5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f0a633e3-5c63-4d21-8ffa-d4e7dc43a536-14_549_958_221_493}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 2 shows a sketch of part of the curve \(C\) with equation
$$y = 2 \ln ( 2 x + 5 ) - \frac { 3 x } { 2 } , \quad x > - 2.5$$
The point \(P\) with \(x\) coordinate - 2 lies on \(C\).
- Find an equation of the normal to \(C\) at \(P\). Write your answer in the form \(a x + b y = c\), where \(a\), \(b\) and \(c\) are integers.
The normal to \(C\) at \(P\) cuts the curve again at the point \(Q\), as shown in Figure 2
- Show that the \(x\) coordinate of \(Q\) is a solution of the equation
$$x = \frac { 20 } { 11 } \ln ( 2 x + 5 ) - 2$$
The iteration formula
$$x _ { n + 1 } = \frac { 20 } { 11 } \ln \left( 2 x _ { n } + 5 \right) - 2$$
can be used to find an approximation for the \(x\) coordinate of \(Q\).
- Taking \(x _ { 1 } = 2\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.