# My textbook sets $\ln \left\{ 1 + \left[ \frac{L}{L_0} - 1 \right] \right\}$ equal to $\frac{L}{L_0} - 1$. What's the justification for this?

Sat to study physics. I started to study this new chapter Heat and Thermodynamics. It included these steps: $$\int_{L_0}^L \frac{dL}{L} = \int_{\theta_0}^\theta \alpha d \theta; \text{ or } \ln \left[ \frac{L}{L_0} \right] = \alpha \Delta \theta$$ (where $$\Delta \theta = \theta - \theta_0$$) $$\text{or } \ln \left\{ 1 + \left[ \frac{L}{L_0} - 1 \right] \right\} = \alpha \Delta \theta \qquad \text{or} \qquad \frac{L}{L_0} - 1 = \alpha \Delta \theta$$

This deduction they made is intriguing me. What do they mean by this? Does natural log have such a property?

• $\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ so if $x \ll 1$ we get the approximation $\ln(1+x) \approx x$ – John Rennie Nov 12 '18 at 16:35
• Alternatively, if $L$ is close to $L_0$, then $\displaystyle \int_{L_0}^L dL = \int_{\theta_0}^\theta L \alpha d\theta \approx \int_{\theta_0}^\theta L_0 \alpha d\theta,$ and so $L - L_0 \approx L_0 \alpha \Delta \theta$ without introducing logs at all. – alephzero Nov 12 '18 at 17:27

The equation is using the approximation $$\ln(1 + x) \approx x$$, which is the first term of the Taylor series of $$f(x) = \ln (1 + x)$$: $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots.$$ If $$x$$ is small much less than 1, then all of the terms except the first will be negligible.