## Algebra and Trigonometry 10th Edition

$36$
From the previous part, we have the equation for a learning curve: $N= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$ Plug in $N=25$ to find $t$. Therefore, $25= 30 (1-e^{\frac{t}{20} \ln \frac{11}{30}})$ Simplify: $\dfrac{5}{6}=1- e^{\frac{t}{20} \ln \frac{11}{30}}$ or, $\ln \dfrac{1}{6}=\ln [1- e^{\frac{t}{20} \ln \frac{11}{30}}]$ or, $\dfrac{1}{6}=\dfrac{t}{20} \ln \dfrac{11}{30}$ This gives: $t \approx 36$