F7PureStretch

Exponential Models and Log-Linear Data

Q: How do I recover a from the intercept of a log-linear graph?

The intercept equals a (or a). Raise 10 (or e) to that power to get a itself — never treat the intercept as a.

Q: Why is finding k simpler than finding b?

Because in y= a+kx, k is already a coefficient (not a base), so the gradient of the straight line IS k with no further conversion. In y= a+x b, the gradient is b, which still needs to be un-logged to find b.

A scientist plots population against time and gets a curve, not a straight line — hard to read exact values from. But plot the LOGARITHM of the population against time instead, and an exponential model suddenly becomes a straight line, with the gradient and intercept revealing the model's constants directly.

32 min Video by TLMaths Exponentials and Logarithms

The big picture

This final lesson in the chapter connects everything: exponential models describe real growth and decay ( $y = a b^{x}$ or $y = a e^{k x}$ ), and taking logarithms of both sides turns these curved relationships into straight lines. This "linearising" trick is exactly how scientists fit models to messy real data — plot $lo g y$ against $x$ , read off the gradient and intercept from the straight line, then convert back to find $a$ and $b$ (or $k$ ). It is the single most practical application of everything this chapter has built.

A-Level Maths F7-01: Exponential Growth & Decay: Long-Term BehaviourWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Use exponential models $y = a b^{x}$ and $y = a e^{k x}$ to represent growth and decay
Take logarithms of an exponential model to obtain a linear relationship
Interpret the gradient and intercept of a log-linear graph in terms of the original model
Fit a model from given or plotted data

Linearising $y = ab^x$

Taking $lo g$ of both sides of $y = a b^{x}$ and applying the laws of logarithms turns a curved relationship into a straight-line one in terms of $lo g y$ and $x$ .

lo g y = lo g a + x lo g b

Compare to

Y = c + m X

: plotting

lo g y

against

x

gives a straight line with gradient

lo g b

and intercept

lo g a

1Intercept is

lo g a = 1

, so

a = 1 0^{1} = 10

2Gradient is

lo g b = 0.301

, so

b = 1 0^{0.301} \approx 2

Answer

a = 10

b \approx 2

(so

y = 10 \times 2^{x}

Tip — The intercept of the straight-line graph gives $lo g a$ directly — you must "undo" the log (raise 10 to that power, or $e$ if using $ln$ ) to recover $a$ itself.

Linearising $y = ae^{kx}$

For a model built on $e$ (the natural choice whenever the rate of change is proportional to the current value), take $ln$ of both sides instead. The power law simplifies $ln (e^{k x})$ to just $k x$ , since $ln$ and $e$ are inverses.

ln y = ln a + k x

Plotting

ln y

against

x

gives gradient

k

and intercept

ln a

1Gradient equals

k

directly:

k = - 0.05

2Intercept is

ln a = 4.6

, so

a = e^{4.6} \approx 99.5

3Since

k < 0

, the model represents decay — consistent with radioactive decay.

Answer

a \approx 99.5

k = - 0.05

(decay, since

k

is negative).

Why this works

Notice $ln y = ln a + k x$ needs no "convert the gradient back" step for $k$ — because the base is already $e$ , the gradient of the straight line IS $k$ itself, unlike the $y = a b^{x}$ case where you must un-log the gradient to get $b$ .

Interpreting real data

In practice, you are given a table of $(x, y)$ data believed to follow an exponential model. Take logs of every $y$ -value, plot against $x$ , and if the points lie close to a straight line, this supports the exponential model — a technique examiners test directly by giving you two data points on the log-linear line and asking you to find the model.

1At

x = 0

lo g a = 2.30

, so

a = 1 0^{2.30} \approx 200

2Gradient (change in

lo g y

per unit

x

\frac{2.90 - 2.30}{2 - 0} = \frac{0.60}{2} = 0.30 = lo g b

3So

b = 1 0^{0.30} \approx 2.0

Answer

a \approx 200

b \approx 2.0

Think like an examiner

Choose $lo g$ (base 10) to linearise $y = a b^{x}$ and $ln$ (base $e$ ) to linearise $y = a e^{k x}$ — matching the log base to the model's base keeps the algebra clean.
The intercept of a log-linear graph gives $lo g a$ (or $ln a$ ), NOT $a$ itself — always exponentiate to recover the original constant.
For $y = a e^{k x}$ , the gradient of the $ln y$ vs $x$ graph IS $k$ directly — no extra conversion step needed there.
A negative gradient or negative $k$ signals decay; positive signals growth.

Common misconceptions

Reading the intercept of a log-linear graph as the value of

a

itself.

The intercept equals

lo g a

(or

ln a

) — you must raise 10 (or

e

) to that power to find

a

Forgetting to un-log the gradient when finding

b

y = a b^{x}

The gradient of

lo g y

x

equals

lo g b

, so

b = 1 0^{gradient}

— not the gradient itself.

Using

ln

to linearise

y = a b^{x}

and treating the gradient as

b

directly.

Whichever log base you use, the gradient always equals

lo g_{(that base)} b

— it must still be un-logged to recover

b

(this step is only skipped for

k

in the

e^{k x}

case, since

k

is already the exponent coefficient, not a base to be recovered).

Linearising exponential models

lo g y = lo g a + x lo g b

From y = ab^x; gradient log b, intercept log a.

ln y = ln a + k x

From y = ae^{kx}; gradient k, intercept ln a.

Stretch yourself

A bacterial culture follows $N = a e^{k t}$ . When $ln N$ is plotted against $t$ , the line passes through $(0, 3.0)$ and $(4, 5.8)$ . Find $a$ and $k$ , and predict $N$ when $t = 10$ (to 3 s.f.).

Hint — Use the intercept for $a$ and the gradient between the two given points for $k$ , then substitute $t = 10$ into the full model.

Questions students ask

Why do we take logs of an exponential model?

Because it turns a curved relationship into a straight line, which is far easier to fit to real data — you can read the gradient and intercept directly off a straight-line graph.

How do I recover

a

from the intercept of a log-linear graph?

The intercept equals

lo g a

(or

ln a

). Raise 10 (or

e

) to that power to get

a

itself — never treat the intercept as

a

Why is finding

k

simpler than finding

b

Because in

ln y = ln a + k x

k

is already a coefficient (not a base), so the gradient of the straight line IS

k

with no further conversion. In

lo g y = lo g a + x lo g b

, the gradient is

lo g b

, which still needs to be un-logged to find

b

Key takeaways

Taking logs of $y = a b^{x}$ gives the straight line $lo g y = lo g a + x lo g b$ .
Taking $ln$ of $y = a e^{k x}$ gives the straight line $ln y = ln a + k x$ .
Read the intercept as $lo g a$ (or $ln a$ ) and un-log it to find $a$ .
For the $a b^{x}$ model, the gradient is $lo g b$ (un-log to find $b$ ); for the $a e^{k x}$ model, the gradient IS $k$ directly.

How this fits the course

Build on

Straight-Line Graphs

Leads to

Differentiation from First Principles

Test yourself

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Exponential Models and Log-Linear Data

32 min Video by TLMaths Exponentials and Logarithms