Solve exponential equation via iteration

4

1. \includegraphics[max width=\textwidth, alt={}, center]{4ce3208e-8ceb-4848-a9c7-fcda166319f4-05_753_944_278_630} The diagram shows the graph of \(y = 3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x }\).
   On the diagram, sketch the graph of \(y = | 5 x - 4 |\), and show that the equation \(3 - e ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) has exactly two real roots. It is given that the two roots of \(3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) are denoted by \(\alpha\) and \(\beta\), where \(\alpha < \beta\).
2. Show by calculation that \(\alpha\) lies between 0.36 and 0.37 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 7 - \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } } \right)\) to find \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4

1. \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-05_753_944_278_630} The diagram shows the graph of \(y = 3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x }\).
   On the diagram, sketch the graph of \(y = | 5 x - 4 |\), and show that the equation \(3 - e ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) has exactly two real roots. It is given that the two roots of \(3 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } = | 5 x - 4 |\) are denoted by \(\alpha\) and \(\beta\), where \(\alpha < \beta\).
2. Show by calculation that \(\alpha\) lies between 0.36 and 0.37 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 7 - \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } } \right)\) to find \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4 Using the substitution \(u = 3 ^ { x }\), or otherwise, solve, correct to 3 significant figures, the equation $$3 ^ { x } = 2 + 3 ^ { - x }$$

2 Solve, correct to 3 significant figures, the equation $$\mathrm { e } ^ { x } + \mathrm { e } ^ { 2 x } = \mathrm { e } ^ { 3 x }$$

2 Solve the equation $$\ln ( 1 + x ) = 1 + \ln x$$ giving your answer correct to 2 significant figures.

2 Solve the equation $$\ln \left( 1 + x ^ { 2 } \right) = 1 + 2 \ln x$$ giving your answer correct to 3 significant figures.

2 Find the real root of the equation \(\frac { 2 \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } { 2 + \mathrm { e } ^ { x } } = 3\), giving your answer correct to 3 decimal places. Your working should show clearly that the equation has only one real root.

1 Solve the equation \(\ln \left( \mathrm { e } ^ { 2 x } + 3 \right) = 2 x + \ln 3\), giving your answer correct to 3 decimal places.

5

1. By sketching a suitable pair of graphs, show that the equation \(2 + \mathrm { e } ^ { - 0.2 x } = \ln ( 1 + x )\) has only one root.
2. Show by calculation that this root lies between 7 and 9 . \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-08_2716_40_109_2009}
3. Use the iterative formula $$x _ { n + 1 } = \exp \left( 2 + \mathrm { e } ^ { - 0.2 x _ { n } } \right) - 1$$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. \(\left[ \exp ( x ) \right.\) is an alternative notation for \(\left. \mathrm { e } ^ { x } .\right]\) \includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-10_481_789_262_639} The diagram shows the curve \(y = \sin 2 x ( 1 + \sin 2 x )\), for \(0 \leqslant x \leqslant \frac { 3 } { 4 } \pi\), and its minimum point \(M\). The shaded region bounded by the curve that lies above the \(x\)-axis and the \(x\)-axis itself is denoted by \(R\).

3. $$f ( x ) = 2 ^ { x - 1 } - 4 + 1.5 x \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \frac { 1 } { 3 } \left( 8 - 2 ^ { x } \right)$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\), where \(\alpha = 1.6\) to one decimal place.
2. Starting with \(x _ { 0 } = 1.6\), use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 8 - 2 ^ { x _ { n } } \right)$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. By choosing a suitable interval, prove that \(\alpha = 1.633\) to 3 decimal places.

3. $$\mathrm { f } ( x ) = \ln ( x + 2 ) - x + 1 , \quad x > - 2 , x \in \mathbb { R } .$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \(2 < x < 3\).
2. Use the iterative formula $$x _ { n + 1 } = \ln \left( x _ { n } + 2 \right) + 1 , x _ { 0 } = 2.5$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) giving your answers to 5 decimal places.
3. Show that \(x = 2.505\) is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

7. $$f ( x ) = 3 x e ^ { x } - 1$$ The curve with equation \(y = \mathrm { f } ( x )\) has a turning point \(P\).

1. Find the exact coordinates of \(P\). The equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 0.25\) and \(x = 0.3\)
2. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 3 } \mathrm { e } ^ { - x _ { n } }$$ with \(x _ { 0 } = 0.25\) to find, to 4 decimal places, the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\).
3. By choosing a suitable interval, show that a root of \(\mathrm { f } ( x ) = 0\) is \(x = 0.2576\) correct to 4 decimal places.

2. $$\mathrm { g } ( x ) = \mathrm { e } ^ { x - 1 } + x - 6$$

1. Show that the equation \(\mathrm { g } ( x ) = 0\) can be written as $$x = \ln ( 6 - x ) + 1 , \quad x < 6$$ The root of \(\mathrm { g } ( x ) = 0\) is \(\alpha\).
   The iterative formula $$x _ { n + 1 } = \ln \left( 6 - x _ { n } \right) + 1 , \quad x _ { 0 } = 2$$ is used to find an approximate value for \(\alpha\).
2. Calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 4 decimal places.
3. By choosing a suitable interval, show that \(\alpha = 2.307\) correct to 3 decimal places.

7. (a) Sketch the curve with equation \(y = \ln x\).
(b) Show that the tangent to the curve with equation \(y = \ln x\) at the point ( \(\mathrm { e } , 1\) ) passes through the origin.
(c) Use your sketch to explain why the line \(y = m x\) cuts the curve \(y = \ln x\) between \(x = 1\) and \(x = \mathrm { e }\) if \(0 < m < \frac { 1 } { \mathrm { e } }\). Taking \(x _ { 0 } = 1.86\) and using the iteration \(x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }\),
(d) calculate \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }\) and \(x _ { 5 }\), giving your answer to \(x _ { 5 }\) to 3 decimal places. The root of \(\ln x - \frac { 1 } { 3 } x = 0\) is \(\alpha\).
(e) By considering the change of sign of \(\ln x - \frac { 1 } { 3 } x\) over a suitable interval, show that your answer for \(x _ { 5 }\) is an accurate estimate of \(\alpha\), correct to 3 decimal places.

6 Find the solution to this equation, correct to 3 significant figures. $$\left( 2 ^ { x } \right) \left( 2 ^ { x + 1 } \right) = 10 .$$

6. \(\quad f ( x ) = 2 x ^ { 2 } + 3 \ln ( 2 - x ) , \quad x \in \mathbb { R } , \quad x < 2\).

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = 2 - \mathrm { e } ^ { k x ^ { 2 } }$$ where \(k\) is a constant to be found. The root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\) is 1.9 correct to 1 decimal place.
2. Use the iterative formula $$x _ { n + 1 } = 2 - \mathrm { e } ^ { k x _ { n } ^ { 2 } }$$ with \(x _ { 0 } = 1.9\) and your value of \(k\), to find \(\alpha\) correct to 3 decimal places.
   You should show the result of each iteration.
3. Solve the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\).

4 You are given the equation \(( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0\).

1. Verify that 0 is a root of the equation. There are also two other roots, \(\alpha\) and \(\beta\), where \(0 < \alpha < \beta\).
2. The iterative formula \(x _ { r + 1 } = \ln \left( 2 x _ { r } - 1 \right) ^ { 2 }\) is to be used to find a root of the equation.
   1. Sketch the line \(y = x\) and the curve \(y = \ln ( 2 x - 1 ) ^ { 2 }\) on the same axes, showing the roots \(0 , \alpha\) and \(\beta\).
   2. By drawing a 'staircase' diagram on your sketch, starting with a value of \(x\) that is between \(\alpha\) and \(\beta\), show that this iteration does not converge to \(\alpha\).
   3. Using this iterative formula with \(x _ { 1 } = 3.75\), find the value of \(\beta\) correct to 3 decimal places.
   4. Using the Newton-Raphson method with \(x _ { 1 } = 1.6\), find the root \(\alpha\) of the equation \(( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0\) correct to 5 significant figures. Show the result of each iteration.

1

1. Use Simpson's rule with 7 ordinates (6 strips) to find an estimate for \(\int _ { 0 } ^ { 3 } 4 ^ { x } \mathrm {~d} x\).
2. A curve is defined by the equation \(y = 4 ^ { x }\). The curve intersects the line \(y = 8 - 2 x\) at a single point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1.2 and 1.3.
   2. The equation \(4 ^ { x } = 8 - 2 x\) can be rearranged into the form \(x = \frac { \ln ( 8 - 2 x ) } { \ln 4 }\). Use the iterative formula \(x _ { n + 1 } = \frac { \ln \left( 8 - 2 x _ { n } \right) } { \ln 4 }\) with \(x _ { 1 } = 1.2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
      (2 marks)

1 The curve \(y = 3 ^ { x }\) intersects the curve \(y = 10 - x ^ { 3 }\) at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 1 and 2 .
2. 1. Show that the equation \(3 ^ { x } = 10 - x ^ { 3 }\) can be rearranged into the form $$x = \sqrt [ 3 ] { 10 - 3 ^ { x } }$$
   2. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { 10 - 3 ^ { x _ { n } } }\) with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.

2 For \(0 < x \leqslant 2\), the curves with equations \(y = 4 \ln x\) and \(y = \sqrt { x }\) intersect at a single point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.5.
2. Show that the equation \(4 \ln x = \sqrt { x }\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \sqrt { x _ { n } } } { 4 } \right) }$$ with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. Figure 1, on the page 3, shows a sketch of parts of the graphs of \(y = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d3c66c34-b09c-4223-8383-cf0a68419bf9-3_1285_1543_356_296}
   \end{figure}

3

1. The equation \(\mathrm { e } ^ { - x } - 2 + \sqrt { x } = 0\) has a single root, \(\alpha\).
   Show that \(\alpha\) lies between 3 and 4 .
2. Use the recurrence relation \(x _ { n + 1 } = \left( 2 - e ^ { - x _ { n } } \right) ^ { 2 }\), with \(x _ { 1 } = 3.5\), to find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
3. The diagram below shows parts of the graphs of \(y = \left( 2 - \mathrm { e } ^ { - x } \right) ^ { 2 }\) and \(y = x\), and a position of \(x _ { 1 }\). On the diagram, draw a staircase or cobweb diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis. \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-03_1100_1402_881_367}

2 A curve has equation \(y = 2 \ln ( 2 \mathrm { e } - x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find an equation of the normal to the curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) at the point on the curve where \(x = \mathrm { e }\).
   [0pt] [4 marks]
3. The curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) intersects the line \(y = x\) at a single point, where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1 and 3 .
   2. Use the recurrence relation $$x _ { n + 1 } = 2 \ln \left( 2 \mathrm { e } - x _ { n } \right)$$ with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   3. Figure 1, on the opposite page, shows a sketch of parts of the graphs of \(y = 2 \ln ( 2 \mathrm { e } - x )\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
      [0pt] [2 marks] \section*{(c)(iii)} \begin{figure}[h]
      \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-05_864_1284_1802_386}
      \end{figure}

2 The curve with equation \(y = x ^ { x }\), where \(x > 0\), intersects the line \(y = 5\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2 and 3 .
2. Show that the equation \(x ^ { x } = 5\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \ln 5 } { x } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \ln 5 } { x _ { n } } \right) }$$ with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. 1. Use Simpson's rule with 7 ordinates ( 6 strips) to find an approximation to $$\int _ { 0.5 } ^ { 1.7 } \left( 5 - x ^ { x } \right) \mathrm { d } x$$ giving your answer to three significant figures.
   2. Hence find an approximation to \(\int _ { 0.5 } ^ { 1.7 } x ^ { x } \mathrm {~d} x\).

4 [Figure 1, printed on the insert, is provided for use in this question.]

1. Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to \(\int _ { 1 } ^ { 2 } 3 ^ { x } \mathrm {~d} x\), giving your answer to three significant figures.
2. The curve \(y = 3 ^ { x }\) intersects the line \(y = x + 3\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0.5 and 1.5.
   2. Show that the equation \(3 ^ { x } = x + 3\) can be rearranged into the form $$x = \frac { \ln ( x + 3 ) } { \ln 3 }$$
   3. Use the iteration \(x _ { n + 1 } = \frac { \ln \left( x _ { n } + 3 \right) } { \ln 3 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\) to two significant figures.
   4. The sketch on Figure 1 shows part of the graphs of \(y = \frac { \ln ( x + 3 ) } { \ln 3 }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.

The root of the equation \(f(x) = 0\), where $$f(x) = x + \ln 2x - 4$$ is to be estimated using the iterative formula \(x_{n+1} = 4 - \ln 2x_n\), with \(x_0 = 2.4\).

1. Showing your values of \(x_1, x_2, x_3, \ldots\), obtain the value, to 3 decimal places, of the root. [4]
2. By considering the change of sign of \(f(x)\) in a suitable interval, justify the accuracy of your answer to part (a). [2]