## Calculus (3rd Edition)

$$t=\frac{1}{4}\ln3$$
$6e^{-4t}=2$ Dividing both sides by 6: $e^{-4t}=\frac{2}{6}=\frac{1}{3}$ Taking the natural log of both sides: $-4t=\ln{\frac{1}{3}}=\ln({3^{-1})}=-\ln3$ Solving for $t$: $t=\frac{1}{4}\ln3$