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.
\begin{enumerate}[label=(\alph*)]
\item 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.
\item 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.
\begin{enumerate}[label=(\roman*)]
\item 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.
\item 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)
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C4 2012 Q4 [9]}}