Exponential Equations in A Level Physics: Using e, ln and Straight-Line Graphs

Use this resource

Work through it in three passes

1

Understand the shape

Learn why the original physics graph is curved and why taking ln makes a straight line.

2

Choose the right plot

Match each Module 6 equation to the graph you should draw, including background correction and capacitor charging.

3

Do the graph task

Calculate logged values, plot them, check the best-fit line, then use the gradient to find the physics constant.

By the end you should be able to

Recognise equations of the form y = Ae^kx and y = Ae^-kx.
Use ln to solve for an unknown in an exponential equation.
Choose what to plot so an exponential relationship becomes a straight line.
Use the gradient and intercept to find physical constants such as RC, λ and μ.

01 Core idea

What an exponential equation is really saying

A Level Physics exponentials usually describe a quantity that changes at a rate proportional to how much of it is left. That is why the graph is curved: the value changes quickly at first, then more slowly as the value gets smaller.

Decay form

y = Ae^-kx

Used when a quantity falls towards zero: capacitor discharge, radioactive activity, photon intensity through an absorber.

Approach-to-final form

y = y_final(1 - e^-kx)

Used when a quantity rises towards a maximum: capacitor voltage while charging.

The exponent must be dimensionless

In e^-t/RC, the quantity t/RC has no unit because RC is a time constant in seconds. In e^-λt, λ has unit s^-1, so λt has no unit.

02 e and ln

The rules you actually use

ln means natural logarithm. It is the inverse of e^x, so it is the operation that "undoes" an exponential.

ln(e^x) = x so ln(e^-λt) = -λt

ln(ab) = ln(a) + ln(b) so ln(Ae^-kx) = ln(A) - kx

ln(a/b) = ln(a) - ln(b) so ln(y/A) = -kx

e^ln(a) = a so if ln(A) = 2.48, then A = e^2.48

Half left

e^-kx = 1/2

kx = ln(2)

37% left

e^-1 = 0.368

After one time constant, a decaying quantity is about 37% of its initial value.

63% reached

1 - e^-1 = 0.632

After one time constant, a charging capacitor has reached about 63% of its final voltage.

03 Linearising

Turning an exponential curve into a straight line

The straight-line equation is y = mx + c. Your job is to make an exponential look like that. For a simple decay equation:

y = Ae^-kx

ln(y) = ln(A) - kx

This means if you plot ln(y) on the vertical axis against x on the horizontal axis, the graph should be a straight line with gradient -k and intercept ln(A).

Original graph: curved decay

Linearised graph: straight line

1

Rearrange first. If there is a background or final value, subtract it before taking logs.

2

Take natural logs. Use ln, not log₁₀, because OCR exponential formulae use e.

3

Plot the transformed variable. The horizontal axis stays as time, thickness or distance; the vertical axis becomes a natural log.

4

Use the line. The gradient gives the physical constant; the intercept gives the initial value after applying e^intercept.

About units inside ln

Strictly, logs are taken of dimensionless ratios, such as ln(V / 1 V) or ln(A / 1 Bq). In exam tables you will often see ln(V) for the logged numerical value. The gradient is unchanged, so the physics result is unchanged.

04 Module 6 map

Where this appears in Module 6

Capacitor discharge

V = V₀e^-t/RC

Plot ln(V) against t. Gradient = -1/RC, so RC = -1/gradient.

Capacitor charging

V_C = E(1 - e^-t/RC)

Rearrange to E - V_C = Ee^-t/RC. Plot ln(E - V_C) against t.

Radioactive decay

A = A₀e^-λt

Plot ln(A) against t. Gradient = -λ. Half-life T_1/2 = ln(2)/λ.

Count rate with background

C = C_bg + C₀e^-λt

Subtract background first. Plot ln(C - C_bg) against t.

X-ray or gamma attenuation

I = I₀e^-μx

Plot ln(I) against x. Gradient = -μ. Or plot ln(I₀/I) against x for gradient +μ.

Medical tracers and PET

A = A₀e^-λt

Radiopharmaceutical activity decays exponentially. The same graphing method helps find activity, decay constant or effective half-life.

05 Assessment task

Plot the graph yourself, then check it

Your task

A capacitor is discharged through a resistor. Use the data to test whether the discharge is exponential. Calculate ln(V), plot ln(V) against t, then use the straight-line graph to find RC.

Fill in the missing ln(V) values to 3 decimal places.
Press Plot my points to see your transformed graph.
Press Show expected graph only after you have tried.
Use the gradient to calculate RC and use the intercept to find V₀.

What success looks like

Your plotted points should lie close to a straight decreasing line.
The gradient should be about -0.025 s^-1.
The time constant should be about 40 s.
The intercept should give an initial voltage of about 12 V.

t / s	V / V	Your ln(V)
0	12.00
20	7.28
40	4.41
60	2.68
80	1.62

Fill the table first. The graph uses your ln(V) values.

When your ln(V) values are entered, plot the points and look for a straight line.

Gradient from your line / s^-1 Time constant RC / s Initial voltage V₀ / V

After plotting, estimate the gradient from your straight line and answer the three questions.

Reveal full answer and marking guide

t / s	V / V	ln(V)
0	12.00	2.485
20	7.28	1.985
40	4.41	1.484
60	2.68	0.986
80	1.62	0.482

gradient = (0.482 - 2.485) / (80 - 0) = -0.0250 s^-1

For V = V₀e^-t/RC, gradient = -1/RC, so RC = -1 / -0.0250 = 40 s.

intercept = ln(V₀) = 2.485, so V₀ = e^2.485 = 12.0 V.

06 Worked examples

Worked examples with graph choices

Example 1: capacitor discharge

A 12.0 V capacitor discharge gives the data below. Show how to find the time constant.

t / s	V / V	ln(V)
0	12.0	2.485
20	7.28	1.985
40	4.41	1.484
60	2.68	0.986
80	1.62	0.482

Plot ln(V) against t. The gradient is about:

gradient = (0.482 - 2.485) / (80 - 0) = -0.0250 s^-1

For discharge, gradient = -1/RC.

RC = -1 / gradient = -1 / -0.0250 = 40 s

The intercept is ln(V₀) = 2.485, so V₀ = e^2.485 = 12.0 V.

Example 2: capacitor charging needs rearranging

A capacitor charges from a 6.0 V supply. You cannot plot ln(V_C) against time and expect a straight line.

Start with the charging equation:

V_C = E(1 - e^-t/RC)

Rearrange so the exponential is on its own:

E - V_C = Ee^-t/RC

ln(E - V_C) = ln(E) - t/RC

t / s	V_C / V	E - V_C / V	ln(E - V_C)
0	0.00	6.00	1.792
10	1.98	4.02	1.391
20	3.30	2.70	0.993
30	4.19	1.81	0.593
40	4.79	1.21	0.191

gradient = (0.191 - 1.792) / 40 = -0.0400 s^-1, so RC = 25 s

Example 3: nuclear count rate with background

A radioactive source is measured with a background count rate of 18 counts s^-1.

t / s	Measured C / counts s^-1	Corrected C - C_bg	ln(C - C_bg)
0	238	220	5.394
75	174	156	5.049
150	128	110	4.700
225	96	78	4.357
300	73	55	4.007

Plot ln(C - C_bg) against t. Do not log the uncorrected measured count rate.

gradient = (4.007 - 5.394) / 300 = -0.00462 s^-1

λ = 0.00462 s^-1, so T_1/2 = ln(2) / λ = 0.693 / 0.00462 = 150 s

Example 4: attenuation in medical imaging

For X-ray or gamma photons through tissue, intensity often follows I = I₀e^-μx.

x / cm	I / % of incident intensity	ln(I / I₀)
0	100	0.000
2	69.8	-0.360
4	48.7	-0.720
6	34.0	-1.079

Plot ln(I/I₀) against x. The line passes through the origin because ln(1) = 0.

gradient = -0.180 cm^-1, so μ = 0.180 cm^-1

half-value thickness = ln(2) / μ = 0.693 / 0.180 = 3.85 cm

This is the same mathematics as radioactive decay: distance through material replaces time, and attenuation coefficient replaces decay constant.

07 Common traps

Common mistakes and quick fixes

Logging before rearranging

For charging capacitors, ln(V_C) is not linear. Use ln(E - V_C).

Forgetting background

For radioactivity, subtract background count rate before taking logs. Otherwise the graph curves and the half-life is wrong.

Losing the minus sign

If the quantity decays, the gradient is negative. The physical constant RC, λ or μ is positive.

Using two rounded points badly

Use a best-fit line over the full plotted range. Two rounded table values can shift the gradient noticeably.

08 Practice

Self-check questions

1. A graph of ln(V) against t for a discharging capacitor has gradient -0.0125 s^-1. Find RC.

gradient = -1/RC, so RC = -1 / -0.0125 = 80 s.

2. Which graph should you plot for V_C = E(1 - e^-t/RC)?

Plot ln(E - V_C) against t. The gradient is -1/RC.

3. A corrected count rate halves every 4.0 hours. What is the decay constant?

λ = ln(2) / T_1/2 = 0.693 / 4.0 = 0.173 h^-1.

4. Intensity follows I = I₀e^-μx. What does the gradient of ln(I/I₀) against x equal?

The gradient is -μ. If you plot ln(I₀/I) against x instead, the gradient is +μ.

09 Exam method

Final checklist for graphing exponentials

Identify the physical equation and the independent variable: time, thickness or distance.
Rearrange so the exponential term is isolated. Subtract background or final value where needed.
Take natural logs and compare with y = mx + c.
Label the vertical axis as the logged quantity, for example ln(V) or ln(C - C_bg).
Use the gradient to find the constant, keeping the sign clear.
Convert the intercept back with e^intercept if you need the initial value.

Exponential Equations: using e, ln and straight-line graphs

Work through it in three passes

Understand the shape

Choose the right plot

Do the graph task

What an exponential equation is really saying

Decay form

Approach-to-final form

The rules you actually use

Half left

37% left

63% reached

Turning an exponential curve into a straight line

Original graph: curved decay

Linearised graph: straight line

Where this appears in Module 6

Capacitor discharge

Capacitor charging

Radioactive decay

Count rate with background

X-ray or gamma attenuation

Medical tracers and PET

Plot the graph yourself, then check it

Your task

What success looks like

Worked examples with graph choices

Example 1: capacitor discharge

Example 2: capacitor charging needs rearranging

Example 3: nuclear count rate with background

Example 4: attenuation in medical imaging

Common mistakes and quick fixes

Logging before rearranging

Forgetting background

Losing the minus sign

Using two rounded points badly

Self-check questions

Final checklist for graphing exponentials