## Algebra and Trigonometry 10th Edition

$N= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$
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 $N= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$