Answer
$s(t)=60$ $then$ $t\approx0.549$
$s(t)=90$ $then$ $t\approx1.099$
Work Step by Step
Solving the equation $s(t)=y$, then we have:
$s(t)=y⇔\frac{120}{1+3e^{-2t}}=y⇔\frac{1+3e^{-2t}}{120}=\frac{1}{y}⇔$
$⇔1+3e^{-2t}=\frac{120}{y}⇔3e^{-2t}=\frac{120}{y}-1⇔e^{-2t}=((\frac{120}{y}-1):3)⇔$
$⇔-2t=\ln((\frac{120}{y}-1):3)⇔t=\frac{\ln((\frac{120}{y}-1):3)}{(-2)}$
For $y=60$ $then$ $t\approx0.549$ and for $y=90$ $then$ $t\approx1.099$