How do you tell whether a model represents growth or decay?

For N = N0 a to the power t the quantity grows when the base a is greater than 1 and decays when a is between 0 and 1. For N = N0 e to the power (kt) it grows when k is greater than 0 and decays when k is less than 0. A decay factor such as 0.85 means the quantity keeps 85 per cent and loses 15 per cent each period.

How do you find the time for a model to reach a given value?

Substitute the target value, divide by the initial amount to isolate the power, then take a logarithm of both sides to bring the exponent down. For example 2000 times 1.05 to the power t = 4000 gives 1.05 to the power t = 2, so t = log 2 divided by log 1.05, about 14.2 years.

How do you tell whether a model represents growth or decay?

For N = N0 a to the power t the quantity grows when the base a is greater than 1 and decays when a is between 0 and 1. For N = N0 e to the power (kt) it grows when k is greater than 0 and decays when k is less than 0. A decay factor such as 0.85 means the quantity keeps 85 per cent and loses 15 per cent each period.

How do you find the time for a model to reach a given value?

Substitute the target value, divide by the initial amount to isolate the power, then take a logarithm of both sides to bring the exponent down. For example 2000 times 1.05 to the power t = 4000 gives 1.05 to the power t = 2, so t = log 2 divided by log 1.05, about 14.2 years.

Year 12 Methods (Unit 3 & 4) Exponential and logarithmic functions

Applications of exponential functions

Q: What is an exponential model?

An exponential model has the form N = N0 times a to the power t, or N = N0 e to the power (kt), where N0 is the initial amount at time t = 0, a is the per-period growth or decay factor, and k is the continuous rate constant. The same quantity is multiplied by a fixed factor over each equal time step, which is what makes the growth or decay exponential.

Q: What is half-life and doubling time?

The half-life is the time for a decaying quantity to halve, and the doubling time is the time for a growing quantity to double. For an exponential model these times are constant - they do not depend on the starting amount - so after two half-lives a quarter remains, after three half-lives an eighth, and so on.

Q: How do compound interest and depreciation use exponentials?

Compound interest at rate r per period grows an amount as A = P times (1 + r) to the power t, so 6 per cent gives a factor 1.06. Depreciation that loses r each period is decay: V = P times (1 - r) to the power t, so losing 12 per cent gives a factor 0.88. Both are exponential models with a per-period factor.

Q: What is Newton's law of cooling?

Newton's law of cooling models an object cooling toward its surroundings as T = T_room + (T0 minus T_room) times e to the power (minus kt). The temperature falls exponentially and levels off at the room temperature, which is a positive horizontal asymptote, not zero.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), exponential functions model real-world change — population and bacterial growth, radioactive decay and half-life, compound interest, depreciation and Newton's law of cooling. Each uses $N=N_0\,a^{t}$ or $N=N_0\,e^{kt}$, where $N_0$ is the initial amount and the base or the sign of $k$ sets growth or decay. You read off values, interpret the parameters, and solve for a rate or a time using logarithms.

An exponential model describes a quantity that is multiplied by a fixed factor over each equal step of time. It is written $N=N_0\,a^{t}$ (a per-period factor $a$) or $N=N_0\,e^{kt}$ (a continuous rate $k$). Here $N_0$ is the initial amount — the value when $t=0$, because $a^{0}=1$ and $e^{0}=1$.

The model shows growth when the base $a>1$ (or $k>0$) and decay when $0growth factor of \(1.05$ means the quantity increases by $5\%$ each period; a decay factor of $0.88$ means it keeps $88\%$ and loses $12\%$ each period. Compound interest, depreciation, population growth and radioactive decay are all this one idea with different factors.

Two special times describe the pace of change: the doubling time (for growth) and the half-life (for decay). For an exponential model these are constant — they do not depend on the starting amount. Many decay models level off at a non-zero value: Newton's law of cooling approaches the surrounding room temperature, a positive horizontal asymptote.

Key idea. \(N=N_0\,a^{t}\) or \(N=N_0\,e^{kt}\): \(N_0\) is the initial amount; \(a>1\) or \(k>0\) is growth, \(0
    

Growth: $N=N_0\,a^{t}$ ($a>1$), rising from $N_0$

Decay / cooling: levels off at a positive asymptote

Per-period model (factor $a$) and continuous model (rate $k$):

\[N=N_0\,a^{t} \qquad N=N_0\,e^{kt} \qquad (N_0=\text{initial amount})\]

N = N_{0} a^{t}

Compound interest (rate $r$ per period) and depreciation:

\[A=P\,(1+r)^{t} \qquad V=P\,(1-r)^{t}\]

A = P {(1 + r)}^{t}

Newton's law of cooling toward a room temperature $T_{\text{room}}$:

\[T=T_{\text{room}}+(T_0-T_{\text{room}})\,e^{-kt}\]

T = T_{room} + (T_{0} - T_{room}) e^{- k t}

Solving for time. Isolate the power, then take a logarithm: from \((1.05)^{t}=2\), \(t=\dfrac{\log 2}{\log 1.05}\approx 14.2\). Any base of log works, as long as it is the same on both sides.

How to set up and solve an exponential model

Identify $N_0$ and the factor. The initial amount is the value at $t=0$. A growth of $r$ gives factor $1+r$; a loss of $r$ gives $1-r$; "doubles" gives base $2$, "halves" gives base $\tfrac12$.
Find an unknown rate from data. Substitute a second reading and solve for $a$, $b$ or $k$ — take a root for a per-period factor, or a logarithm for a continuous $k$.
Evaluate a value. Substitute the time $t$ and compute with technology.
Solve for a time. Set the model equal to the target, divide by $N_0$ to isolate the power, then take a logarithm of both sides to bring $t$ down.

Half-life and doubling time. Set the model to \(\tfrac12 N_0\) (or \(2N_0\)) and solve for \(t\). For \(N=N_0 e^{kt}\) the doubling time is \(\dfrac{\ln 2}{k}\); it is the same whatever the starting amount.

Example 1 — Evaluate a growth model

A colony of bacteria triples each hour: $N=500\times 3^{\,t}$ after $t$ hours. Find the number present at the start and after $2$ hours.

Solution

$N(0)$	$=$	$500\times 3^{0}=500$
$N(2)$	$=$	$500\times 3^{2}$
	$=$	$500\times 9=4500$

Start: $500$; after $2$ hours: $\mathbf{4500}$. Base $3>1$, so this is growth.

500 \times 3^{2} = 4500

Example 2 — Depreciation (decay factor)

A van bought for $\$30000$ loses $12\%$ of its value each year. Write the model for its value $V$ after $t$ years, and find $V$ after $3$ years.

Solution

$\text{factor}$	$=$	$1-0.12=0.88$
$V$	$=$	$30000\,(0.88)^{t}$
$V(3)$	$=$	$30000\,(0.88)^{3}$
	$\approx$	$\$20444$

Losing $12\%$ keeps $88\%$, so the factor is $0.88$ (decay).

30000 {(0.88)}^{3}

Example 3 — Solve for the time with logs

A $\$2000$ deposit grows as $A=2000\times(1.05)^{t}$. After how many years does it first exceed $\$4000$?

Solution

$2000\,(1.05)^{t}$	$>$	$4000$
$(1.05)^{t}$	$>$	$2$
$t$	$>$	$\dfrac{\log 2}{\log 1.05}\approx 14.2$

So it first exceeds $\$4000$ after $\mathbf{15}$ whole years.

t = \frac{\log 2}{\log 1.05}

Example 4 — Newton's law of cooling

Coffee cools as $T=20+70\,e^{-0.05t}$ degrees Celsius after $t$ minutes. Find the initial temperature, the temperature after $10$ minutes, and the temperature it approaches.

Solution

$T(0)$	$=$	$20+70\,e^{0}=90$
$T(10)$	$=$	$20+70\,e^{-0.5}$
	$\approx$	$62.46$

As $t\to\infty$, $e^{-0.05t}\to 0$, so $T\to 20$: initial $90^\circ$C, after $10$ min $\approx 62.46^\circ$C, approaching the room temperature $\mathbf{20^\circ}$C.

20 + 70 e^{- 0.5} \approx 62.46

Common pitfalls

The growth factor is not the percentage. A factor of $1.03$ is a $3\%$ increase (it is $1+0.03$), not $103\%$ or $1.03\%$. A decay factor of $0.85$ is a $15\%$ decrease.

You must take a logarithm to solve for time. Once the power is isolated, an unknown in the exponent comes down only with a logarithm — you cannot just divide. From $(1.05)^{t}=2$, $t=\log 2/\log 1.05$, not $2/1.05$.

Cooling levels off at the room temperature, not zero. In $T=T_{\text{room}}+(T_0-T_{\text{room}})e^{-kt}$ the horizontal asymptote is $T_{\text{room}}$. Radioactive decay $M=M_0e^{-kt}$ does approach $0$; cooling does not.

Frequently asked questions

What is an exponential model?

$N=N_0\,a^{t}$ or $N=N_0\,e^{kt}$: $N_0$ is the initial amount and the quantity is multiplied by a fixed factor over each equal time step.

How do you tell growth from decay?

Growth when $a>1$ or $k>0$; decay when \(0

How do you find the time for a model to reach a value?

Isolate the power (divide by $N_0$) then take a logarithm of both sides, e.g. $(1.05)^{t}=2$ gives $t=\log 2/\log 1.05\approx 14.2$.

What is half-life and doubling time?

The time to halve (decay) or double (growth). For an exponential model it is constant, so after two half-lives a quarter remains.

How do compound interest and depreciation use exponentials?

Interest grows as $A=P(1+r)^{t}$ (factor $1+r$); depreciation decays as $V=P(1-r)^{t}$ (factor $1-r$).

What is Newton's law of cooling?

$T=T_{\text{room}}+(T_0-T_{\text{room}})e^{-kt}$: the temperature falls exponentially and levels off at the room temperature, a positive asymptote.

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\(N(0)\)	\(=\)	\(500\times 3^{0}=500\)
\(N(2)\)	\(=\)	\(500\times 3^{2}\)
	\(=\)	\(500\times 9=4500\)

\(\text{factor}\)	\(=\)	\(1-0.12=0.88\)
\(V\)	\(=\)	\(30000\,(0.88)^{t}\)
\(V(3)\)	\(=\)	\(30000\,(0.88)^{3}\)
	\(\approx\)	\(\$20444\)

\(2000\,(1.05)^{t}\)	\(>\)	\(4000\)
\((1.05)^{t}\)	\(>\)	\(2\)
\(t\)	\(>\)	\(\dfrac{\log 2}{\log 1.05}\approx 14.2\)

\(T(0)\)	\(=\)	\(20+70\,e^{0}=90\)
\(T(10)\)	\(=\)	\(20+70\,e^{-0.5}\)
	\(\approx\)	\(62.46\)