The current, 1 amps, in an electric circuit at time t seconds is given by I=16-16(0.5)^t, t≥0 What is the value of dI/dt when t = 3?

Solution: $(\mathrm{a}) ; \frac{\mathrm{dI}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[16-16(0.5)^{\mathrm{t}}\right]=-16(0.5)^{\mathrm{t}} \times \ln (0.5)$ at $\mathrm{t}=3, \frac{\mathrm{dI}}{\mathrm{dt}}=-16(0.5)^{3} \times \ln (0.5)=-2 \ln (0.5)=2 \ln 2=\ln 4$