3$months
(c) see graph.
