Exponential equations - Year 10 Mathematics

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.)

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Year 10 core - answers

Fluency

Tier 1: basic skills

Initial value $a = 7$ , base $b = 2$ .
$x = 0$ : $y = 1$ . $x = 1$ : $y = 4$ . $x = 3$ : $y = 64$ .
Decay (since $0 < 0.6 < 1$ ).
$y = 100 \times 1.2^t$ .
$81 = 3^4$ , so $x = 4$ .
$\tfrac{1}{8} = 2^{-3}$ , so $x = -3$ .
$10000 = 10^4$ , so $x = 4$ .
Number of half-lives: $\tfrac{12}{4} = 3$ . Mass: $160 \times \left(\tfrac{1}{2}\right)^3 = 160 \times \tfrac{1}{8} = 20$ g.

Reasoning

Tier 2: mixed practice

$64 = 4^3$ , so $x = 3$ .
$25 = 5^2$ , $125 = 5^3$ . $(5^2)^x = 5^3$ . $2x = 3$ . $x = \tfrac{3}{2}$ .
Model: $N = 800 \times 3^{t/5}$ . After $15$ days: $N = 800 \times 3^3 = 800 \times 27 = 21600$ insects.
Half-lives: $\tfrac{18}{6} = 3$ . Mass: $500 \times \left(\tfrac{1}{2}\right)^3 = 500 \times \tfrac{1}{8} = 62.5$ g. Model: $M = 500 \times \left(\tfrac{1}{2}\right)^{t/6}$ .
Rule of 70: $\tfrac{70}{8} = 8.75$ years. Check: $1000 \times 1.08^9 = 1000 \times 1.999 \approx 1999$ dollars. Confirms doubling in about $9$ years.
$5^x = 125 = 5^3$ . $x = 3$ .

Reasoning

Tier 3: explain and apply

Since $b > 0$ , raising $b$ to any power gives a positive result: $b^x > 0$ for all $x$ . Multiplying by $a > 0$ keeps it positive. So $y = a \times b^x > 0$ always. The curve approaches the $x$ -axis (its asymptote) but never touches or crosses it.
Colony A: $N_A = 100 \times 2^{t/2}$ . Colony B: $N_B = 3200 \times \left(\tfrac{1}{2}\right)^{t/2}$ . Set equal: $100 \times 2^{t/2} = 3200 \times 2^{-t/2}$ . Multiply both sides by $2^{t/2}$ : $100 \times 2^t = 3200$ . $2^t = 32 = 2^5$ . $t = 5$ . They are equal after $\mathbf{10}$ hours. Wait — let $u = t/2$ : $100 \times 2^u = 3200 \times 2^{-u}$ , $100 \times 2^{2u} = 3200$ , $2^{2u} = 32 = 2^5$ , $2u = 5$ , $u = 2.5$ , $t = 5$ hours. Check: $N_A = 100 \times 2^{2.5} \approx 566$ , $N_B = 3200 \times 2^{-2.5} \approx 566$ . Equal after 5 hours.
$V = 30000 \times 0.88^t$ . Need $V < 10000$ : $0.88^t < \tfrac{1}{3}$ . By trial: $0.88^8 \approx 0.3596$ , $0.88^9 \approx 0.3165$ , $0.88^{10} \approx 0.2785$ . After $9$ years: $V \approx 9494 < 10000$ . First worth less than $10000 after 9 whole years.
Half-life $= 5$ years (the denominator in the exponent). After $20$ years: $M = 800 \times \left(\tfrac{1}{2}\right)^4 = 800 \times \tfrac{1}{16} = 50$ g. Below $100$ g: $\left(\tfrac{1}{2}\right)^{t/5} < \tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3$ , so $\tfrac{t}{5} > 3$ , $t > 15$ years.

Reasoning

Challenge

$8 = 2^3$ and $4 = 2^2$ . So $2^{3(2x+1)} = 2^{2(3x-2)}$ . $6x + 3 = 6x - 4$ . $3 = -4$ , which is false. No solution.
Let $P_n$ be the population after year $n$ . $P_0 = 10000$ . $P_{n+1} = 0.9 P_n + 500$ . Year 1: $0.9(10000) + 500 = 9500$ . Year 2: $0.9(9500) + 500 = 9050$ . Year 3: $0.9(9050) + 500 = 8645$ . The population is declining. (It would stabilise at $P = 0.9P + 500$ , so $0.1P = 500$ , $P = 5000$ , but it has not reached that level yet.)
Need $5000 \times 1.06^t > 8000 \times 1.03^t$ . $\left(\tfrac{1.06}{1.03}\right)^t > 1.6$ . $1.02913^t > 1.6$ . By trial: $t = 16$ : $1.02913^{16} \approx 1.585$ ; $t = 17$ : $\approx 1.631$ . Investment A first exceeds B after 17 whole years.

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