1.06g Equations with exponentials: solve a^x = b

3 Use logarithms to solve the equation \(7 ^ { x } = 2 ^ { x + 1 }\), giving the value of \(x\) correct to 3 significant figures.

8

1. Use logarithms to solve the equation \(5 ^ { 3 w - 1 } = 4 ^ { 250 }\), giving the value of \(w\) correct to 3 significant figures.
2. Given that \(\log _ { x } ( 5 y + 1 ) - \log _ { x } 3 = 4\), express \(y\) in terms of \(x\).

8 \includegraphics[max width=\textwidth, alt={}, center]{4d03f4e3-ae6c-4a0d-ae4d-d89258d2919a-4_417_931_255_607} The diagram shows the curve \(y = 2 ^ { x } - 3\).

1. Describe the geometrical transformation that transforms the curve \(y = 2 ^ { x }\) to the curve \(y = 2 ^ { x } - 3\).
2. State the \(y\)-coordinate of the point where the curve \(y = 2 ^ { x } - 3\) crosses the \(y\)-axis.
3. Find the \(x\)-coordinate of the point where the curve \(y = 2 ^ { x } - 3\) crosses the \(x\)-axis, giving your answer in the form \(\log _ { a } b\).
4. The curve \(y = 2 ^ { x } - 3\) passes through the point ( \(p , 62\) ). Use logarithms to find the value of \(p\), correct to 3 significant figures.
5. Use the trapezium rule, with 2 strips each of width 0.5 , to find an estimate for \(\int _ { 3 } ^ { 4 } \left( 2 ^ { x } - 3 \right) \mathrm { d } x\). Give your answer correct to 3 significant figures.

8 \includegraphics[max width=\textwidth, alt={}, center]{b2c1188d-881e-4fb5-bece-5a51006543c7-4_524_822_274_609} The diagram shows the curves \(y = a ^ { x }\) and \(y = 4 b ^ { x }\).

1. (a) State the coordinates of the point of intersection of \(y = a ^ { x }\) with the \(y\)-axis.
   (b) State the coordinates of the point of intersection of \(y = 4 b ^ { x }\) with the \(y\)-axis.
   (c) State a possible value for \(a\) and a possible value for \(b\).
2. It is now given that \(a b = 2\). Show that the \(x\)-coordinate of the point of intersection of \(y = a ^ { x }\) and \(y = 4 b ^ { x }\) can be written as $$x = \frac { 2 } { 2 \log _ { 2 } a - 1 } .$$

5 Solve the equation \(2 ^ { 4 x - 1 } = 3 ^ { 5 - 2 x }\), giving your answer in the form \(x = \frac { \log _ { 10 } a } { \log _ { 10 } b }\).

8

1. Use logarithms to solve the equation $$2 ^ { n - 3 } = 18000$$ giving your answer correct to 3 significant figures.
2. Solve the simultaneous equations $$\log _ { 2 } x + \log _ { 2 } y = 8 , \quad \log _ { 2 } \left( \frac { x ^ { 2 } } { y } \right) = 7$$

4

1. Express \(2 \log _ { 3 } x - \log _ { 3 } ( x + 4 )\) as a single logarithm.
2. Hence solve the equation \(2 \log _ { 3 } x - \log _ { 3 } ( x + 4 ) = 2\).

8

1. The curve \(y = 3 ^ { x }\) can be transformed to the curve \(y = 3 ^ { x - 2 }\) by a translation. Give details of the translation.
2. Alternatively, the curve \(y = 3 ^ { x }\) can be transformed to the curve \(y = 3 ^ { x - 2 }\) by a stretch. Give details of the stretch.
3. Sketch the curve \(y = 3 ^ { x - 2 }\), stating the coordinates of any points of intersection with the axes.
4. The point \(P\) on the curve \(y = 3 ^ { x - 2 }\) has \(y\)-coordinate equal to 180 . Use logarithms to find the \(x\)-coordinate of \(P\), correct to 3 significant figures.
5. Use the trapezium rule, with 2 strips each of width 1.5, to find an estimate for \(\int _ { 1 } ^ { 4 } 3 ^ { x - 2 } \mathrm {~d} x\). Give your answer correct to 3 significant figures.

6 Use logarithms to solve the equation \(235 \times 5 ^ { x } = 987\), giving your answer correct to 3 decimal places.

6

1. On the same axes, sketch the curves \(y = 3 ^ { x }\) and \(y = 3 ^ { 2 x }\), identifying clearly which is which.
2. Given that \(3 ^ { 2 x } = 729\), find in either order the values of \(3 ^ { x }\) and \(x\).

2 It is given that \(p = \mathrm { e } ^ { 280 }\) and \(q = \mathrm { e } ^ { 300 }\).

1. Use logarithm properties to show that \(\ln \left( \frac { \mathrm { e } \mathrm { p } ^ { 2 } } { q } \right) = 261\).
2. Find the smallest integer \(n\) which satisfies the inequality \(5 ^ { n } > p q\).

5 \includegraphics[max width=\textwidth, alt={}, center]{6d15cb4d-f540-488b-b94e-7a494f192ba5-2_469_721_1932_662} The diagram shows the curves \(y = \mathrm { e } ^ { 2 x }\) and \(y = 8 \mathrm { e } ^ { - x }\). The shaded region is bounded by the curves and the \(y\)-axis. Without using a calculator,

1. solve an appropriate equation to show that the curves intersect at a point for which \(x = \ln 2\),
2. find the area of the shaded region, giving your answer in simplified form.

1 Solve the equation \(\mathrm { e } ^ { 2 x } - 5 \mathrm { e } ^ { x } = 0\).

10

1. Express \(\frac { 16 + 5 x - 2 x ^ { 2 } } { ( x + 1 ) ^ { 2 } ( x + 4 ) }\) in partial fractions.
2. It is given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( 16 + 5 x - 2 x ^ { 2 } \right) y } { ( x + 1 ) ^ { 2 } ( x + 4 ) }$$ and that \(y = \frac { 1 } { 256 }\) when \(x = 0\). Find the exact value of \(y\) when \(x = 2\). Give your answer in the form \(A \mathrm { e } ^ { n }\).

7 Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0 , x = 0.5\), and as \(t\) increases, \(x\) approaches 1 . \begin{figure}[h] \end{figure} One model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ( 1 - x )$$

1. Verify that \(x = \frac { 1 } { 1 + \mathrm { e } ^ { - t } }\) satisfies this differential equation, including the initial condition.
2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. An alternative model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ^ { 2 } ( 1 - x ) ,$$ with \(x = 0.5\) when \(t = 0\) as before.
3. Find constants \(A , B\) and \(C\) such that \(\frac { 1 } { x ^ { 2 } ( 1 - x ) } = \frac { A } { x ^ { 2 } } + \frac { B } { x } + \frac { C } { 1 - x }\).
4. Hence show that \(t = 2 + \ln \left( \frac { x } { 1 - x } \right) - \frac { 1 } { x }\).
5. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value.

7 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
5. Which of the two models fits the data better?

6 A mobile phone company records their annual sales on \(31 ^ { \text {st } }\) December every year.
Paul thinks that the annual sales, \(S\) million, can be modelled by the equation \(S = a b ^ { t }\), where \(a\) and \(b\) are both positive constants and \(t\) is the number of years since \(31 ^ { \text {st } }\) December 2015. Paul tests his theory by using the annual sales figures from \(31 ^ { \text {st } }\) December 2015 to \(31 { } ^ { \text {st } }\) December 2019. He plots these results on a graph, with \(t\) on the horizontal axis and \(\log _ { 10 } S\) on the vertical axis.

1. Explain why, if Paul's model is correct, the results should lie on a straight line of best fit on his graph. The results lie on a straight line of best fit which has a gradient of 0.146 and an intercept on the vertical axis of 0.583 .
2. Use these values to obtain estimates for \(a\) and \(b\), correct to 2 significant figures.
3. Use this model to predict the year in which, on the \(31 { } ^ { \text {st } }\) December, the annual sales would first be recorded as greater than 200 million.
4. Give a reason why this prediction may not be reliable.

5

1. The graph of \(y = 2 ^ { x }\) can be transformed to the graph of \(y = 2 ^ { x + 4 }\) either by a translation or by a stretch.
   1. Give full details of the translation.
   2. Give full details of the stretch.
2. In this question you must show detailed reasoning. Solve the equation \(\log _ { 2 } ( 8 x ) = 1 - \log _ { 2 } ( 1 - x )\).

2

1. 1. Show that \(\frac { 1 } { 3 - 2 \sqrt { x } } + \frac { 1 } { 3 + 2 \sqrt { x } }\) can be written in the form \(\frac { a } { b + c x }\), where \(a , b\) and \(c\) are constants to be determined.
   2. Hence solve the equation \(\frac { 1 } { 3 - 2 \sqrt { x } } + \frac { 1 } { 3 + 2 \sqrt { x } } = 2\).
2. In this question you must show detailed reasoning. Solve the equation \(2 ^ { 2 y } - 7 \times 2 ^ { y } - 8 = 0\).

6. \begin{figure}[h] \end{figure} A company makes a particular type of children's toy.
The annual profit made by the company is modelled by the equation $$P = 100 - 6.25 ( x - 9 ) ^ { 2 }$$ where \(P\) is the profit measured in thousands of pounds and \(x\) is the selling price of the toy in pounds. A sketch of \(P\) against \(x\) is shown in Figure 1.
Using the model,

1. explain why \(\pounds 15\) is not a sensible selling price for the toy. Given that the company made an annual profit of more than \(\pounds 80000\)
2. find, according to the model, the least possible selling price for the toy. The company wishes to maximise its annual profit.
   State, according to the model,
3. 1. the maximum possible annual profit,
   2. the selling price of the toy that maximises the annual profit.

13. \begin{figure}[h] \end{figure} The value of a rare painting, \(\pounds V\), is modelled by the equation \(V = p q ^ { t }\), where \(p\) and \(q\) are constants and \(t\) is the number of years since the value of the painting was first recorded on 1st January 1980. The line \(l\) shown in Figure 3 illustrates the linear relationship between \(t\) and \(\log _ { 10 } V\) since 1st January 1980. The equation of line \(l\) is \(\log _ { 10 } V = 0.05 t + 4.8\)

1. Find, to 4 significant figures, the value of \(p\) and the value of \(q\).
2. With reference to the model interpret
   1. the value of the constant \(p\),
   2. the value of the constant \(q\).
3. Find the value of the painting, as predicted by the model, on 1st January 2010, giving your answer to the nearest hundred thousand pounds.

1. In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

1. Solve the equation $$x \sqrt { 2 } - \sqrt { 18 } = x$$ writing the answer as a surd in simplest form.
2. Solve the equation $$4 ^ { 3 x - 2 } = \frac { 1 } { 2 \sqrt { 2 } }$$

1. The temperature, \(\theta ^ { \circ } \mathrm { C }\), of a cup of tea \(t\) minutes after it was placed on a table in a room, is modelled by the equation

$$\theta = 18 + 65 \mathrm { e } ^ { - \frac { t } { 8 } } \quad t \geqslant 0$$ Find, according to the model,

1. the temperature of the cup of tea when it was placed on the table,
2. the value of \(t\), to one decimal place, when the temperature of the cup of tea was \(35 ^ { \circ } \mathrm { C }\).
3. Explain why, according to this model, the temperature of the cup of tea could not fall to \(15 ^ { \circ } \mathrm { C }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bcbd842f-b2e2-4587-ab4c-15a57a449e5d-16_675_951_973_573} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The temperature, \(\mu ^ { \circ } \mathrm { C }\), of a second cup of tea \(t\) minutes after it was placed on a table in a different room, is modelled by the equation $$\mu = A + B \mathrm { e } ^ { - \frac { t } { 8 } } \quad t \geqslant 0$$ where \(A\) and \(B\) are constants.
   Figure 2 shows a sketch of \(\mu\) against \(t\) with two data points that lie on the curve.
   The line \(l\), also shown on Figure 2, is the asymptote to the curve.
   Using the equation of this model and the information given in Figure 2
4. find an equation for the asymptote \(l\).

1. In this question you must show all stages of your working.

\section*{Solutions relying entirely on calculator technology are not acceptable.} The air pressure, \(P \mathrm {~kg} / \mathrm { cm } ^ { 2 }\), inside a car tyre, \(t\) minutes from the instant when the tyre developed a puncture is given by the equation $$P = k + 1.4 \mathrm { e } ^ { - 0.5 t } \quad t \in \mathbb { R } \quad t \geqslant 0$$ where \(k\) is a constant.
Given that the initial air pressure inside the tyre was \(2.2 \mathrm {~kg} / \mathrm { cm } ^ { 2 }\)

1. state the value of \(k\). From the instant when the tyre developed the puncture,
2. find the time taken for the air pressure to fall to \(1 \mathrm {~kg} / \mathrm { cm } ^ { 2 }\) Give your answer in minutes to one decimal place.
3. Find the rate at which the air pressure in the tyre is decreasing exactly 2 minutes from the instant when the tyre developed the puncture.
   Give your answer in \(\mathrm { kg } / \mathrm { cm } ^ { 2 }\) per minute to 3 significant figures.

1. Using the laws of logarithms, solve the equation

$$2 \log _ { 5 } ( 3 x - 2 ) - \log _ { 5 } x = 2$$