## College Algebra (6th Edition)

$120$ decibels
Substitute $I$ with $10^{12}I_{0}$ in the given formula: $D=10\displaystyle \log\frac{10^{12}I_{0}}{I_{0}}=10\log 10^{12}$ $...$Basic Logarithmic Properties: $\log_{b}b^{x}=x$... $D=10\cdot 12=120$ decibels