Exponential equations

What you will learn

recognise and write exponential functions in the form $y = a \times b^x$ ,
distinguish growth ( $b > 1$ ) from decay ( $0 < b < 1$ ),
solve simple exponential equations by expressing both sides as powers of the same base,
calculate doubling time and half-life,
apply exponential models to real-world contexts such as populations and radioactive decay.

Worked example 0 Real-world example: bacterial growth

A petri dish starts with $500$ bacteria. The population doubles every $3$ hours. How many bacteria after $12$ hours?

Each $3$ -hour period the population multiplies by $2$ .
In $12$ hours there are $\dfrac{12}{3} = 4$ doubling periods.
Population: $500 \times 2^4 = 500 \times 16 = 8000$ bacteria.
General model: $N = 500 \times 2^{t/3}$ , where $t$ is time in hours.

Key idea: the base $2$ tells us the quantity doubles; the exponent $t/3$ counts how many doublings have occurred.

1. What is an exponential function?

An exponential function has the variable in the exponent:

Formula reference

y = a \times b^x

$a$ = initial value (the $y$ -value when $x = 0$ , since $b^0 = 1$ ).
$b$ = base or growth/decay factor.
$x$ = independent variable (often time).

Worked example 1 Identifying parts of an exponential function

For $y = 3 \times 5^x$ , state the initial value and the base. Find $y$ when $x = 0$ , $x = 1$ , and $x = 2$ .

Initial value $a = 3$ , base $b = 5$ .
$x = 0$ : $y = 3 \times 1 = 3$ . $x = 1$ : $y = 3 \times 5 = 15$ . $x = 2$ : $y = 3 \times 25 = 75$ .
Each step multiplies by $5$ — this is exponential growth.

2. Growth vs decay

Formula reference

Growth: $b > 1$ . Each period the quantity is multiplied by a factor greater than $1$ .
Decay: $0 < b < 1$ . Each period the quantity is multiplied by a factor less than $1$ (it shrinks).

A percentage increase of $r\%$ gives $b = 1 + \dfrac{r}{100}$ .

A percentage decrease of $r\%$ gives $b = 1 - \dfrac{r}{100}$ .

Worked example 2 Writing a decay model

A car is worth $40000 and loses $15\%$ of its value each year. Write an exponential model for its value $V$ after $t$ years.

Initial value: $a = 40000$ .
Decay factor: $b = 1 - 0.15 = 0.85$ .
Model: $V = 40000 \times 0.85^t$ .
After $3$ years: $V = 40000 \times 0.85^3 = 40000 \times 0.6141 \approx 24565$ dollars.

3. Solving simple exponential equations

When both sides can be written as powers of the same base, set the exponents equal.

Worked example 3 Solving by matching bases

Solve $2^x = 32$ .

Write $32$ as a power of $2$ : $32 = 2^5$ .
So $2^x = 2^5$ , giving $x = 5$ .

Worked example 4 Solving with a common base

Solve $9^x = 27$ .

Write both as powers of $3$ : $9 = 3^2$ and $27 = 3^3$ .
$(3^2)^x = 3^3$ , so $3^{2x} = 3^3$ .
$2x = 3$ , giving $x = \tfrac{3}{2}$ .

Worked example 5 Solving with a coefficient

Solve $5 \times 3^x = 405$ .

Divide both sides by $5$ : $3^x = 81$ .
$81 = 3^4$ , so $x = 4$ .

4. Half-life and doubling time

Formula reference

Doubling time: the time for a quantity to double. If $y = a \times 2^{t/d}$ , then $d$ is the doubling time.

Half-life: the time for a quantity to halve. If $y = a \times \left(\tfrac{1}{2}\right)^{t/h}$ , then $h$ is the half-life.

Worked example 6 Using half-life

A radioactive sample has a mass of $200$ g and a half-life of $10$ years. Find the mass after $30$ years.

Number of half-lives: $\dfrac{30}{10} = 3$ .
Mass: $200 \times \left(\tfrac{1}{2}\right)^3 = 200 \times \tfrac{1}{8} = 25$ g.
General model: $M = 200 \times \left(\tfrac{1}{2}\right)^{t/10}$ .

Worked example 7 Finding doubling time

A town’s population is $12000$ and grows at $5\%$ per year. After how many years will it double?

Model: $P = 12000 \times 1.05^t$ . We need $P = 24000$ .
$1.05^t = 2$ .
By trial: $1.05^{14} \approx 1.980$ , $1.05^{15} \approx 2.079$ .
The population doubles between year $14$ and year $15$ ; approximately $14.2$ years.

Quick estimate: the Rule of 70 says doubling time $\approx \dfrac{70}{r}$ where $r$ is the percentage rate. Here $\tfrac{70}{5} = 14$ years.

Practice

Fluency

Tier 1: basic skills

For $y = 7 \times 2^x$ , state the initial value and the base.
Evaluate $y = 4^x$ when $x = 0$ , $x = 1$ , $x = 3$ .
State whether $y = 0.6^x$ represents growth or decay.
A quantity starts at $100$ and increases by $20\%$ each year. Write the exponential model.
Solve $3^x = 81$ .
Solve $2^x = \tfrac{1}{8}$ .
Solve $10^x = 10000$ .
A substance has a half-life of $4$ hours. If you start with $160$ g, how much remains after $12$ hours?

Reasoning

Tier 2: mixed practice

Solve $4^x = 64$ .
Solve $25^x = 125$ .
A population of $800$ insects triples every $5$ days. Write a model and find the population after $15$ days.
A sample of $500$ g has a half-life of $6$ years. Find the mass after $18$ years. Write the general model.
A savings account starts with $1000 and earns $8\%$ per year (compounded annually). Use the Rule of 70 to estimate the doubling time. Then calculate $1000 \times 1.08^9$ to check.
Solve $2 \times 5^x = 250$ .

Reasoning

Tier 3: explain and apply

Explain why an exponential decay function $y = a \times b^x$ (with $0 < b < 1$ ) never reaches zero, no matter how large $x$ is.
Two bacteria colonies start at the same time. Colony A has $100$ bacteria and doubles every $2$ hours. Colony B has $3200$ bacteria and halves every $2$ hours. After how many hours do they have the same population?
A car worth $30000 depreciates at $12\%$ per year. After how many whole years is it first worth less than $10000?
The mass of a radioactive isotope is modelled by $M = 800 \times \left(\tfrac{1}{2}\right)^{t/5}$ . Find the half-life, the mass after $20$ years, and the time when the mass first drops below $100$ g.

Challenge

Reasoning

Harder reasoning

Solve $8^{2x+1} = 4^{3x-2}$ . (Hint: express both sides as powers of $2$ .)
A lake contains $10000$ fish. Due to overfishing the population drops by $10\%$ each year, but each year $500$ fish are also added by restocking. Write a recurrence relation and find the population after $3$ years. Is the population stabilising, growing, or declining?
Two investments start at the same time. Investment A is $5000 growing at $6\%$ p.a. Investment B is $8000 growing at $3\%$ p.a. After how many whole years does Investment A first exceed Investment B? (Use trial and improvement.)