## University Calculus: Early Transcendentals (3rd Edition)

$\ln (\dfrac{8}{9})$
We have $1+e^b+e^{2b}=\dfrac{1}{1-e^b}=9$ and $\dfrac{1}{9}=1-e^b$ or, $e^b=\dfrac{8}{9}$ Thus, $b=\ln (\dfrac{8}{9})$