lean_workbook_plus_48801

Letting $n$ be the original number of bacteria and $h$ be the number of hours it takes for the initial number of bacteria to decrease by $50\%$, we have $n \cdot (0.9)^h = \frac{1}{2}n$, so $(0.9)^h = \frac{1}{2}$. Taking the logarithm, we have $h = \log_{0.9}{0.5}$. By the Change of Base Formula, this is $\frac{\log{0.5}}{\log{0.9}}$. Plugging this into a calculator we get $\boxed{6.6}$ hours.