University Calculus: Early Transcendentals (3rd Edition)

$\approx 92.1 s$
Given: When the voltage is 10% of its original value, that is, $V =0.1 V_0$ Now, we have $V(t)=V_0e^{-t/40}$ ...(1) This implies that $0.1 V_0=V_0e^{-t/40} \implies t=-40 (\ln (0.1))$ This implies that $t \approx 92.1 s$ Hence, $t \approx 92.1 s$