📈 Exponential Functions
From capacitor discharge to bacterial growth, exponential functions model any quantity that changes at a rate proportional to itself. Work through each tab: rules first, then the interactive tools, then worked examples and practice.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
📈Growth & Decayy = Ae^(kt)
🔍Solving EquationsUsing ln to isolate x
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Model growth and decay with y = Aekt; evaluate exponential expressions; and solve equations where the unknown sits in the exponent, using natural logarithms.
💡 e ≈ 2.71828 is not a rounded convenience — it's the unique base for which the rate of change of eˣ is eˣ itself. That's why it appears throughout engineering.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
📈
Exponential Growth & Decay
One equation, two behaviours — determined entirely by the sign of k.
🔧Growth & Decay ExplorerEnter values, then press Calculate
Enter A, k and t for y = Aekt, then press Calculate. Use negative k for decay, positive k for growth…
📖 Key Vocabulary
Euler's number, e — an irrational constant ≈ 2.71828, the base of natural growth.
Rate constant, k — controls how fast growth or decay happens; larger |k| means faster change.
Asymptote — the line y=0 that a decay curve approaches but never touches.
Time constant, τ = 1/k — the time for decay to fall to 1/e (≈37%) of its start value.
👁️ Visual aid: both curves start at the same point (t=0, y=A) but head in opposite directions — up without limit, or down towards (never reaching) zero.
✏️Worked Example 1 — Evaluating an exponentialType: Direct evaluation
Question Evaluate y = 5e0.2×3 to 3 s.f.
Exponent
0.2 × 3 = 0.6Compute
y = 5 × e^0.6 = 5 × 1.8221Answer y ≈ 9.11
📉Worked Example 2 — Decay to a fraction of the start valueType: Fractional decay
Question A quantity decays as y = Ae−0.1t. Find the time for y to fall to 25% of A.
Set up
0.25A = Ae−0.1t → 0.25 = e−0.1tTake ln
ln 0.25 = −0.1t → −1.386 = −0.1tAnswer t = 13.9 (3 s.f.) time units
📊Worked Example 3 — Comparing two growth ratesType: Comparison
Question Population A grows as 200e0.03t, population B as 150e0.05t. After how long does B overtake A?
Set equal
200e0.03t = 150e0.05tRearrange
200/150 = e0.02t → ln(1.333) = 0.02tAnswer
t = 0.2877/0.02 ≈ 14.4 time units — despite starting smaller, B's faster rate lets it overtake A.⚙️Worked Example 4 — Engineering context (capacitor discharge)Type: Engineering
Question A capacitor discharges as V = 12e−t/RC with RC = 2 s. Find V after 5 s.
Substitute
V = 12e−5/2 = 12 × 0.0821Answer V ≈ 0.985 V — after 2.5 time constants, less than 10% of the original voltage remains.
🧠 Common slip: confusing the rate k with the time constant τ = 1/k — they are reciprocals, not the same number.
🔍
Solving Exponential Equations
When the unknown sits in the exponent, taking logarithms of both sides brings it down to ground level.
🔧Exponential Equation SolverSolve A·ekx = c for x
Enter A, k and c for A·ekx = c, then press Solve to see the full working…
📖 Key Vocabulary
Natural logarithm (ln) — the inverse of eˣ; the standard tool for isolating an exponent.
Exponent — the power x in eˣ; the unknown being solved for.
Extraneous solution — a value produced algebraically that fails to satisfy the original equation (rare here, but always worth checking).
Domain restriction — since eˣ > 0 always, A·ekx = c has no solution if c/A ≤ 0.
👁️ Visual aid: ln turns the exponential equation into a linear one in x — solve it exactly like any other linear equation.
✏️Worked Example 1 — Direct application of lnType: Basic solve
Question Solve e3x = 50 for x.
Take ln of both sides
ln(e3x) = ln 50Simplify left side
3x = ln 50 = 3.912Answer x = 3.912 ÷ 3 = 1.304 (3 d.p.)
🔢Worked Example 2 — Isolating the exponential firstType: Isolate then solve
Question Solve 3e2x = 45 for x.
Isolate
e2x = 15Take ln
2x = ln 15 = 2.708Answer x = 1.354 (3 d.p.)
🧮Worked Example 3 — Equation with an added constantType: Rearrange first
Question Solve 2ex + 3 = 19 for x.
Isolate the exponential
2ex = 16 → ex = 8Take ln
x = ln 8Answer x = 2.079 (3 d.p.) — always strip off any added or multiplying terms before taking logs.
⚙️Worked Example 4 — Engineering context (time to a target voltage)Type: Engineering
Question A capacitor discharges as V = 12e−t/2. Find the time for V to fall to 1 V.
Set up
1 = 12e−t/2 → e−t/2 = 1/12Take ln
−t/2 = ln(1/12) = −2.485Answer t = 4.97 s — solving for time-to-target is one of the most common exponential problems in circuit design.
🧠 Common slip: taking ln before isolating the exponential — ln(2ex+3) cannot be simplified with a single log law, so always tidy the equation into A·ekx = c form first.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔋 RC Circuit — Capacitor Charging
A charging capacitor follows V = V_s(1 − e−t/RC), approaching the supply voltage but never quite reaching it.
V(t) = V_s (1 − e^(−t/RC))
📟 Try it — Capacitor Charging Voltage
Enter supply voltage, R, C and time, then press Calculate…
☢️ Radioactive Decay — Half-Life
N = N₀e−λt models the number of undecayed nuclei remaining after time t, where λ is the decay constant.
N = N₀ e^(−λt)
📟 Try it — Remaining Quantity
Enter N₀, λ and t, then press Calculate…
🌡️ Newton's Law of Cooling
An object's temperature difference from its surroundings decays exponentially: θ = θ₀e−kt, where θ is the excess temperature.
θ(t) = θ₀ e^(−kt)
📟 Try it — Cooling Temperature
Enter θ₀, k and t, then press Calculate…
🦠 Bacterial Growth — Doubling Time
Under ideal conditions, bacteria grow as N = N₀ekt, doubling at regular intervals determined by k.
Doubling time = ln(2) ÷ k
📟 Try it — Doubling Time Calculator
Enter the growth rate k, then press Calculate…
🎮
Quick-Fire Exponential Game
12 questions on growth, decay and solving. How many can you get?
⚡ Exponential Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Evaluate e² to 3 s.f.
e² = 7.39
Question 2 🥉2 marks
Evaluate 3e−1 to 3 s.f.
3 × 0.3679 = 1.10
Question 3 🥉2 marks
Solve eˣ = 20 for x.
x = ln20 = 3.00 (3 s.f.)
Question 4 🥉2 marks
State whether y = 4e−0.2t represents growth or decay, and why.
The exponent's coefficient is −0.2 (negative) → decay — y falls towards zero as t increases.
Question 5 🥈3 marks
Solve 3e2x = 45 for x.
e^2x = 15
2x = ln15 = 2.708
x = 1.354 (3 d.p.)
2x = ln15 = 2.708
x = 1.354 (3 d.p.)
Question 6 🥈4 marks
A quantity decays as y = 60e−0.08t. Find the time for y to reach 15.
15 = 60e^(−0.08t) → 0.25 = e^(−0.08t)
ln0.25 = −0.08t → −1.386 = −0.08t
t ≈ 17.3
ln0.25 = −0.08t → −1.386 = −0.08t
t ≈ 17.3
Question 7 🥈3 marks
Solve 5eˣ − 2 = 33 for x.
5eˣ = 35 → eˣ = 7
x = ln7 = 1.946 (3 d.p.)
x = ln7 = 1.946 (3 d.p.)
Question 8 🥈4 marks
Two exponential models 100e0.04t and 80e0.06t — find when they're equal.
100/80 = e^(0.02t) → ln1.25 = 0.02t
0.2231 = 0.02t
t ≈ 11.2
0.2231 = 0.02t
t ≈ 11.2
Question 9 🥇4 marks
A radioactive sample decays as N = N₀e−0.03t (t in days). Find the half-life.
0.5 = e^(−0.03t)
ln0.5 = −0.03t → −0.6931 = −0.03t
t ≈ 23.1 days
ln0.5 = −0.03t → −0.6931 = −0.03t
t ≈ 23.1 days
Question 10 🥇5 marks
A capacitor charges as V = 10(1−e−t/5). Find the time to reach 9 V.
9 = 10(1−e^(−t/5)) → 0.9 = 1−e^(−t/5)
e^(−t/5) = 0.1 → −t/5 = ln0.1 = −2.303
t ≈ 11.5 s
e^(−t/5) = 0.1 → −t/5 = ln0.1 = −2.303
t ≈ 11.5 s
Question 11 🥇5 marks
Bacteria grow as N = N₀e0.4t (t in hours). Find the doubling time.
2N₀ = N₀e^(0.4t) → 2 = e^(0.4t)
ln2 = 0.4t → 0.6931 = 0.4t
t ≈ 1.73 hours
ln2 = 0.4t → 0.6931 = 0.4t
t ≈ 1.73 hours
Question 12 🥇5 marks
An engine cools as θ = 90e−0.05t (θ in °C excess, t in minutes). Find the excess temperature after 20 min, and the time for it to fall below 10°C.
θ(20) = 90e^(−1) = 90×0.3679 = 33.1°C
10 = 90e^(−0.05t) → e^(−0.05t)=1/9
−0.05t = ln(1/9) = −2.197
t ≈ 43.9 min
10 = 90e^(−0.05t) → e^(−0.05t)=1/9
−0.05t = ln(1/9) = −2.197
t ≈ 43.9 min