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

$2.745$
Using the properties of logarithms, the given expression, $\log_b 45 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_b (3^2\cdot5) \\\\= \log_b 3^2+\log_b5 \\\\= 2\log_b 3+\log_b5 .\end{array} Since it is given that $\log_b 3=0.792$ and $\log_b5=1.161$, the expression above, $2\log_b 3+\log_b5 ,$ evaluates to \begin{array}{l}\require{cancel} 2(0.792)+1.161 \\\\= 2.745 .\end{array}