Precalculus (10th Edition)

$42$
We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$. Hence, $5^{\log_5{6}+\log_5 {7}}=5^{\log_5 {(6\cdot 7)}}=5^{\log_5{42}}$. We also know that $a^{\log_a {x}}=x$. Thus, $5^{\log_5 {42}}=42$