## College Algebra (11th Edition)

$\approx1.7993$
Using $\log_b a=\dfrac{\log a}{\log b}$ or the Change-of-Base Formula and the laws of logarithms, the given expression, $\log 63 ,$ evaluates to \begin{array}{l}\require{cancel} \dfrac{\log 63}{\log 10} \\\\\approx \dfrac{1.7993}{1} \\\\ \approx1.7993 .\end{array}