4 A scientist is testing models for the growth and decay of colonies of bacteria. For a particular colony, which is growing, the model is \(P = A \mathrm { e } ^ { \frac { 1 } { 8 } t }\), where \(P\) is the number of bacteria after a time \(t\) minutes and \(A\) is a constant.

1. This growing colony consists initially of 500 bacteria. Calculate the number of bacteria, according to the model, after one hour. Give your answer to the nearest thousand.
2. For a second colony, which is decaying, the model is \(Q = 500000 \mathrm { e } ^ { - \frac { 1 } { 8 } t }\), where \(Q\) is the number of bacteria after a time \(t\) minutes. Initially, the growing colony has 500 bacteria and, at the same time, the decaying colony has 500000 bacteria.
   1. Find the time at which the populations of the two colonies will be equal, giving your answer to the nearest 0.1 of a minute.
   2. The population of the growing colony will exceed that of the decaying colony by 45000 bacteria at time \(T\) minutes. Show that $$\left( \mathrm { e } ^ { \frac { 1 } { 8 } T } \right) ^ { 2 } - 90 \mathrm { e } ^ { \frac { 1 } { 8 } T } - 1000 = 0$$ and hence find the value of \(T\), giving your answer to one decimal place.
      (4 marks)