Answer
$N= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$
Work Step by Step
We need to write the equation for a learning curve.
$N= 30 (1-e^{kt})$
Set $(t,N)= (20,19)$ to compute $a$
$19= 30 (1-e^{20k})$
or, $e^{20k}= \dfrac{11}{30}$
Take the $\log$ on each side.
$\ln e^{20 k} =\ln \dfrac{11}{30}$
Simplify: $k=\dfrac{1}{20} \ln \dfrac{11}{30}$
So, we have $N= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$
