← A-Level Mathematics · OCR · A-Level Maths revision

OCR A-level Mathematics A (H240) · 1.06 Exponentials and Logarithms

Mini-Lesson

Exponentials and logarithms

3 missions. Guess, learn, then check. Your place is saved.

Mission 1Compound 100% interest ever more often. Does £1 grow without limit?
Mission 2How many times must you fold paper for it to reach the Moon?
Mission 3Tea cools from 90 °C to 60 °C in 10 minutes. When will it be 40 °C?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Compound 100% interest ever more often. Does £1 grow without limit?

Make a guess: Compound 100% interest ever more often. Does £1 grow without limit?

Exponentials and logarithms · exponential functions and e

The graph of $y = a^{x}$ y = a^x

For $a > 0$ a > 0, $y = a^{x}$ y = a^x is an exponential function: the variable is in the power.
Every graph passes through $(0, 1)$ (0, 1), because $a^{0} = 1$ a⁰ = 1.
$y > 0$ y > 0 for all $x$ x. The $x$ x-axis, $y = 0$ y = 0, is an asymptote.
$a > 1$ a > 1: increasing (growth). $0 < a < 1$ 0 < a < 1: decreasing (decay).
$y = a^{- x} = (\frac{1}{a})^{x}$ y = a^(−x) = (1/a)^x is the reflection of $y = a^{x}$ y = a^x in the $y$ y-axis.

Quick check

?Which statement about the curve

y = 5 \times 0. 8^{x}

is true?

Worked example Exponentials and logarithms · exponential functions and e

Compounding more often

Question

£1 earns 100% interest a year. Find its value after one year if the interest is paid once, twice, 4 times and 12 times, each time at an equal share of the rate.

1
Paid $n$ n times, each payment multiplies by $(1 + \frac{1}{n})$ (1 + 1/n), so £1 becomes $(1 + \frac{1}{n})^{n}$ (1 + 1/n)ⁿ
2
Once: $(1 + 1)^{1} = 2$ (1 + 1)¹ = 2. Twice: $1. 5^{2} = 2.25$ 1.5² = 2.25
3
4 times: $1.2 5^{4} = 2.441 \dots$ 1.25⁴ = 2.441…. 12 times: $(1 + \frac{1}{12})^{12} = 2.613 \dots$ (1 + 1/12)¹² = 2.613…

Answer

£2, £2.25, £2.44, £2.61

⚠️ Watch out

twice a year is not 100% twice. Each payment is 50%, so the multiplier is 1.5, not 2.

Calculate

?The same £1 at 100% a year is compounded daily, 365 times. Find its value after one year. Give your answer to the nearest penny.

Exponentials and logarithms · exponential functions and e

The number e

As $n$ n grows, $(1 + \frac{1}{n})^{n}$ (1 + 1/n)ⁿ gets closer to a fixed number, e $= 2.71828 \dots$ = 2.71828… (irrational, like $π$ π).
Interest added continuously at rate $r$ r per year turns $P$ P into $P e^{r t}$ Pe^rt after $t$ t years: continuous compound interest.
$y = e^{x}$ y = e^x has the shape of every $a^{x}$ a^x with $a > 1$ a > 1: through $(0, 1)$ (0, 1), asymptote $y = 0$ y = 0.
$y = e^{a x + b} + c$ y = e^(ax+b) + c has asymptote $y = c$ y = c and meets the $y$ y-axis at $e^{b} + c$ e^b + c.

Exponentials and logarithms · exponential functions and e

Why e is special: its gradient

The gradient of $y = e^{x}$ y = e^x at any point equals its $y$ y-value. At $(1, e)$ (1, e) the gradient is e.
More generally:

\frac{d}{d x} (e^{k x}) = k e^{k x}

So for $y = A e^{k x}$ y = Ae^kx, the rate of change is $k y$ ky: proportional to $y$ y.
That is why exponential models suit populations, money, radioactive decay and cooling: the bigger the amount, the faster it changes.

Worked example Exponentials and logarithms · exponential functions and e

A rate of growth

Question

A population is modelled by

P = 200 e^{0.03 t}

, where

t

is in years. Find the rate at which the population is growing when

t = 10

1
Differentiate using $k = 0.03$ k = 0.03: $\frac{d P}{d t} = 0.03 \times 200 e^{0.03 t} = 6 e^{0.03 t}$ dP/dt = 0.03 × 200e^0.03t = 6e^0.03t
2
Substitute $t = 10$ t = 10: $6 e^{0.3} = 8.099 \dots$ 6e^0.3 = 8.099…
3
Check with "rate $= k P$ = kP": $P = 200 e^{0.3} = 269.97 \dots$ P = 200e^0.3 = 269.97… and $0.03 \times 269.97 = 8.10$ 0.03 × 269.97 = 8.10

Answer

about 8.10 per year

⚠️ Watch out

the derivative is

6 e^{0.03 t}

, not

200 e^{0.03 t}

. Forgetting the factor

k

gives 270, which is the population, not its rate of change.

Calculate

?£500 is invested at 4% a year, compounded continuously. Find its value after 10 years. Give your answer to the nearest penny.

Calculate

?Find the gradient of the curve

y = e^{- 0.5 x}

at the point where

x = 2

. Give your answer to 3 significant figures.

Mission check

?£1 earns 100% interest a year, compounded more and more often. Does it grow without limit?

Mission 2

How many times must you fold paper for it to reach the Moon?

Make a guess: How many times must you fold paper for it to reach the Moon?

Exponentials and logarithms · logarithms and their laws

A logarithm is a power

$lo g_{a} b$ log_a b is the power you raise $a$ a to in order to get $b$ b:

a^{x} = b ⟺ x = lo g_{a} b (a > 0, a \neq = 1, b > 0)

Example: $2^{5} = 32$ 2⁵ = 32, so $lo g_{2} 32 = 5$ log₂ 32 = 5.
Always: $lo g_{a} a = 1$ log_a a = 1 and $lo g_{a} 1 = 0$ log_a 1 = 0.
The natural logarithm $ln x = lo g_{e} x$ ln x = log_e x. So $ln e^{x} = x$ ln e^x = x and $e^{l n x} = x$ e^(ln x) = x: they are inverse functions.
$y = ln x$ y = ln x is the reflection of $y = e^{x}$ y = e^x in $y = x$ y = x: through $(1, 0)$ (1, 0), asymptote $x = 0$ x = 0, defined only for $x > 0$ x > 0.

Quick check

?What is the value of

lo g_{3} (\frac{1}{9})

Exponentials and logarithms · logarithms and their laws

The laws of logarithms

Because logs are powers, the index laws become:

$lo g_{a} x + lo g_{a} y = lo g_{a} (x y) lo g_{a} x - lo g_{a} y = lo g_{a} (\frac{x}{y})$ log_a x + log_a y = log_a (xy) log_a x − log_a y = log_a(x/y) $k lo g_{a} x = lo g_{a} (x^{k}), e.g. - lo g_{a} x = lo g_{a} \frac{1}{x}, - \frac{1}{2} lo g_{a} x = lo g_{a} \frac{1}{x}$ klog_a x = log_a(x^k), e.g. −log_a x = log_a 1/x, −1/2log_a x = log_a 1/(√x)

Proof of the first: let $x = a^{p}$ x = a^p, $y = a^{q}$ y = a^q. Then $x y = a^{p + q}$ xy = a^(p+q), so $lo g_{a} (x y) = p + q$ log_a(xy) = p + q.
Power law: $x^{k} = (a^{p})^{k} = a^{k p}$ x^k = (a^p)^k = a^kp, so $lo g_{a} (x^{k}) = k p = k lo g_{a} x$ log_a(x^k) = kp = klog_a x.
There is no law for $lo g_{a} (x + y)$ log_a(x + y).

Worked example Exponentials and logarithms · logarithms and their laws

Combining logs to solve an equation

Question

Solve

lo g_{2} x + lo g_{2} (x + 2) = 3

1
Combine into one log: $lo g_{2} (x (x + 2)) = 3$ log₂(x(x + 2)) = 3
2
Undo the log: $x (x + 2) = 2^{3} = 8$ x(x + 2) = 2³ = 8
3
Solve the quadratic: $x^{2} + 2 x - 8 = 0$ x² + 2x − 8 = 0, so $(x + 4) (x - 2) = 0$ (x + 4)(x − 2) = 0
4
Check each root: $x = - 4$ x = −4 makes $lo g_{2} x$ log₂ x undefined, so reject it

Answer

x = 2

⚠️ Watch out

always check roots in the original equation. You cannot take the log of a negative number or zero.

Calculate

?Solve

lo g_{2} (x + 10) - lo g_{2} (x - 4) = 3

Exponentials and logarithms · logarithms and their laws

Solving $a^{x} = b$ a^x = b by taking logs

Take $ln$ ln (or $lo g_{10}$ log₁₀) of both sides; the power law brings $x$ x down:

a^{x} = b ⟹ x ln a = ln b ⟹ x = \frac{ln b}{ln a}

For $e^{a x + b} = p$ e^(ax+b) = p, take $ln$ ln: $a x + b = ln p$ ax + b = ln p. For $ln (a x + b) = q$ ln (ax + b) = q, use e: $a x + b = e^{q}$ ax + b = e^q.
Inequalities: if $0 < a < 1$ 0 < a < 1 then $ln a < 0$ ln a < 0, so dividing by $ln a$ ln a reverses the inequality sign.
For OCR, solving inequalities like this (for example in compound-interest and other sequence models) is A-level only content, not AS.

Worked example Exponentials and logarithms · logarithms and their laws

An exponential equation

Question

Solve

3^{2 x - 1} = 20

, giving

x

to 3 significant figures.

1
Take ln of both sides: $(2 x - 1) ln 3 = ln 20$ (2x − 1)ln 3 = ln 20
2
Divide by $ln 3$ ln 3: $2 x - 1 = \frac{ln 20}{ln 3} = 2.7268 \dots$ 2x − 1 = (ln 20)/(ln 3) = 2.7268…
3
Solve the linear equation: $2 x = 3.7268 \dots$ 2x = 3.7268…, so $x = 1.8634 \dots$ x = 1.8634…

Answer

x = 1.86

⚠️ Watch out

\frac{ln 20}{ln 3}

is not

ln \frac{20}{3}

(that is 1.90, and it gives

x = 1.45

Calculate

?A ball is dropped and each bounce reaches 0.9 times the height of the one before. After

n

bounces the height is

0. 9^{n}

of the original. Find the smallest whole number

n

for which

0. 9^{n} < 0.01

Calculate

?Solve

ln (3 x - 2) = 4

. Give

x

to 2 decimal places.

Mission check

?A sheet of paper is 0.1 mm thick and each fold doubles its thickness. The Moon is on average 384,400 km away. How many folds until the stack reaches the Moon?

Mission 3

Tea cools from 90 °C to 60 °C in 10 minutes. When will it be 40 °C?

Make a guess: Tea cools from 90 °C to 60 °C in 10 minutes. When will it be 40 °C?

Exponentials and logarithms · modelling with exponentials and logs

Exponential growth and decay models

$y = A e^{k t}$ y = Ae^kt: growth if $k > 0$ k > 0, decay if $k < 0$ k < 0. Equivalently $y = A b^{t}$ y = Ab^t.
$A$ A is the initial value, the value when $t = 0$ t = 0.
The rate of change is $k y$ ky: proportional to the amount present.
To find the constants: put $t = 0$ t = 0 to get $A$ A, then use a second reading and take ln to get $k$ k.

Worked example Exponentials and logarithms · modelling with exponentials and logs

Drug concentration

Question

The concentration of a drug in the blood is modelled by

C = 8 e^{- k t}

mg/l,

t

hours after a dose. After 3 hours

C = 5

. Find

k

, and the time when

C

first falls to 2 mg/l.

1
Substitute the reading: $5 = 8 e^{- 3 k}$ 5 = 8e^(−3k), so $e^{- 3 k} = 0.625$ e^(−3k) = 0.625
2
Take ln: $- 3 k = ln 0.625$ −3k = ln 0.625, so $k = \frac{1}{3} ln 1.6 = 0.1567$ k = 1/3ln 1.6 = 0.1567 (4 s.f.)
3
Set $C = 2$ C = 2: $e^{- k t} = 0.25$ e^(−kt) = 0.25, so $t = \frac{ln 4}{k} = 8.848 \dots$ t = (ln 4)/k = 8.848…

Answer

k = 0.157

(3 s.f.);

C = 2

after about 8.85 hours

⚠️ Watch out

keep

k

unrounded in step 3. Using

k = 0.16

gives 8.66 hours.

Calculate

?A population of 500 grows to 800 in 4 years. It is modelled by

P = A e^{k t}

, with

t

in years. Use the model to predict the population after 10 years. Give your answer to the nearest whole number.

Exponentials and logarithms · modelling with exponentials and logs

Models that level off

Some quantities decay towards a value other than 0. Cooling tea approaches room temperature.
Newton's law of cooling (a model): the excess temperature above the room decays exponentially. In a 20 °C room:

θ = 20 + A e^{- k t}

As $t \to \infty$ t → ∞, $e^{- k t} \to 0$ e^(−kt) → 0, so $θ \to 20$ θ → 20: a horizontal asymptote. Initially $θ = 20 + A$ θ = 20 + A.
Limitations: unlimited growth is impossible (food, space); a constant room temperature is an assumption. Refine the model when its predictions disagree with data.

Calculate

?A dish from an oven cools in a kitchen according to

θ = 20 + 160 e^{- 0.05 t}

, where

θ

°C is its temperature

t

minutes after leaving the oven. Find the time when its temperature is 60 °C. Give your answer to 1 decimal place.

minutes

Quick check

?The number of rabbits on an island is modelled by

R = 50 e^{0.4 t}

, where

t

is in years. Which is the best criticism of the model?

Exponentials and logarithms · modelling with exponentials and logs

Turning curves into straight lines with logs

$y = a x^{n}$ y = axⁿ: take logs, $lo g y = n lo g x + lo g a$ log y = nlog x + log a. Plot $lo g y$ log y against $lo g x$ log x: a straight line, gradient $n$ n, intercept $lo g a$ log a.
$y = k b^{x}$ y = kb^x: take logs, $lo g y = x lo g b + lo g k$ log y = xlog b + log k. Plot $lo g y$ log y against $x$ x: a straight line, gradient $lo g b$ log b, intercept $lo g k$ log k.
A straight line on the log graph tells you which model fits. Undo the logs to recover the constants: $a = 1 0^{intercept}$ a = 10^intercept (or $e^{intercept}$ e^intercept for ln).

Worked example Exponentials and logarithms · modelling with exponentials and logs

Reading a log-linear graph

Question

For some data, the graph of

lo g_{10} y

against

x

is a straight line through

(0, 0.48)

and

(5, 1.48)

. Find

k

and

b

in the model

y = k b^{x}

1
Compare with $lo g_{10} y = x lo g_{10} b + lo g_{10} k$ log₁₀ y = xlog₁₀ b + log₁₀ k
2
Gradient: $\frac{1.48 - 0.48}{5 - 0} = 0.2$ (1.48 − 0.48)/(5 − 0) = 0.2, so $lo g_{10} b = 0.2$ log₁₀ b = 0.2 and $b = 1 0^{0.2} = 1.585$ b = 10^0.2 = 1.585 (4 s.f.)
3
Intercept: $lo g_{10} k = 0.48$ log₁₀ k = 0.48, so $k = 1 0^{0.48} = 3.020$ k = 10^0.48 = 3.020 (4 s.f.)

Answer

y = 3.02 \times 1.5 8^{x}

(3 s.f.)

⚠️ Watch out

the intercept is

lo g_{10} k

, not

k

. Writing

k = 0.48

forgets to undo the log.

Calculate

?For some data, the graph of

ln y

against

ln x

is a straight line through

(0, 1.2)

and

(2, 4.2)

. The data follow

y = a x^{n}

. Use the model to estimate

y

when

x = 4

. Give your answer to 3 significant figures.

Mission check

?Tea cools from 90 °C to 60 °C in 10 minutes in a 20 °C room. Using

θ = 20 + 70 e^{- k t}

, when will it be 40 °C?

Recap

Big ideas

$y = a^{x}$ y = a^x ( $a > 0$ a > 0) passes through $(0, 1)$ (0, 1), stays above the asymptote $y = 0$ y = 0, grows if $a > 1$ a > 1 and decays if $0 < a < 1$ 0 < a < 1.
$(1 + \frac{1}{n})^{n} \to e = 2.718 \dots$ (1 + 1/n)ⁿ → e = 2.718…; continuous compounding gives $P e^{r t}$ Pe^rt.
The gradient of $e^{k x}$ e^kx is $k e^{k x}$ ke^kx: rate of change proportional to size, the signature of exponential growth and decay.
$a^{x} = b ⟺ x = lo g_{a} b$ a^x = b ⇔ x = log_a b; $ln x$ ln x is $lo g_{e} x$ log_e x, the inverse of $e^{x}$ e^x.
$lo g_{a} x + lo g_{a} y = lo g_{a} x y$ log_a x + log_a y = log_a xy, $lo g_{a} x - lo g_{a} y = lo g_{a} \frac{x}{y}$ log_a x − log_a y = log_a x/y, $k lo g_{a} x = lo g_{a} x^{k}$ klog_a x = log_a x^k; check that every log in your answer is defined.
To solve $a^{x} = b$ a^x = b, take logs of both sides: $x = \frac{l n b}{l n a}$ x = (ln b)/(ln a); dividing by a negative log reverses an inequality.
$y = A e^{k t}$ y = Ae^kt: $A$ A is the initial value; find $k$ k from a second reading by taking ln. Models like $θ = 20 + A e^{- k t}$ θ = 20 + Ae^(−kt) level off at 20.
Every model has limits: check long-term predictions and refine when they become unrealistic.
$y = a x^{n}$ y = axⁿ gives a straight line of $lo g y$ log y against $lo g x$ log x (gradient $n$ n); $y = k b^{x}$ y = kb^x gives a straight line of $lo g y$ log y against $x$ x (gradient $lo g b$ log b).

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Exponentials and logarithms for OCR A-level Mathematics A (H240). 🎉

Mission 1Compound 100% interest ever more often. Does £1 grow without limit?
Mission 2How many times must you fold paper for it to reach the Moon?
Mission 3Tea cools from 90 °C to 60 °C in 10 minutes. When will it be 40 °C?

Your stars: 0 / 0

Next: test yourself in the Verify stage.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to A-Level Mathematics · OCR

Specification points and sources for this lesson are listed at the bottom of the page.

Sources (4)

The specification points and sources behind this lesson.

1 OCR A Level Mathematics A H240 specification, 1.06a, 1.06b, 1.06i (Stage 1)
2 OCR A Level Mathematics A H240 specification, 1.06c–1.06g (Stage 1); 1.04k guidance on exponential inequalities
3 OCR A Level Mathematics A H240 specification, 1.06h, 1.06i (Stage 1)
4 NASA Space Place, "How Far Away Is the Moon?": average distance 384,400 km (238,855 miles). https://spaceplace.nasa.gov/s/moon-distance

Exponentials and logarithms

Compound 100% interest ever more often. Does £1 grow without limit?

The graph of y=axy = a^x

Compounding more often

The number e

Why e is special: its gradient

A rate of growth

How many times must you fold paper for it to reach the Moon?

A logarithm is a power

The laws of logarithms

Combining logs to solve an equation

Solving ax=ba^x = b by taking logs

An exponential equation

Tea cools from 90 °C to 60 °C in 10 minutes. When will it be 40 °C?

Exponential growth and decay models

Drug concentration

Models that level off

Turning curves into straight lines with logs

Reading a log-linear graph

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

The graph of $y = a^{x}$ y = a^x

Solving $a^{x} = b$ a^x = b by taking logs