Exponential models for HSC Maths Standard 2

Syllabus dot point

“Model practical problems with exponential functions of the form $y = a b^x$ and interpret growth, decay and asymptotes”

HSCMaths Standard 2Year 12: Algebra13 min readUpdated 20 Jun 2026

Quick answer

An exponential model $y = a b^x$ has initial value $a$ (the $y$ -intercept) and per-period multiplier $b = 1 + r/100$ for growth or $1 - r/100$ for decay; the line $y = 0$ is a horizontal asymptote. A constant ratio between consecutive table values is the signature of an exponential. Solve for the exponent by taking logs of both sides.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to spot when an exponential model applies. It applies whenever the change over each period is a fixed percentage of the current size, such as bacterial growth, radioactive decay, compound interest, or depreciation (the steady loss in value of an asset over time). You then set up $y = a b^x$ with the right base, evaluate it, and solve exponential equations using logarithms. The deeper idea is that exponential change is multiplicative: each step multiplies by the same number $b$ . That is exactly what separates it from linear change, where each step instead adds the same number.

The answer

The standard form

y = a b^x

$a$ is the initial value at $x = 0$ (since $b^0 = 1$ ).
$b$ is the base, the per-unit multiplier.
$b > 1$ : growth. $0 < b < 1$ : decay.

Reading a model is mostly reading these two numbers. The $a$ tells you where the curve starts on the vertical axis; the $b$ tells you, in one number, both the direction (up if $b > 1$ , down if $b < 1$ ) and the speed of the change.

From percentage rate to base $b$

Growth at $r\%$ per period: $b = 1 + \frac{r}{100}$ . So $7\%$ growth gives $b = 1.07$ .
Decay at $r\%$ per period: $b = 1 - \frac{r}{100}$ . So $12\%$ decay gives $b = 0.88$ .

This is the same multiplier used in straight compound interest or declining-balance depreciation. The base is always close to $1$ for ordinary percentage rates, which is a quick sanity check: a base of $0.12$ or $1.5$ for a " $12\%$ " rate is almost certainly a slip.

The asymptote

The graph of $y = a b^x$ (for $a > 0$ ) gets closer and closer to the $x$ -axis but never touches it. The line $y = 0$ is a horizontal asymptote, which means a line a curve keeps approaching but never reaches. So a decaying quantity heads towards zero without ever reaching exactly zero. That is why a model never predicts a population, balance or mass of precisely $0$ . For a growth curve the same line $y = 0$ is the asymptote on the left: as $x$ becomes very negative the curve flattens towards zero, while to the right it climbs without limit.

Solving for the exponent

If $y = a b^x$ and you know $y$ , $a$ and $b$ and want $x$ :

\frac{y}{a} = b^x \implies x = \frac{\log(y / a)}{\log b} = \frac{\ln(y / a)}{\ln b}.

Either log base works, because the bases cancel in the ratio. Use whichever your calculator gives easily. This is the standard "how long until..." calculation. It tells you how long until the investment doubles, the population reaches a target, or the drug concentration drops to a safe level.

Reading off worded problems

"Doubles every $T$ time units" implies $b = 2$ per $T$ , so per unit of time the base is $2^{1/T}$ .
"Half-life of $T$ " implies $b = 0.5$ per $T$ , so per unit $b = 0.5^{1/T}$ .
"Grows at $r\%$ continuously" is technically $e^{r t}$ , but Standard 2 uses discrete compounding, so use $b = 1 + r/100$ .

Spotting an exponential from a table: the constant ratio

When a table is supplied and you must name the model, the test for an exponential is the ratio of consecutive values.

Key fact

An exponential model $y = a b^x$ has initial value $a$ (the $y$ -intercept) and per-period multiplier $b = 1 + \frac{r}{100}$ for growth or $1 - \frac{r}{100}$ for decay. From a table of evenly spaced $x$ -values, the data is exponential exactly when consecutive $y$ -values have a constant ratio (you multiply by the same $b$ each step). A constant difference is linear instead; a constant second difference is quadratic. Solve for an unknown exponent by taking logs of both sides.

Take the decay table $A = 8000, 4000, 2000, 1000, 500$ . The ratios are $\frac{4000}{8000} = \frac{2000}{4000} = \frac{1000}{2000} = \frac{500}{1000} = 0.5$ : constant, so the model is exponential with $b = 0.5$ , and the first value $8000$ is $a$ . The full model is $A = 8000(0.5)^t$ . Contrast a linear table, where you would instead subtract the same amount each step.

Sketch the exponential curve, stage by stage

Graph an exponential model the same way you graph any curve: tabulate, plot, then join with a smooth curve. The build below is a decay, an asset worth $A = 8000(0.5)^t$ dollars after $t$ years (it halves annually), with table values $A = 8000, 4000, 2000, 1000, 500$ at $t = 0, 1, 2, 3, 4$ .

Stage 1, draw and scale the axes. Time $t$ runs along the horizontal axis (0 to 4 years) and the amount $A$ up the vertical axis (0 to 8000 dollars, the starting value). The starting value sets the top of the scale.

Stage 2, plot the table points. Plot $(0, 8000)$ , $(1, 4000)$ , $(2, 2000)$ , $(3, 1000)$ and $(4, 500)$ . Each value is exactly half the one before it, the constant ratio $0.5$ that marks the data as exponential rather than linear.

Stage 3, join the points with a smooth curve. Join the dots with one smooth curve that falls steeply at first and then flattens. Decay curves keep dropping but the drop gets smaller and smaller; the curve never turns back up.

Stage 4, mark the intercept and the asymptote. The curve cuts the vertical axis at $(0, 8000)$ : that intercept is the value of $a$ , the starting amount, because $b^0 = 1$ . The dashed horizontal line $A = 0$ is the asymptote: the curve falls towards it forever but never reaches it, so the asset's modelled value gets tiny but never hits exactly zero.

A growth curve is built the same way, but the points climb (multiply up each step) and the curve rises ever more steeply to the right, with the asymptote $y = 0$ now sitting under the flat left-hand tail.

How exam questions ask about exponential models

The phrasing changes but each version maps to a familiar step:

"Set up / write an equation for..." or "write a model for..." Identify $a$ (the starting value) and $b$ (use $1 + r/100$ for growth, $1 - r/100$ for decay), then write $y = a b^x$ with the variable named.
"Find the value / population / amount after $n$ periods." Substitute $x = n$ and evaluate, rounding sensibly for the context (whole dollars, whole organisms).
"How long until it reaches / doubles / halves / falls below...?" Set $y$ to the target, divide by $a$ , and take logs to solve for $x$ .
"What does the model predict in the long run / does it ever reach zero?" Point to the asymptote $y = 0$ : a decaying quantity approaches but never reaches it.
"Which model fits this table?" Check the ratio of consecutive values: a constant ratio means exponential.
"By what percentage does it change each period?" Read it off the base: $b = 1.07$ is $7\%$ growth, $b = 0.88$ is $12\%$ decay.
"Sketch the graph." Mark the $y$ -intercept $a$ , show the correct direction (rising for $b > 1$ , falling for $b < 1$ ), and draw the curve flattening towards the asymptote $y = 0$ .

Four exponential models: interest, population, half-life and depreciation

Four contexts are worked through: compound interest on a $5000 investment, Sydney's projected population, a half-life decay, and the declining value of a laptop. Each uses the exponential model $y = a \cdot b^t$ with a different base.

Compound interest on a $5000 investment

When a quantity grows by a fixed percentage each period, the multiplier for one period is $b = 1 + r/100$ . Starting with $5000 at $6\%$ per annum compounded annually, the accumulated amount after $t$ years is:

A = 5000(1.06)^t

Evaluate the investment after 10 years: Substituting $t = 10$ into the model gives the balance at that point. We use the known approximation $(1.06)^{10} \approx 1.7908$ . The amount is in dollars.

A = 5000 \times 1.7908 \approx 8954

Find the doubling time using logarithms: To find when $A = 2 \times 5000 = 10\,000$ , that is $ $10\,000$ , we solve $(1.06)^t = 2$ . Taking the natural logarithm of both sides and isolating $t$ gives the doubling time.

t = \frac{\ln 2}{\ln 1.06} \approx \frac{0.6931}{0.0583} \approx 11.9 \text{ years}

Project Sydney's population

Sydney's population was approximately $5.26$ million in $2021$ . Using the ABS estimate of $1.3\%$ annual growth, the multiplier per year is $1.013$ , so the model (with $t$ measured in years from 2021) is:

P = 5.26(1.013)^t \text{ million}

Project the population to 2051: Setting $t = 30$ corresponds to the year 2051. Computing $(1.013)^{30}$ and multiplying by the initial population gives the projected figure.

(1.013)^{30} \approx 1.475

P \approx 5.26 \times 1.475 \approx 7.76 \text{ million}

Decay of a medical isotope from its half-life

A medical isotope used in a Sydney hospital has a half-life of $6$ hours: every $6$ hours, half of it remains. Starting from $80$ mg, the model in hours is $M = 80 (0.5)^{t/6}$ , because raising $0.5$ to the power $t/6$ counts how many $6$ -hour half-lives have elapsed. After $18$ hours, that is three half-lives:

M = 80(0.5)^{18/6} = 80(0.5)^3 = 80 \times 0.125 = 10 \text{ mg}

Depreciation of a laptop

Depreciation at a fixed percentage each year is a decay process: each year the value is multiplied by $b = 1 - 0.20 = 0.80$ . The model for a laptop purchased for $1800 is:

V = 1800(0.80)^t

Find the laptop value after 4 years: Substituting $t = 4$ and using $(0.80)^4 = 0.4096$ gives the laptop's book value at that point, in dollars.

V = 1800 \times 0.4096 \approx 737.28

Find when the laptop value falls below $500: We need the smallest whole number of years for which $V < 500$ . Set up the equation $1800(0.80)^t = 500$ , divide, and take logs.

(0.80)^t = \frac{500}{1800} \approx 0.2778 \implies t = \frac{\log 0.2778}{\log 0.80} \approx \frac{-0.5560}{-0.0969} \approx 5.74

Since $t$ must be a whole number of years, the laptop drops below $500 during the sixth year.

Final answer: the investment reaches approximately $8954 after 10 years and doubles in about $11.9$ years; Sydney's projected population in 2051 is approximately $7.76$ million; the isotope decays to $10$ mg after $18$ hours; and the laptop is worth approximately $737.28 after 4 years, falling below $500 during the sixth year.

Common traps

Using $r$ instead of $1 + r$ as the base: $7\%$ growth uses base $1.07$ , not $0.07$ . The base is always close to $1$ for typical rates.
Sign error on decay: $12\%$ decay is base $0.88$ . Easy to write $1.12$ by reflex.
Forgetting that $a$ is the initial value: At $x = 0$ the function equals $a$ , since $b^0 = 1$ . Use this to read off the starting amount or the $y$ -intercept.
Trying to solve $b^x = c$ without logs: Standard 2 expects you to take logs and rearrange. Use $\log$ or $\ln$ , your choice; the bases cancel.
Treating the asymptote as a value the curve reaches: $y = 0$ is approached forever but never reached, so the model never predicts exactly zero.
Mishandling a half-life exponent: A half-life of $T$ means the base is $0.5$ and the exponent is (time) $/T$ , not just the time. Write $0.5^{t/T}$ .

Exam technique

State the model $y = a b^x$ with the variable named before substituting; markers reward seeing the base chosen correctly ( $1 + r/100$ or $1 - r/100$ ). For a "how long until" part, show the line where you take logs, because that is the method mark, and round only at the end. Keep the rounding sensible for the context: whole dollars for money, whole organisms for populations, and round a "falls below" answer up to the next whole period. If a table is given and you must name the model, write the ratios down to justify "exponential". And remember the asymptote in any "in the long run" interpretation: the quantity tends to zero without ever reaching it.

Note

Exponential change means a quantity gets multiplied by the same number every time period, instead of having the same amount added. That multiplier is the base $b$ . If $b$ is bigger than 1 the quantity grows; if $b$ is between 0 and 1 it shrinks (we call shrinking "decay"). The starting amount is $a$ , the value when time is 0. To pick out this pattern from a table, divide each value by the one before it: if you always get the same answer, the model is exponential.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style4 marks

The population of bacteria in a culture is modelled by

P = 200 (1.15)^t

, where

t

is time in hours. Find the population after

8

hours and the time taken to reach

1000

Show worked answer →

At $t = 8$ : $P = 200 (1.15)^8$ .

$(1.15)^8 \approx 3.0590$ .

$P \approx 200 \times 3.0590 \approx 611.8$ , round to $612$ bacteria.

For $P = 1000$ : $1000 = 200 (1.15)^t$ , so $(1.15)^t = 5$ .

Take logs of both sides: $t \log 1.15 = \log 5$ , so $t = \frac{\log 5}{\log 1.15} \approx \frac{0.6990}{0.0607} \approx 11.5$ hours.

Markers reward the substitution at $t = 8$ rounded sensibly, taking logs to solve the exponential equation, and the time to the nearest tenth of an hour.

2021 HSC-style3 marks

A car valued at $45000 depreciates at

12\%

per annum. Find its value after

5

years using an exponential model.

Show worked answer →

Depreciation at $12\%$ means the value is multiplied by $0.88$ each year.

Model: $V = 45000 (0.88)^t$ .

At $t = 5$ : $V = 45000 (0.88)^5$ .

$(0.88)^5 \approx 0.5277$ .

$V \approx 45000 \times 0.5277 \approx 23747.94$ , i.e. $23747.94.

Markers reward the multiplier $0.88$ (not $0.12$ ), the correct exponent, and an answer rounded sensibly. Lose half a mark for using $1.12$ instead of $0.88$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

The population of a town is modelled by

P = 8000(1.03)^t

, where

t

is the number of years from now. Find the population the model predicts after

4

years, to the nearest whole person.

Show worked solution →

Substitute $t = 4$ . Put the value straight into the model:

P = 8000(1.03)^4.

Compute the growth factor. Raise the base to the power $4$ :

(1.03)^4 = 1.1255 \text{ (to 4 d.p.)}.

Multiply and round. A population is a whole number, so

P = 8000 \times 1.1255\ldots = 9004.07\ldots \approx 9004.

Answer: the model predicts about $9004$ people after $4$ years.

foundation2 marks

A table of an exponential model

P = a b^t

is shown. |

t

| 0 | 1 | 2 | 3 | | --- | --- | --- | --- | --- | |

P

| 6 | 18 | 54 | 162 | (a) State the value of

a

. (b) Find the base

b

Show worked solution →

Part (a), read off $a$ . The initial value $a$ is the value of $P$ when $t = 0$ , because $b^0 = 1$ . From the table $P = 6$ at $t = 0$ , so

a = 6.

Part (b), find $b$ from the constant ratio. For an exponential model each value is the previous one multiplied by $b$ , so divide consecutive values:

\frac{18}{6} = 3, \quad \frac{54}{18} = 3, \quad \frac{162}{54} = 3.

The ratio is a constant $3$ , which is the base.

Answer: $a = 6$ and $b = 3$ , so the model is $P = 6(3)^t$ .

foundation3 marks

A colony starts with

300

bacteria and grows at

20\%

per hour. Let

P

be the number of bacteria after

t

hours. (a) Write an exponential model of the form

P = a b^t

. (b) Find the number of bacteria after

5

hours, to the nearest whole bacterium.

Show worked solution →

Part (a), set up the model. The starting value is $a = 300$ . Growth at $20\%$ per hour gives the base $b = 1 + \frac{20}{100} = 1.2$ , so

P = 300(1.2)^t.

Part (b), evaluate at $t = 5$ . Substitute $t = 5$ and compute the growth factor:

(1.2)^5 = 2.4883 \text{ (to 4 d.p.)}.

P = 300 \times 2.4883 = 746.49\ldots

Round for context. Bacteria are whole, so $P \approx 746$ .

Check. The base $1.2$ is just above $1$ , as expected for a $20\%$ growth rate, and $746$ is larger than the starting $300$ , which is right for growth.

foundation3 marks

A drone bought for $1200 depreciates at

15\%

per annum. Let

V

be its value in dollars after

t

years. (a) State the base

b

for the exponential model. (b) Write the model

V = a b^t

. (c) Find the drone's value after

3

years, correct to the nearest cent.

Show worked solution →

Part (a), find the base. Depreciation at $15\%$ per year multiplies the value by $b = 1 - \frac{15}{100} = 0.85$ each year (not $0.15$ ).

Part (b), write the model. The initial value is $a = 1200$ , so

V = 1200(0.85)^t.

Part (c), evaluate at $t = 3$ . Compute the decay factor, then multiply:

(0.85)^3 = 0.6141 \text{ (to 4 d.p.)}.

V = 1200 \times 0.6141\ldots = 736.95.

State the answer. After $3$ years the drone is worth $736.95.

Check. The base $0.85$ is just below $1$ , correct for decay, and $736.95 is less than the $1200 paid.

core3 marks

After a coffee, the mass of caffeine in a person's body is

200

mg and then decreases by

13\%

each hour. Let

C

be the mass of caffeine in mg after

t

hours. (a) Write a model of the form

C = a b^t

. (b) Find the caffeine remaining after

6

hours, correct to

1

decimal place. (c) Find the caffeine remaining after

10

hours, correct to

1

decimal place.

Show worked solution →

Part (a), write the model. Decreasing by $13\%$ each hour multiplies the mass by $b = 1 - \frac{13}{100} = 0.87$ (not $0.13$ ). With initial mass $a = 200$ ,

C = 200(0.87)^t.

Part (b), evaluate at $t = 6$ . Raise the base to the power $6$ , then multiply:

(0.87)^6 = 0.4336 \text{ (to 4 d.p.)},

C = 200 \times 0.4336\ldots = 86.7 \text{ mg}.

Part (c), evaluate at $t = 10$ . Same method with exponent $10$ :

(0.87)^{10} = 0.2484 \text{ (to 4 d.p.)},

C = 200 \times 0.2484\ldots = 49.7 \text{ mg}.

Answer: about $86.7$ mg remains after $6$ hours and about $49.7$ mg after $10$ hours. The base $0.87$ is just below $1$ , correct for decay, and the later value is smaller, as expected.

core4 marks

$8000 is invested at

4.5\%

per annum compounded annually. (a) Write a model for the value

A

in dollars after

t

years. (b) Find the value after

6

years, correct to the nearest cent. (c) By what percentage does the investment grow each year?

Show worked solution →

Part (a), write the model. The multiplier for one year is $b = 1 + \frac{4.5}{100} = 1.045$ , and the initial amount is $a = 8000$ , so

A = 8000(1.045)^t.

Part (b), evaluate at $t = 6$ . Compute the growth factor to enough places to keep the cents accurate:

(1.045)^6 = 1.302260 \text{ (to 6 d.p.)}.

A = 8000 \times 1.302260 = 10418.08.

So after $6$ years the investment is worth $10418.08.

Part (c), read the percentage change. The base $1.045$ means a per-year multiplier of $1.045$ , i.e. growth of $4.5\%$ each year.

Check. $10418.08 is more than the $8000 invested, as expected for growth at a positive rate.

core4 marks

A spreadsheet records the number of members

P

of a new club at the end of each year: |

t

(years) | 0 | 1 | 2 | 3 | | --- | --- | --- | --- | --- | |

P

| 5000 | 5500 | 6050 | 6655 | (a) Show that an exponential model fits this data. (b) Write the model

P = a b^t

. (c) Use it to predict the membership after

5

years.

Show worked solution →

Part (a), test the ratio of consecutive values. For an exponential model the ratio of one value to the previous must be constant:

\frac{5500}{5000} = 1.1, \quad \frac{6050}{5500} = 1.1, \quad \frac{6655}{6050} = 1.1.

The ratio is a constant $1.1$ , so the data is exponential (a constant ratio, not a constant difference).

Part (b), write the model. The initial value is $a = 5000$ (the value at $t = 0$ ) and the per-year multiplier is $b = 1.1$ , so

P = 5000(1.1)^t.

Part (c), predict at $t = 5$ . Compute the growth factor and multiply:

(1.1)^5 = 1.6105 \text{ (to 4 d.p.)}.

P = 5000 \times 1.6105 = 8052.55.

Membership is a whole number, so the model predicts about $8053$ members after $5$ years.

Check. Each step multiplies by $1.1$ ( $10\%$ growth), and $8053$ continues the rising pattern of the table.

core4 marks

A radioactive isotope has a half-life of

8

days. A sample starts with a mass of

240

mg. (a) Write a model for the mass

M

mg remaining after

t

days. (b) Find the mass remaining after

24

days. (c) Find the mass remaining after

20

days, correct to

1

decimal place.

Show worked solution →

Part (a), write the model. A half-life means the base is $0.5$ and the exponent counts how many half-lives have passed, which is $\frac{t}{8}$ . With initial mass $a = 240$ ,

M = 240(0.5)^{t/8}.

Part (b), evaluate at $t = 24$ . Here $\frac{24}{8} = 3$ , exactly three half-lives:

M = 240(0.5)^3 = 240 \times 0.125 = 30 \text{ mg}.

Part (c), evaluate at $t = 20$ . Now $\frac{20}{8} = 2.5$ , so

(0.5)^{2.5} = 0.1768 \text{ (to 4 d.p.)}.

M = 240 \times 0.1768\ldots = 42.4 \text{ mg}.

Check. Halving from $240$ gives $120, 60, 30$ at $8, 16, 24$ days, so $30$ mg at $24$ days is correct, and $42.4$ mg at $20$ days sensibly lies between the $60$ mg at $16$ days and the $30$ mg at $24$ days.

exam4 marks

The population of a country town is modelled by

P = 12000(1.025)^t

, where

t

is the number of years after

2025

. Use the model to find the year during which the population first reaches

20000

. Show how you use logarithms.

Show worked solution →

Set up the equation. Put $P = 20000$ into the model:

12000(1.025)^t = 20000.

Isolate the power. Divide both sides by $12000$ :

(1.025)^t = \frac{20000}{12000} = 1.6667 \text{ (to 4 d.p.)}.

Take logarithms of both sides. This is the method step:

t \log 1.025 = \log 1.6667,

t = \frac{\log 1.6667}{\log 1.025} = \frac{0.22185}{0.010724} = 20.69 \text{ (to 2 d.p.)}.

Interpret for the year. Since $t = 20.69$ years, the population passes $20000$ during the $21$ st year after $2025$ , that is during the year $2046$ .

Check. At $t = 20$ , $P = 12000(1.025)^{20} = 19663$ , still under $20000$ ; at $t = 21$ , $P = 12000(1.025)^{21} = 20155$ , just over, confirming the crossing is during year $21$ .

exam5 marks

A factory machine is bought for $60000 and depreciates at

18\%

per annum. (a) Write a model for the value

V

in dollars after

t

years. (b) Find the value after

4

years, to the nearest dollar. (c) Find the first whole year at the end of which the value has fallen below $20000. Show your use of logarithms.

Show worked solution →

Part (a), write the model. Depreciation at $18\%$ gives base $b = 1 - \frac{18}{100} = 0.82$ , with $a = 60000$ :

V = 60000(0.82)^t.

Part (b), evaluate at $t = 4$ . Compute the decay factor and multiply:

(0.82)^4 = 0.4521 \text{ (to 4 d.p.)}.

V = 60000 \times 0.4521\ldots = 27127.

So after $4$ years the value is about $27127.

Part (c), solve for the value below $20000. Set $V = 20000$ and isolate the power:

(0.82)^t = \frac{20000}{60000} = 0.3333 \text{ (to 4 d.p.)}.

Take logarithms of both sides:

t = \frac{\log 0.3333}{\log 0.82} = \frac{-0.4771}{-0.086186} = 5.54 \text{ (to 2 d.p.)}.

Since $t$ must be a whole number of years, round up: the value first drops below $20000 at the end of the $6$ th year.

Check. At $t = 5$ , $V = 60000(0.82)^5 = 22244$ (still above $20000); at $t = 6$ , $V = 60000(0.82)^6 = 18240$ (below), confirming year $6$ .

exam5 marks

$15000 is invested at

5.5\%

per annum compounded annually, giving

A = 15000(1.055)^t

dollars after

t

years. (a) Find the value after

8

years, to the nearest cent. (b) Find the doubling time of the investment, correct to

1

decimal place, using logarithms.

Show worked solution →

Part (a), evaluate at $t = 8$ . Compute the growth factor, then multiply:

(1.055)^8 = 1.5347 \text{ (to 4 d.p.)}.

A = 15000 \times 1.5347\ldots = 23020.30.

So after $8$ years the investment is worth $23020.30.

Part (b), set up the doubling equation. Doubling means $A = 2 \times 15000 = 30000$ , so

15000(1.055)^t = 30000 \implies (1.055)^t = 2.

Take logarithms. This is the method step:

t = \frac{\ln 2}{\ln 1.055} = \frac{0.693147}{0.053541} = 12.9 \text{ (to 1 d.p.)}.

So the investment doubles in about $12.9$ years. Notice the doubling time does not depend on the starting amount, only on the rate.

Check. At $t = 12.9$ , $(1.055)^{12.9} = 1.9951$ , just under $2$ , and the value keeps rising, so doubling occurs just after $12.9$ years as found.

exam6 marks

An invasive water weed is first noticed covering

50

^2

of a wetland, and the area it covers grows by

35\%

each month. (a) Write a model for the area

A

in m

^2

after

t

months. (b) Find the area covered after

6

months, to the nearest m

^2

. (c) Find the first whole month at the end of which the weed covers more than

1000

^2

. Show your use of logarithms.

Show worked solution →

Part (a), write the model. Growth of $35\%$ each month gives base $b = 1 + \frac{35}{100} = 1.35$ , with initial area $a = 50$ , so

A = 50(1.35)^t.

Part (b), evaluate at $t = 6$ . Raise the base to the power $6$ , then multiply:

(1.35)^6 = 6.0534 \text{ (to 4 d.p.)},

A = 50 \times 6.0534\ldots = 302.67\ldots \approx 303 \text{ m}^2.

Part (c), set up the equation. Put $A = 1000$ and isolate the power:

50(1.35)^t = 1000 \implies (1.35)^t = \frac{1000}{50} = 20.

Take logarithms of both sides. This is the method step:

t = \frac{\log 20}{\log 1.35} = \frac{1.3010}{0.1303} = 9.98 \text{ (to 2 d.p.)}.

Since $t$ must be a whole number of months, round up: the weed first covers more than $1000$ m $^2$ at the end of the $10$ th month.

Answer: the area is about $303$ m $^2$ after $6$ months, and it first passes $1000$ m $^2$ at the end of month $10$ .

Check. At $t = 9$ , $A = 50(1.35)^9 = 744.69$ (still under $1000$ ); at $t = 10$ , $A = 50(1.35)^{10} = 1005.33$ (over), confirming month $10$ .

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA

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