logarithmic and exponential applications

Question

For a certain strain of bacteria, k=0.783 when t is measured in hours. how long will it take 10 bacteria to increase to100? Please can you in depth explain it.

Answer 1

Let A(t) = number of bacteria after t hours

 

If we assume exponential growth where A(t) = 10e^0.783t, then we are looking for the value of t such that:

 

100 = 10e^0.783t

 

10 = e^0.783t

 

ln10  = ln(e^0.783t)

 

ln10 = 0.783t

 

t = (ln10)/(0.783) = 2.94 hours

Answer 2

f(t) = ae^kt, where a = starting amount, e = Euler's constant, k = rate, t = time

 

100 = 10e^0.783t                   Divide both sides by 10

10 = e^0.783t                          Write as the natural logarithm

ln 10 = ln e^0.783t             Use the law of exponents in logarithms

ln 10 = 0.783t ln e           Determine the ln

2.30 ≈ 0.783t                  Divide both sides by 0.783

2.94 ≈ t

 

It take just under three hours.