Exponential functions have many applications in growth and decay such as population growth, investment growth, radioactive decay and depreciation. Before looking at some applications in further detail, let's look at how to solve some exponential equations.
Exponential equations
We want to look at solving problems where the unknown is in the exponent–these are called exponential equations. They look like:
$3^x=81$3x=81, $5\times2^x=40$5×2x=40 and $7^{5x-2}=20$75x−2=20
We could use technology to solve these or algebraically using logarithms, which we will explore in further detail later in this course. However, in many cases, algebraic manipulation using our index laws will allow us to solve exponential equations without a calculator. To achieve this we need to write both sides of the equation with the same base, then we can equate the indices.
When both sides of an exponential equation are written with the same base, we can equate the indices:
If $a^x=a^y$ax=ay, then $x=y$x=y.
To write both sides with the same base it is helpful to be familiar with low powers of prime numbers, so you can recognise them and rewrite them in index form. It is also vital to be confident with the index laws.
Worked examples
Example 1
Solve $\left(27\right)^{x+1}=\frac{1}{81}$(27)x+1=181, for $x$x.
Think: Can you spot a common base that both sides could be written in? Both $27$27 and $81$81 are powers of $3$3.
Do: Use index laws to write both sides as a single power of three and then equate the indices.
| $\left(27\right)^{x+1}$(27)x+1 | $=$= | $\frac{1}{81}$181 |
Writing down the equation |
| $\left(3^3\right)^{x+1}$(33)x+1 | $=$= | $\frac{1}{3^4}$134 |
Expressing both sides using powers of $3$3 |
| $3^{3x+3}$33x+3 | $=$= | $3^{-4}$3−4 |
Using the index law $A^{-n}=\frac{1}{A^n}$A−n=1An |
| Hence $3x+3$3x+3 | $=$= | $-4$−4 |
Equating exponents |
| $3x$3x | $=$= | $-7$−7 |
Moving the constant terms to one side |
| $\therefore$∴ $x$x | $=$= | $-\frac{7}{3}$−73 |
Solving for $x$x |
Example 2
Solve $5\times16^y=40\times\sqrt[3]{32}$5×16y=40×3√32, for $y$y.
Think: On the left-hand side, we have a power of $2$2, but it's multiplied by a $5$5. If we first divide both sides by the factor of $5$5, can we then write both sides as powers of $2$2?
Do:
| $5\times16^y$5×16y | $=$= | $40\times\sqrt[3]{32}$40×3√32 |
| $\left(16\right)^y$(16)y | $=$= | $8\times\sqrt[3]{32}$8×3√32 |
We now have an equation with $16$16, $8$8 and $32$32 which can all be written as powers of $2$2. Proceed with index laws and remember $\sqrt[n]{x}=x^{\frac{1}{n}}$n√x=x1n.
| $\left(16\right)^y$(16)y | $=$= | $8\times\sqrt[3]{32}$8×3√32 |
Writing down the equation |
| $\left(2^4\right)^y$(24)y | $=$= | $2^3\times\left(2^5\right)^{\frac{1}{3}}$23×(25)13 |
Expressing both sides using powers of $2$2 |
| $2^{4y}$24y | $=$= | $2^{\left(3+\frac{5}{3}\right)}$2(3+53) |
Using index laws |
| $2^{4y}$24y | $=$= | $2^{\frac{14}{3}}$2143 |
Combining terms in the exponent |
| Hence $4y$4y | $=$= | $\frac{14}{3}$143 |
Equating exponents |
| $y$y | $=$= | $\frac{14}{12}$1412 |
Solving for $y$y
|
| $y$y | $=$= | $\frac{7}{6}$76 |
Simplifying |
Practice questions
question 1
Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)x+7=23 for $x$x.
Question 2
Solve the equation $8^{x+5}=\frac{1}{32\sqrt{2}}$8x+5=132√2.
Applications
Exponential growth and decay models arise in many real-world situations. Here are the general functions for growth and decay, their appearance and some real-world examples:
| Growth | Decay |
|---|---|
 |
 |
|
Examples:
|
Examples:
|
The base model for growth and decay is $y=A_0a^x$y=A0ax, where $A_0>0$A0>0 and is the initial amount. When the multiplier $a$a is larger than one, we are in a growth situation and when the multiplier $a$a is between zero and one we are in a decay situation.
The models above have been drawn for $x\ge0$x≥0, as most models will have a horizontal axis representing time and start at an initial time of zero. However, do read questions carefully as this may not always be the case.
Notation: The variables $y$y and $x$x may change to highlight the variables in the model. Function notation is common such as $P\left(t\right)=3000\left(1.05\right)^t$P(t)=3000(1.05)t, to indicate the population as a function of time. Another common notation is sequence notation, such as $H_n=3\left(0.65\right)^n$Hn=3(0.65)n, where $n$n indicates the number of times the multiplier has been applied. For example $H_3$H3 may indicate the rebound height of a ball after $3$3 bounces. Sequence notation is particularly common for situations where $n$n can only take on integer values.
Exponential growth models:
- Have the form $y=A_0a^x$y=A0ax, where $A_0>0$A0>0 and is the initial amount and $a>0$a>0.
- The growth rate as a percentage per unit of $x$x is given by: $r=\left(a-1\right)\times100%$r=(a−1)×100%
- An alternate form is: $y=A_0\left(1+r\right)^x$y=A0(1+r)x
Exponential decay models:
- Have the form $y=A_0a^x$y=A0ax, where $A_0>0$A0>0 and is the initial amount and $0
- The decay rate as a percentage per unit of $x$x is given by: $r=\left(1-a\right)\times100%$r=(1−a)×100%
- An alternate form is: $y=A_0\left(1-r\right)^x$y=A0(1−r)x
Note: The rates here describe an average rate of change over one unit of time expressed as a percentage and not the instantaneous rate of change at a given point.
Let's look at some examples more closely and how the formula arises in practical situations.
Worked examples
Example 3
Sarah buys a piece of artwork for $\$1500$$1500 that is expected to appreciate (increase in value) by $8%$8% each year.
(a) Find a model for $V_n$Vn, the value of the artwork after $n$n years.
Think: Identify the initial amount and rate of increase and use appropriate variables in the formula $y=A_0\left(1+r\right)^x$y=A0(1+r)x
Do: $A_0=\$1500$A0=$1500, $r=8%$r=8%, hence: $V_n=1500\left(1+\frac{8}{100}\right)^n$Vn=1500(1+8100)n or $V_n=1500\left(1.08\right)^n$Vn=1500(1.08)n
(b) Find the estimated value of the artwork in $6$6 years' time, to the nearest cent.
Substitute $n=6$n=6 into the model and round to $2$2 decimal places.
| $V_6$V6 | $=$= | $1500\left(1.08\right)^6$1500(1.08)6 |
| $\approx$≈ | $\$2380.31$$2380.31 |
Example 4
The population of Tasmanian devils is declining due to disease. A model for the population, $P$P, in a specified area, $t$t years after observation began, is modelled by the equation: $P\left(t\right)=800\left(0.85\right)^t$P(t)=800(0.85)t.
(a) How many devils are there initially?
Think: The initial population is given by the number or coefficient out the front of the exponential term.
Do: So there are $800$800 devils.
(b) At what rate does the model suggest the population is decreasing at per year?
Think: We are multiplying by $0.85$0.85, this is equivalent to $85%$85%.
Do: We we are decreasing by $15%$15% each year. This can also be found by the formula:
| $\text{Rate of increase}$Rate of increase | $=$= | $\left(1-a\right)\times100%$(1−a)×100% |
| $=$= | $\left(1-0.85\right)\times100%$(1−0.85)×100% | |
| $=$= | $15%$15% |
c) When is the population in the area predicted to reach a critical level of $200$200 remaining devils?
Think: This would be found by solving the equation $200=800\left(0.85\right)^t$200=800(0.85)t. If we divide both sides by $800$800 we get the equation:
$\left(\frac{1}{4}=\left(0.85\right)^t\right)$(14=(0.85)t)
We can see that solving for $t$t is equivalent to finding how many times we multiply $0.85$0.85 by itself to get $\frac{1}{4}$14. By trial and error we can see that $t$t must be between $8$8 and $9$9 since $0.85^8\approx0.272$0.858≈0.272 and $0.85^9\approx0.232$0.859≈0.232. But we could find $t$t more accurately with the help of technology.
Do: We can use technology to solve this by graphing both sides of the equation and finding the $x$x-coordinate of the point of intersection. That is, draw $y_1=800\left(0.85\right)^t$y1=800(0.85)t and $y_2=200$y2=200. This will provide a nice visualisation of the problem. Use your knowledge of exponential functions and a practical domain to obtain an appropriate view window.
From the graph, we can see the model predicts the population will reach a critical level in approximately $8.53$8.53 years' time. We can also solve this using an equation solver function in the calculator.
Practice questions
Question 3
Luigi purchased a sculpture for $\$2900$$2900, and it is expected to increase in value by $9%$9% per year.
Write a function $y$y to represent the value of the sculpture after $x$x years.
Find the value of the sculpture after $8$8 years, rounding to the nearest cent.
Question 4
Consider the function $f\left(t\right)=\frac{8}{7}\left(\frac{3}{8}\right)^t$f(t)=87(38)t, where $t$t represents time.
What is the initial value of the function?
Express the function in the form $f\left(t\right)=\frac{8}{7}\left(1-r\right)^t$f(t)=87(1−r)t, where $r$r is a decimal.
Does the function represent growth or decay of an amount over time?
decay
Agrowth
BWhat is the rate of decay per time period? Give the rate as a percentage.
Question 5
The frequency $f$f (Hz) of the $n$nth key of an $88$88-key piano is given by $f\left(n\right)=440\left(2^{\frac{1}{12}}\right)^{n-49}$f(n)=440(2112)n−49.
Determine the frequency of the forty-ninth key.
Determine the frequency of the $40$40th key.
Give your answer to the nearest whole number.
Solve for the value of $n$n that corresponds to the key with a frequency of $1760$1760 Hz.
Further applications and base $e$e
All exponential models can be written with any base, however, it is common practice to use the base $e$e as you will see, this has interesting properties when applying calculus. Also, when modelling exponential growth (or decay), we are often interested in populations, investments and similar variables which are growing at continuous rate proportional to the current value. As seen in our investigation, this leads us to naturally use the base $e$e in such circumstances and form the following general form for exponential models:
$y=A_0e^{kx}$y=A0ekx
The basic model of the form $y=A_0a^x$y=A0ax can be written in the form $y=A_0e^{kx}$y=A0ekx. Where $A_0>0$A0>0 and gives the initial amount, and the multiplier $a=e^k$a=ek determines the growth/decay rate. As before, if $a>1$a>1 we are in a growth situation, this will occur for $k>0$k>0. If $0
$k$k can be interpreted as the continuous rate of growth expressed as a percentage of the current value of $y$y for $k>0$k>0 or the rate of decay for $k<0$k<0. This is an example of an instantaneous rate of change.
Worked example
Example 5
A model for a population, $P$P, of foxes after $t$t years is given by $P=600e^{0.05t}$P=600e0.05t.
(a) Is the population increasing or decreasing?
Think: For a model $P=A_0e^{kt}$P=A0ekt, with $A_0>0$A0>0 the population will increase if $k>0$k>0 and decrease if $k<0$k<0.
Do: The population is increasing.
(b) How many foxes are there predicted to be after $18$18 months?
Think: The variable $t$t is in years. Hence, make the substitution $t=1.5$t=1.5 in the model and evaluate, giving the answer to the nearest fox.
Do:
| $P$P | $=$= | $600e^{0.05t}$600e0.05t |
| $=$= | $600e^{0.05\times1.5}$600e0.05×1.5 | |
| $\approx$≈ | $647$647 foxes |
(c) If the population continues to grow at this rate, how long will it take for the population to double? Round answer to two decimal places.
Think: The initial population of foxes was $600$600, for the population to double we need to solve for the time when $P=1200$P=1200.
Do:
| $P$P | $=$= | $600e^{0.05t}$600e0.05t |
| $1200$1200 | $=$= | $600e^{0.05t}$600e0.05t |
| $2$2 | $=$= | $e^{0.05t}$e0.05t |
| $\therefore t$∴t | $\approx$≈ | $13.86$13.86 years |
We can see that solving for when the population reaches $1200$1200 is equivalent to solving when we are multiplying the initial population by $2$2. We can use technology to solve this by graphing the situation or using the solve features of the calculator.
(d) What percentage is the population growing by each year? Round your answer to two decimal places.
Think: We can use the calculator to evaluate $a=e^k$a=ek. We can then find the rate of of growth as $rate=\left(a-1\right)\times100%$rate=(a−1)×100%.
Do:
| Rate | $=$= | $\left(a-1\right)\times100%$(a−1)×100% |
| $=$= | $\left(e^{0.05}-1\right)\times100%$(e0.05−1)×100% | |
| $\approx$≈ | $\left(1.0513-1\right)\times100%$(1.0513−1)×100% | |
| $=$= | $5.13%$5.13% |
Hence, the population is growing at approximately $5.13%$5.13% per year.
Note: We can also interpret $k$k as the continuous growth rate. That is the population is growing at a continuous rate of $5%$5%, like an account that continuously compounds.
Many other applications can involve an exponential term as part of a model. Such as:
- Logistic models, which can be used to model populations with a limiting aspect such as predators or limited food resources
- Surge models, which can be used to model concentrations of a drug in the bloodstream
- Terminal velocity models, which can model the speed of a falling object
For these models, make effective use of technology to graph and solve problems in context.
Practice questions
Question 6
Does the graph of $f\left(x\right)=1.5e^{-0.35x}$f(x)=1.5e−0.35x illustrate exponential growth or exponential decay?
Exponential growth
AExponential decay
B
Question 7
The weight, in grams, of a radioactive substance left after $t$t years is given by $W\left(t\right)=250e^{-0.015t}$W(t)=250e−0.015t.
What is the initial weight of the substance?
What percentage of the substance is remaining after $5$5 years?
Round your answer to the nearest percent.
What is the approximate half-life of the substance?
Round your answer to two decimal places.
Does the half life of the substance depend on its initial weight?
Yes
ANo
BHow long will it take for there to be $10%$10% of the substance remaining?
Round your answer to two decimal places.
Question 8
A local council has noticed that the rabbit population has increased rapidly, and begins to monitor their population.
$t$t months after they start monitoring, they approximate that the number of rabbits is given by $N\left(t\right)=\frac{B}{0.15+e^{-t}}$N(t)=B0.15+e−t.
Given that they initially record $53$53 rabbits in the area, solve for $B$B.
What would the council expect the rabbit population to reach $1$1 month after they start recording to the nearest whole number?
What would the council expect the rabbit population to reach $1$1 year after they start recording to the nearest whole number.