## University Calculus: Early Transcendentals (3rd Edition)

$\approx 59.8$ft
Given: $L=\dfrac{L_0}{2}$ and $k=18$ Now, we have $L=L_0e^{-kx}$ ...(1) Then $\dfrac{L_0}{2}=L_0e^{-18x} \implies k=\dfrac{\ln (2)}{18} \approx 0.0385$ Now, when after another t = 14 hours, that is, $t=14+10 =24$ Then equation (1) becomes $\dfrac{L_0}{10}=L_0e^{-(0.0385)x}$ This implies that $\ln (10) =0.0385 x$ Hence, $x \approx 59.8$ft