Chapter 6: Review

The chapter problem, solved

The cooling coffee

A cup of coffee is poured at $80^\circ$ C in a room at $20^\circ$ C, so the excess temperature starts at $60^\circ$ C. The model is

E = 60e^{-kt}.

After $10$ minutes the coffee is at $65^\circ$ C, so $E = 45$ .

Step 1. Find k. Substitute $t = 10$ and $E = 45$ :

45 = 60e^{-10k}, \qquad e^{-10k} = \frac{45}{60} = 0.75.

Take logarithms: $-10k = \ln 0.75$ , so

k = -\frac{\ln 0.75}{10} = \frac{\ln\left(4/3\right)}{10} \approx 0.0288.

Step 2. Find the initial rate of cooling. Differentiate the model:

\frac{dE}{dt} = -60k e^{-kt}.

At $t = 0$ ,

\frac{dE}{dt} = -60k \approx -60 \times 0.0288 = -1.73.

The rate is about $-1.73^\circ$ C per minute. The negative sign says the excess temperature is falling. The rate itself shrinks as $E$ shrinks, so the coffee cools quickly at first and more slowly later.

Step 3. Find the time to reach 15. Set $E = 15$ :

15 = 60e^{-kt}, \qquad e^{-kt} = \frac{15}{60} = 0.25.

Take logarithms: $-kt = \ln 0.25 = -\ln 4$ , so

t = \frac{\ln 4}{k} = \frac{10\ln 4}{\ln\left(4/3\right)} \approx 48.2 \text{ minutes}.

The coffee is ready to drink after about $48$ minutes. The second stage takes longer than the first, because the cooling is slower at a lower temperature.

The mould culture

The counts are $200$ , $300$ , $450$ and $675$ at times $0$ , $1$ , $2$ and $3$ hours.

Step 1. Read the growth factor. Divide each count by the one before it:

\frac{300}{200} = 1.5, \qquad \frac{450}{300} = 1.5, \qquad \frac{675}{450} = 1.5.

The number multiplies by $1.5$ each hour, so $b = 1.5$ and $k = 200$ .

Step 2. Read it from a log graph. For the model $N = kb^t$ ,

\ln N = \ln k + t\ln b.

Using the first and last readings, the gradient is

\frac{\ln 675 - \ln 200}{3 - 0} = \frac{\ln 3.375}{3} = \ln 1.5 \approx 0.405.

So $\ln b = \ln 1.5$ and $b = 1.5$ , which agrees with the ratio. The intercept is $\ln 200$ , so $k = 200$ .

Step 3. Predict the time to 2000. Set $N = 2000$ :

2000 = 200\left(1.5\right)^t, \qquad \left(1.5\right)^t = 10.

Take logarithms: $t\ln 1.5 = \ln 10$ , so

t = \frac{\ln 10}{\ln 1.5} \approx 5.68 \text{ hours}.

Step 4. State the limitation. The exponential model describes the data over the first few hours. It cannot hold forever, because the culture will run out of food and space. Predicting far beyond $5.68$ hours is risky, and a better model would include a maximum population.

Mixed practice — choose the method first

For each question, say which method you will use before you calculate. The items mix this chapter with Chapters 1 to 5 and 7.

Solve $3^x = 45$ , giving $x$ exactly.
Write $2\ln x + \ln y$ as a single logarithm.
Differentiate $y = 7e^{4x}$ .
Simplify $3^{x+2} \div 3^{x-1}$ .
Write $\sqrt{50} + \sqrt{18}$ in the form $k\sqrt{2}$ .
Solve $x^2 - 5x + 6 = 0$ .
Solve $\sin x = \dfrac{\sqrt{3}}{2}$ for $0 \le x < 2\pi$ .
Find the coefficient of $x^2$ in the expansion of $\left(1 + 2x\right)^5$ .
Show that $\left(2n + 1\right)^2 - \left(2n - 1\right)^2$ is a multiple of $8$ .
A radioactive sample halves every $12$ years. Find the decay constant $k$ in $m = m_0e^{-kt}$ .
A graph of $\ln y$ against $x$ for a model $y = kb^x$ has gradient $0.6$ . Find $b$ , to three significant figures.
Solve $\ln\left(x - 1\right) = 2$ , stating the domain you use.
Find the gradient of $y = x^3 - 6x^2$ at $x = 2$ .
Show that $\left(x - 2\right)$ is a factor of $x^3 - 4x^2 + x + 6$ .

Answers and concise solutions

1. The value of $x$

Answer: Take logarithms of both sides: $x = \dfrac{\ln 45}{\ln 3} \approx 3.465$ . Since $45 = 9 \times 5$ , this can also be written as $2 + \dfrac{\ln 5}{\ln 3}$ , which is the same number.

2. The single logarithm

Answer: Use the power law on the first term, then the product law: $2\ln x + \ln y = \ln x^2 + \ln y = \ln\left(x^2y\right)$ .

3. The derivative

Answer: The derivative of $e^{4x}$ is $4e^{4x}$ , and the constant $7$ stays in front: $\dfrac{dy}{dx} = 28e^{4x}$ .

4. The value

Answer: Subtract the indices: $3^{(x+2)-(x-1)} = 3^3 = 27$ . The $x$ cancels, so the result is a number.

5. The value of $k$

Answer: $\sqrt{50} = 5\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$ , so the sum is $8\sqrt{2}$ and $k = 8$ .

6. The solutions

Answer: Factorise: $\left(x - 2\right)\left(x - 3\right) = 0$ , so $x = 2$ or $x = 3$ . The roots are the two numbers that add to $5$ and multiply to $6$ .

7. The solutions

Answer: The principal value is $\dfrac{\pi}{3}$ . The second solution is $\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$ . A sketch of the sine curve shows exactly two solutions in one full turn.

8. The coefficient

Answer: The term in $x^2$ is $\dbinom{5}{2}\left(2x\right)^2 = 10 \times 4x^2 = 40x^2$ . So the coefficient is $40$ .

9. The difference minus $8n$

Answer: Expand: $\left(4n^2 + 4n + 1\right) - \left(4n^2 - 4n + 1\right) = 8n$ . Since $8n$ is $8$ times the whole number $n$ , the difference is always a multiple of $8$ .

10. The value of $k$

Answer: When $t = 12$ the mass has halved, so $e^{-12k} = \dfrac{1}{2}$ . Take logarithms: $-12k = -\ln 2$ , giving $k = \dfrac{\ln 2}{12} \approx 0.0578$ per year.

11. The value of $b$

Answer: The gradient is $\ln b$ , so $\ln b = 0.6$ and $b = e^{0.6} = 1.8221\ldots$ , which is $1.82$ to three significant figures. The model grows because $b > 1$ .

12. The value of $x$

Answer: A logarithm needs a positive input, so $x - 1 > 0$ and the domain is $x > 1$ . Take the exponential of both sides: $x - 1 = e^2$ , so $x = e^2 + 1 \approx 8.39$ , which is in the domain.

13. The gradient

Answer: Differentiate: $\dfrac{dy}{dx} = 3x^2 - 12x$ . At $x = 2$ this is $12 - 24 = -12$ . The curve falls at that point.

14. The remainder

Answer: By the factor theorem, $\left(x - 2\right)$ is a factor exactly when the remainder is $0$ . Dividing gives remainder $0$ , so $\left(x - 2\right)$ is a factor. Indeed $2^3 - 4\left(2^2\right) + 2 + 6 = 8 - 16 + 8 = 0$ .

What you can now do

sketch $y = a^x$ and $y = e^x$ , and state their intercept, range and asymptote;
explain why $e$ is the natural base;
differentiate $e^{kx}$ and $a^{kx}$ , and interpret the result as a rate;
explain the link between a rate proportional to the amount and an exponential model;
convert between exponential and logarithmic form, and use the laws of logarithms;
solve $a^x = b$ , including a quadratic in $a^x$ ;
use a logarithmic graph to estimate the constants in $y = kb^x$ and $y = ax^n$ ;
model growth and decay in context, and find a half-life or a doubling time;
state the limitations of an exponential model.

Your relevance reflection

Return to the prompt in the chapter opener: something that grows or shrinks by a fixed factor. Write down the quantity again. Now add three things you could not have said before this chapter: how to write the quantity as an exponential model, how to find the time for it to double or halve, and how to read its growth factor from a straight line. Finish with one sentence about a growth or decay you would measure from now on.