EL4, EL5 - Straight-line graphs for exponential models

Turn an exponential curve into a straight line using base-ten logarithms. EL4 supplies the rearrangement; EL5 lets you estimate the original constants from the line's gradient and intercept. You will use data, draw a line of best fit and judge what its predictions can reasonably claim.

Which quantities make a straight line?

Can a straight line describe a quantity that multiplies by the same factor at every step? It can, if the vertical coordinate records the logarithm of that quantity. The graph changes the way we measure height, while the underlying model stays exponential.

You already know

A straight line $Y=mX+c$ has gradient $m$ and vertical intercept $c$ . For positive arguments, $\log_{10}(uv)=\log_{10}u+\log_{10}v$ and $\log_{10}(u^r)=r\log_{10}u$ .

For $y=ka^x$ , take a logarithm of the complete expression on each side. First split the product, then bring the exponent down. We require $k>0$ and $a>0$ , so $y>0$ and the logarithms exist. A negative exponential can be studied by a separately specified positive quantity such as $-y$ ; you cannot simply plot $\log_{10}y$ for negative $y$ .

Linear form of an exponential model

\begin{aligned} y&=ka^x\\ \log_{10}y&=\log_{10}k+x\log_{10}a \end{aligned}

Plot $Y=\log_{10}y$ vertically against $X=x$ horizontally. Then $m=\log_{10}a$ , $c=\log_{10}k$ , so $a=10^m$ , $k=10^c$ . Here $k,a>0$ .
Not on the formulae sheet: learn it.

The vertical coordinate is a logarithm, so a rise of $1$ means ten times the original quantity. For example, logarithmic heights $1, 2, 3$ represent $10, 100, 1000$ . Equal vertical gaps can therefore represent enormous changes.

Worked example 1: a general positive base

Question. Show that $y=ka^x$ , for $k,a>0$ , gives a straight-line graph of $\log_{10}y$ against $x$ . State its gradient and intercept.

Plan. Apply the product and power laws, then compare with $Y=mX+c$ .

\begin{aligned} \log_{10}y&=\log_{10}(ka^x) && \text{take logs of both sides}\\ &=\log_{10}k+\log_{10}(a^x) && \text{product law}\\ &=\log_{10}k+x\log_{10}a && \text{power law} \end{aligned}

Why leave $\log_{10}k$ separate? Only $a$ , not $k$ , is raised to $x$ . Making $x$ multiply the entire logarithm would change the equation.

Answer. With $Y=\log_{10}y$ , $X=x$ , the equation is linear, with gradient $\log_{10}a$ and intercept $\log_{10}k$ .

Check. At $x=0$ , $y=k$ , so the transformed height is indeed $\log_{10}k$ .

Change only the base representation: write the exponential factor as $10^{bx}$ . This makes the gradient itself the exponent's coefficient.

Worked example 2: a base-ten model

Question. Show that $p=c\times10^{bt}$ , where $c>0$ , produces a straight line when $\log_{10}p$ is plotted against $t$ . State its gradient and intercept.

Plan. Repeat the logarithm laws, using $\log_{10}10=1$ .

\begin{aligned} \log_{10}p&=\log_{10}(c\times10^{bt}) && \text{take logarithms}\\ &=\log_{10}c+bt\log_{10}10 && \text{product and power laws}\\ &=\log_{10}c+bt && \log_{10}10=1 \end{aligned}

Why is the gradient $b$ , rather than $\log_{10}b$ ? It multiplies $t$ directly in the final line equation.

Answer. The gradient is $b$ , and the intercept is $\log_{10}c$ .

Check. Increasing $t$ by $1$ multiplies $p$ by $10^b$ , so its logarithm increases by $b$ .

Always compare your final expression to a line, rather than memorising a letter as “the gradient”.

Watch out

Wrong: $\log_{10}(ka^x)=x\log_{10}(ka)$ . That would put both factors to the power $x$ . Right: $\log_{10}(ka^x)=\log_{10}k+x\log_{10}a$ .

Change the exponential base again and read the resulting coefficient carefully.

Estimating constants from a plotted line

Measured values rarely fall exactly on a model. Transform the data, plot them accurately, then draw one straight line through the middle of the scatter, with points reasonably balanced on either side.

Use uniform scales on both axes. Horizontal positions remain the original inputs, not their row numbers. Choose two widely separated points on the fitted line for its gradient; they need not be original data points. Extend the line to the vertical axis to estimate its intercept. Taking $10$ to a logarithmic value undoes the logarithm.

Worked example 3: a growing dataset

Question. The following dataset is modelled by $y=ka^x$ . Use a plot of $\log_{10}y$ against $x$ to estimate $k$ and $a$ .

$x$	$0$	$2$	$4$	$6$	$8$
$y$	$4.0$	$6.2$	$10.2$	$15.6$	$25.4$
$\log_{10}y$	$0.602$	$0.792$	$1.009$	$1.193$	$1.405$

Plan. Plot the transformed row, fit a line, then calculate its gradient and undo both logarithms.

The left panel below shows the plotted points and a suitable line. Read two points from that line, approximately $(0, 0.60)$ and $(8, 1.40)$ .

\begin{aligned} m&=\frac{1.40-0.60}{8-0}=0.100 && \text{large gradient triangle}\\ c&\approx0.60 && \text{read vertical intercept}\\ a&\approx10^{0.100}=1.2589\ldots && \text{undo logarithm}\\ k&\approx10^{0.60}=3.9811\ldots && \text{undo logarithm} \end{aligned}

Why use the fitted line rather than joining the dots? The line estimates the common trend; joining consecutive measurements treats measurement scatter as part of the model.

Answer. $k\approx3.98$ , $a\approx1.26$ . The data lie close to a straight line, supporting an exponential model over this range.

Check. The fitted value at $x=4$ is $10^{0.60+0.100(4)}=10$ , close to the measured $10.2$ .

Reverse only the order of the measured heights. The transformed line falls, so its multiplier must lie between $0$ and $1$ .

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Algebra

Enumeration (counting and probability)

Coordinate geometry

Pythagoras' theorem and trigonometry

Calculus

Exponentials and logarithms

Numerical methods

EL4, EL5 - Straight-line graphs for exponential models

Which quantities make a straight line?

Estimating constants from a plotted line

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