6.5 Logarithmic graphs and estimating parameters — Practise

Work through these on your own. Decide first whether the model is exponential or a power, then choose the correct plot. Convert a gradient into a growth factor and an intercept into a constant only at the end.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

For the model $y = kb^x$ , with $k > 0$ and $b > 0$ , show that $\ln y = \ln k + x\ln b$ .
A model $y = kb^x$ has data points $\left(0, 200\right)$ and $\left(1, 300\right)$ . Find the gradient of the graph of $\ln y$ against $x$ .
The gradient of a log-linear graph is $\ln 1.5$ . Write $\ln 1.5$ to three decimal places.
A log-linear graph of $\ln y$ against $x$ has intercept $\ln 200$ . Find the value of $k$ in $y = kb^x$ .
For the model $y = ax^3$ , with $a > 0$ and $x > 0$ , show that plotting $\ln y$ against $\ln x$ gives a straight line of gradient $3$ .
The points $\left(1, 3\right)$ , $\left(2, 12\right)$ and $\left(3, 27\right)$ fit a model $y = ax^n$ . Use the first and last points to find $n$ from the gradient of $\ln y$ against $\ln x$ .
A population is modelled by $N = 200\left(1.5\right)^t$ . Predict $N$ when $t = 5$ .
A population is modelled by $N = 200\left(1.5\right)^x$ . Find the value of $x$ when $N = 1000$ , exactly.
A model gives $\log_{10} y = 2 + 0.5x$ . Find $k$ and $b$ in $y = kb^x$ .
A model has data points $\left(0, 5\right)$ and $\left(2, 20\right)$ . Write the gradient of the graph of $\ln y$ against $x$ to three decimal places.
A straight line fitted to a graph of $\ln y$ against $x$ has gradient $0.4$ and passes through $\left(0, 1.2\right)$ . Find $k$ in the model $y = kb^x$ , to three significant figures.
A graph of $\ln y$ against $\ln x$ for a model $y = ax^n$ has intercept $\ln 4$ . Find $a$ .

Answers and concise solutions

1. The difference

Answer: Take logarithms of both sides. The product law turns $\ln\left(kb^x\right)$ into $\ln k + \ln b^x$ , and the power law brings the $x$ down: $\ln k + x\ln b$ . The result is a straight line in $x$ with gradient $\ln b$ and intercept $\ln k$ .

2. The gradient

Answer: The gradient is the change in $\ln y$ divided by the change in $x$ : $\ln 300 - \ln 200 = \ln\dfrac{300}{200} = \ln 1.5$ . So the gradient is $\ln 1.5$ , and the growth factor is $b = 1.5$ .

3. The value

Answer: $\ln 1.5 = 0.405465\ldots$ , which is $0.405$ to three decimal places. The gradient is positive, so the model grows.

4. The value of $k$

Answer: The intercept is $\ln k$ , so $\ln k = \ln 200$ and $k = 200$ . The exponential undoes the logarithm.

5. The difference

Answer: Take logarithms: $\ln y = \ln a + \ln x^3 = \ln a + 3\ln x$ . This is a straight line when $\ln y$ is plotted against $\ln x$ , with gradient $3$ and intercept $\ln a$ . The same argument with any power $n$ gives gradient $n$ ; the power law is what makes the model linear in the logs.

6. The value of $n$

Answer: The gradient is $\dfrac{\ln 27 - \ln 3}{\ln 3 - \ln 1} = \dfrac{\ln 9}{\ln 3} = \dfrac{2\ln 3}{\ln 3} = 2$ . So $n = 2$ , and the model is a quadratic in disguise: $y = ax^2$ .

7. The value of $N$

Answer: $\left(1.5\right)^5 = 7.59375$ , and $200 \times 7.59375 = 1518.75$ . As a fraction this is $\dfrac{6075}{4}$ .

8. The value of $x$

Answer: Divide by $200$ : $\left(1.5\right)^x = 5$ . Take logarithms: $x\ln 1.5 = \ln 5$ , so $x = \dfrac{\ln 5}{\ln 1.5} \approx 3.97$ .

9. The values of $k$ and $b$

Answer: Compare with $\log_{10} y = \log_{10} k + x\log_{10} b$ . The intercept gives $\log_{10} k = 2$ , so $k = 100$ . The gradient gives $\log_{10} b = 0.5$ , so $b = 10^{0.5} = \sqrt{10} \approx 3.16$ .

10. The gradient

Answer: The gradient is $\dfrac{\ln 20 - \ln 5}{2} = \dfrac{\ln 4}{2} = \ln 2 = 0.693147\ldots$ , which is $0.693$ to three decimal places.

11. The value of $k$

Answer: The intercept is $\ln k = 1.2$ , so $k = e^{1.2} = 3.3201\ldots$ , which is $3.32$ to three significant figures. The gradient gives $\ln b = 0.4$ , so $b = e^{0.4} \approx 1.49$ .

12. The value of $a$

Answer: The intercept is $\ln a$ , so $\ln a = \ln 4$ and $a = 4$ . The power law and the product law together turn the product $ax^n$ into a sum, which is what the straight line reads.