The population of bacteria, PP, after tt hours can be modelled by P=P0(2t)P = P_0(2^t), where P0P_0 is a constant.

  1. (i)

    Find the least number of hours (to the nearest hour) it will take for the population to reach 100 times its original size.

  2. (ii)

    Sketch the graph of PP against tt.

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Worked solution

4(i)

1. Set up the equation for the population reaching 100 times its original size

Let the original population be P0P_0.
We want to find tt when P=100P0P = 100P_0.
100P0=P0(2t)100P_0 = P_0(2^t)
2t=1002^t = 100

2. Solve for $t$ using logarithms and round to the nearest hour

Take logarithms on both sides:
log⁡(2t)=log⁡(100)\log(2^t) = \log(100)
tlog⁡2=2t \log 2 = 2
t=2log⁡2≈6.6439t = \frac{2}{\log 2} \approx 6.6439
Since we need the least number of hours to the nearest hour for the population to reach at least 100 times, we round up to 7 hours.

Answer

7

4(ii)

1. Identify the shape and domain of the graph

The function P=P0(2t)P = P_0(2^t) represents exponential growth since the base 2>12 > 1. Since tt represents time, the domain is t≥0t \ge 0.

2. Determine the $P$-intercept and sketch the curve

When t=0t = 0, P=P0(20)=P0P = P_0(2^0) = P_0. The graph starts at (0,P0)(0, P_0) on the PP-axis and curves upwards with an increasing gradient.

Answer

A curve in the first quadrant starting at (0,P0)(0, P_0) and increasing exponentially as tt increases.

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