Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 456: 26

$ \approx 0.3503$

Apply logarithmic property : $\log m-\log n=\log (\dfrac{m}{n})$ Re-write as: $\log 643-\log 287=\log (\dfrac{687}{243})$ Now, take logarithmic to the base $10$ and evaluate the result. Therefore, our answer is: $\log_{10} (\dfrac{687}{243}) \approx 0.3503$