Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 401: 27

$ \approx 59.8$ft

Consider the equation as: $L=L_0e^{-kx}$ As we are given that $L=\dfrac{L_0}{2}$ and $k=18$ Then, we have $\dfrac{L_0}{2}=L_0e^{-18x}$ or, $k=\dfrac{\ln (2)}{18} \approx 0.0385$ Now, we have $\dfrac{L_0}{10}=L_0e^{-(0.0385)x}$ or, $\ln (10) =0.0385 x \implies x \approx 59.8$ft