## College Algebra (11th Edition)

$\log_{9} 13\approx1.1674$
By using the change of base formula, we can rewrite the given logarithm as: $\log_{9} 13=\frac{\log_{10} 13}{\log_{10} 9}$ This can be easily calculated: $\frac{\log_{10} 13}{\log_{10} 9}\approx1.1674$