## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$1.953$
Using the properties of logarithms, the given expression, $\log_b 15 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_b (3\cdot5) \\\\= \log_b 3+\log_b 5 .\end{array} Since it is given that $\log_b 3=0.792$ and $\log_b5=1.161$, the expression above, $\log_b 3+\log_b 5 ,$ evaluates to \begin{array}{l}\require{cancel} 0.792+1.161 \\\\= 1.953 .\end{array}