6.5 Logarithmic graphs and estimating parameters — Learn

Prerequisites: the laws of logarithms from Section 6.3; straight-line graphs and gradients from Chapter 3; the model $y = kb^x$ from Sections 6.2 and 6.6.

By the end of this section you can: take logarithms of a model to turn it into a straight line; estimate the constants in $y = kb^x$ from a graph of $\ln y$ against $x$; estimate the constants in $y = ax^n$ from a graph of $\ln y$ against $\ln x$; and convert an intercept back into the original constant.

1. Turning a curve into a straight line

A straight line is easier to read than a curve. A gradient and an intercept can be measured with a ruler, and a line makes the trend obvious. Logarithms let us turn an exponential or a power model into a straight line.

Start with the exponential model $y = kb^x$, where $k > 0$ and $b > 0$. Take the natural logarithm of both sides:

This has the form $Y = mX + c$ if we plot $Y = \ln y$ against $X = x$. The graph is a straight line with

So the gradient of the line gives $\ln b$, and the intercept gives $\ln k$. To recover $b$ and $k$, undo the logarithm with the exponential: $b = e^{\text{gradient}}$ and $k = e^{\text{intercept}}$.

Now take the power model $y = ax^n$, where $a > 0$ and $x > 0$. Take logarithms:

Plot $Y = \ln y$ against $X = \ln x$. The graph is a straight line with

The table below shows which plot is needed for each model.

Example 1. A model $y = kb^x$ has data points $(0, 200)$ and $(1, 300)$. Estimate $k$ and $b$.

The gradient of the graph of $\ln y$ against $x$ is

So $\ln b = \ln 1.5$ and $b = 1.5$. The intercept is $\ln k = \ln 200$, so $k = 200$. The model is $y = 200\left(1.5\right)^x$.

Example 2. A graph of $\ln y$ against $\ln x$ for a model $y = ax^n$ has gradient $3$ and intercept $\ln 4$. Find $a$ and $n$.

The gradient is $n$, so $n = 3$. The intercept is $\ln a = \ln 4$, so $a = 4$. The model is $y = 4x^3$.

Example 3. A model satisfies $\log_{10} y = 2 + 0.5x$. Find $k$ and $b$ in $y = kb^x$.

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

2. Try it: straightening a curve

Predict first. The figure shows a set of data points from a model $y = k b^x$. Before you touch the controls, predict the shape of the graph of $\ln y$ against $x$: a curve or a straight line? Then predict the sign of the gradient when $b$ is larger than $1$. Write both predictions down.

Test. Move the slider to change the growth factor $b$. Note the shape of the graph of $y$ against $x$. Then press the button to show $\ln y$ against $x$, and read the gradient. Change $b$ again and read the new gradient.

Explain. Write two sentences. The first explains why plotting $\ln y$ instead of $y$ turns the curve into a straight line. The second explains what the gradient of that line tells you about $b$.

Progress check

Return to the mould culture in the chapter problem. Its counts are $200$, $300$, $450$ and $675$ at times $0$, $1$, $2$ and $3$ hours.

1. Work out the ratio between each count and the one before it.
2. Explain why a model of the form $N = kb^t$ fits the data.
3. Use two readings, at $t = 0$ and $t = 3$, to estimate the gradient of $\ln N$ against $t$.
4. Hence estimate $b$ and predict when $N$ reaches $2000$.

Watch out. These mistakes are common.

Quoting the intercept as $k$. For $y = kb^x$ the intercept is $\ln k$, not $k$. So an intercept of $\ln 200$ means $k = 200$, not $k = \ln 200$. Undo the logarithm with the exponential before you state the constant.

Plotting the wrong quantity for a power model. For $y = ax^n$ both axes need logarithms: plot $\ln y$ against $\ln x$. Plotting $\ln y$ against $x$ does not give gradient $n$.

Using logs of negative data. A logarithm needs a positive input, so $y$ and $x$ must be positive for these plots. If the data contain a zero or a negative value, the log-linear method cannot be used directly.

Re-check: after reading a gradient, compute $e^{\text{gradient}}$ and check that the value is plausible. A gradient of $\ln 1.5 \approx 0.405$ should give a growth factor close to $1.5$.

You should now be able to

• explain why $\ln y = \ln k + x\ln b$ is a straight-line equation;
• explain why $\ln y = \ln a + n\ln x$ is a straight-line equation;
• choose the correct plot for an exponential model and for a power model;
• estimate the gradient and intercept from a log-linear graph;
• convert a gradient into a growth factor and an intercept into a constant;
• predict a future value from the fitted model.