Answer
$$\varepsilon_{t}=\ln \frac{L}{L_{0}}=\ln \frac{L_{0}+\delta}{L_{0}}=\ln \left(1+\frac{\delta}{L_{0}}\right)=\ln (1+\varepsilon)$$
Thus,
$$\varepsilon_{t}=\ln (1+\varepsilon)$$
Work Step by Step
$$\varepsilon_{t}=\ln \frac{L}{L_{0}}=\ln \frac{L_{0}+\delta}{L_{0}}=\ln \left(1+\frac{\delta}{L_{0}}\right)=\ln (1+\varepsilon)$$
Thus,
$$\varepsilon_{t}=\ln (1+\varepsilon)$$
