Experimental Data & Linear Models

Video lesson · from 52:432 worked examples

One lesson video covers all of Exponentials & Logarithms, so it opens at 52:43 for this topic — not from the beginning.

Theory

The results of an experiment may show that there is an exponential relationship between the variables. However, the rapid rise/decrease can often make it difficult to interpret the results.

We can use logarithms to convert these exponential graphs to linear graphs which makes it easier to find the relationship between the variables.

Polynomials equations of the form $y = ax^b$ :

\begin{aligned} y &= ax^b \\ \log_{10} y &= \log_{10}(ax^b) \\ \log_{10} y &= \log_{10} a + \log_{10}(x^b) \\ \log_{10} y &= \log_{10} a + b\log_{10} x \\ \log_{10} y &= b\log_{10} x + \log_{10} a \end{aligned}

The equation is now in the linear form $Y = mX + c$ where $Y = \log_{10} y$ , $m = b$ , $X = \log_{10} x$ and $c = \log_{10} a$ .

Exponential equations of the form $y = ab^x$ :

\begin{aligned} y &= ab^x \\ \log_{10} y &= \log_{10}(ab^x) \\ \log_{10} y &= \log_{10} a + \log_{10}(b^x) \\ \log_{10} y &= \log_{10} a + x\log_{10} b \\ \log_{10} y &= (\log_{10} b)x + \log_{10} a \end{aligned}

The equation is now in the linear form $Y = mX + c$ where $Y = \log_{10} y$ , $m = \log_{10} b$ , $X = x$ and $c = \log_{10} a$ .

⚠️ Common Examiner Traps

Pick the right model. Using a relationship of the wrong form is the usual reason this question is not attempted successfully. Read the axes: if both are logarithmic it is $y = ax^b$ ; if only the vertical axis is, it is $y = ab^x$ .
The gradient is not the constant you want: for $y = ax^b$ the gradient is $b$ and the intercept is $\log a$ , so you must undo the log to recover $a$ .
Reproduce every step: a valid method still loses marks if the steps are not all reproduced accurately. Take logs, apply the laws, then compare with $Y = mX + c$ in full.
Undo the log at the end: an answer left as $\log a = 0.6$ is unfinished — give $a$ .

Worked examples

Example 1

The results from an experiment were noted as follows:

$\log_{10} x$	1.70	2.29	2.70	2.85
$\log_{10} y$	1.33	1.67	1.92	2.01

The relationship between these data can be written in the form $y = ax^b$ . Find the values of $a$ and $b$ , and state the formula for $y$ in terms of $x$ .

Calculate the gradient $m$ using the first and last points:

m = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{2.01 - 1.33}{2.85 - 1.70} = \frac{0.68}{1.15} \approx 0.591

For a polynomial model $y = ax^b$ , $b = m$ , so $b = 0.591$ .

Now find the intercept $c = \log_{10} a$ using $Y - mX = c$ . Using the first point (1.70, 1.33):

\begin{aligned} c &= 1.33 - (0.591)(1.70) \\ c &\approx 1.33 - 1.0047 = 0.325 \end{aligned}

Since $c = \log_{10} a$ :

a = 10^{0.325} \approx 2.11

Formula: $y = 2.11 x^{0.591}$

Example 2

The results from an experiment were noted as follows:

x	1.30	2.00	2.30	2.80
$\log_{10} y$	0.886	1.112	1.207	1.363

The relationship between these data can be written in the form $y = ab^x$ . State the formula for $y$ in terms of $x$ .

Calculate the gradient $m$ :

m = \frac{1.112 - 0.886}{2.00 - 1.30} = \frac{0.226}{0.70} \approx 0.323

For an exponential model $y = ab^x$ , $m = \log_{10} b$ :

\begin{aligned} \log_{10} b &= 0.323 \\ b &= 10^{0.323} \approx 2.10 \end{aligned}

Find the intercept $c = \log_{10} a$ using the first point:

\begin{aligned} c &= Y - mX \\ c &= 0.886 - (0.323)(1.30) \\ c &\approx 0.886 - 0.420 = 0.466 \end{aligned}

Since $c = \log_{10} a$ :

a = 10^{0.466} \approx 2.92

Formula: $y = 2.92 (2.10)^x$