## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$b-a$
With $1.5=\frac{3}{2}$, then $\ln(1.5) = \ln \left(\dfrac{3}{2} \right)$ Recall that $\ln\left(\dfrac{M}{N}\right) = \ln M-\ln N$. Using the rule above gives: $\ln \left(\dfrac{3}{2} \right) = \ln3-\ln2$ With $\ln3 = b \hspace{20pt} \text{and} \hspace{20pt} \ln2=a$, then $\ln3-\ln2 = b-a$ Therefore, $\ln1.5 = \boxed{b-a}$