## Intermediate Algebra: Connecting Concepts through Application

a) $25,600$ bacteria b) $3$ hours
a) $B(h) = 100(2^{h})$ $B(8) = 100(2^{8})$ $= 100(256)$ $= 25,600$ bacteria b) $800 = 100(2^{h})$ $\frac{800}{100} = (2^{h})$ $8 = 2^{h}$ $2^{3} = 2^{h}$ $h = 3$ hours