6.5 Logarithmic graphs and estimating parameters — Practise with help

Work each question closed book first. Decide which plot the model needs, then remember that the intercept is the logarithm of the constant, not the constant.

Warm-up: remember

Write $\ln a + 3\ln x$ as a single logarithm.
Find the gradient of the straight line through $(0, 2)$ and $(4, 10)$ .
Work out $e^{\ln 5}$ .

Answers

$\ln a + 3\ln x = \ln\left(ax^3\right)$ .
The gradient is $\dfrac{10 - 2}{4 - 0} = 2$ .
$e^{\ln 5} = 5$ , because the exponential undoes the logarithm.

If question 1 was hard, revise the laws of logarithms in Section 6.3 before you continue.

Practise with help

1. For the model $y = kb^x$ , what quantity is plotted on each axis to give a straight line, and what do the gradient and intercept represent?

Hint

Take the natural logarithm of both sides of $y = kb^x$ and compare the result with $Y = mX + c$ .

Next step

$\ln y = \ln k + x\ln b$ . Match $X = x$ and $Y = \ln y$ , then read off $m$ and $c$ .

Full working

Plot $Y = \ln y$ against $X = x$ . Then

\ln y = \ln k + x\ln b

is a straight line with gradient $m = \ln b$ and intercept $c = \ln k$ . The gradient gives the growth factor $b = e^{m}$ , and the intercept gives $k = e^{c}$ .

2. A model $y = kb^x$ has data points $(0, 400)$ and $(1, 600)$ . Find $b$ and $k$ .

Hint

The gradient of $\ln y$ against $x$ is the change in $\ln y$ , which is $\ln 600 - \ln 400$ .

Next step

Use the quotient law to simplify the gradient to $\ln 1.5$ . The intercept is $\ln 400$ .

Full working

\text{gradient} = \ln 600 - \ln 400 = \ln\frac{600}{400} = \ln 1.5.

So $\ln b = \ln 1.5$ and $b = 1.5$ . The intercept is $\ln 400$ , so $k = 400$ . The model is $y = 400\left(1.5\right)^x$ .

3. A graph of $\ln y$ against $\ln x$ for a model $y = ax^n$ has gradient $2.5$ and intercept $\ln 3$ . Find $a$ and $n$ .

Hint

For a power model the gradient is the power $n$ , and the intercept is $\ln a$ .

Next step

Read the gradient straight off as $n$ . Undo the logarithm in the intercept with the exponential.

Full working

n = 2.5, \qquad \ln a = \ln 3, \qquad a = 3.

The model is $y = 3x^{2.5}$ .

4. A model satisfies $\log_{10} y = 1 + 0.3x$ . Find $k$ and $b$ in $y = kb^x$ .

Hint

Compare the given equation with $\log_{10} y = \log_{10} k + x\log_{10} b$ .

Next step

The intercept gives $\log_{10} k = 1$ , and the gradient gives $\log_{10} b = 0.3$ .

Full working

k = 10^{1} = 10, \qquad b = 10^{0.3} \approx 2.00.

So the model is $y = 10\left(2.00\right)^x$ to three significant figures.

5. A graph of $\ln y$ against $x$ has gradient $0.7$ . Find the growth factor $b$ in $y = kb^x$ , to three significant figures.

Hint

The gradient equals $\ln b$ . Undo the logarithm with the exponential.

Next step

Work out $e^{0.7}$ and round to three significant figures.

Full working

\ln b = 0.7, \qquad b = e^{0.7} = 2.0137\ldots \approx 2.01.

Because $b > 1$ , the model grows.

6. A model $N = 200\left(1.5\right)^t$ has been fitted to the mould data. Use it to predict $N$ when $t = 5$ .

Hint

Substitute $t = 5$ and work out $\left(1.5\right)^5$ first.

Next step

$\left(1.5\right)^5 = 7.59375$ , so multiply by $200$ .

Full working

N = 200 \times \left(1.5\right)^5 = 200 \times 7.59375 = 1518.75.

The prediction is about $1519$ cells after five hours. It assumes the growth pattern continues, which a model cannot guarantee.

7. A student plots $\ln y$ against $x$ for a model $y = ax^n$ and expects the gradient to be $n$ . Explain why this is wrong, and state the correct plot.

Hint

Take logarithms of $y = ax^n$ and see whether $\ln x$ or $x$ appears on the right-hand side.

Next step

The logarithm gives $\ln y = \ln a + n\ln x$ , so the variable on the horizontal axis must be $\ln x$ , not $x$ .

Full working

Taking logarithms gives

\ln y = \ln a + n\ln x.

The right-hand side contains $\ln x$ , not $x$ . So a graph of $\ln y$ against $x$ is not straight in general. The correct plot is $\ln y$ against $\ln x$ , and only then is the gradient equal to the power $n$ .