Question

help me please!! i need help with exponential growth? whos willing to help me and understand? (attachment below)

Answer 1

\[P(0)=3000=A\]So,\[P=Ae ^{kt}=3000e ^{kt}\]We know that P=3600 at t=10 so,\[3600=3000e ^{10k}\]Now we solve for k,\[1.2=e ^{10k}\]Taking the natural log of both sides: \[\ln 1.2=\ln e ^{10k}=10k\]Thus,\[k=\ln 1.2/10\approx0.0182\]

Answer 2

then you plug in 0.0182 into k right?

Answer 3

The question asked you to solve for k, so the anseer is 0.0182. The equation becomes:\[P=3000e ^{0.0182t}\]This now allows you to solve for the population of bacteria at any other time, t.