## Calculus: Early Transcendentals 8th Edition

$b = ln(3e^a-2)$
$A = \int_{0}^{a}e^x = e^a-e^0 = e^a-1$ $B = \int_{0}^{b}e^x = e^b-e^0 = e^b-1$ It is given that $B = 3A$: $B = 3A$ $e^b-1 = 3(e^a-1)$ $e^b-1 = 3e^a-3$ $e^b = 3e^a-2$ $b = ln(3e^a-2)$