Coaching Centre
Exponential growth and decay
The graphs of y = a^x and y = a^(−x), exponential models y = k·a^x for growth and decay, what k and a mean in context, finding when an amount doubles or halves, and the limits of exponential models.
A linear model adds the same amount each step. An exponential model multiplies by the same number each step. That one difference is huge: a bank balance earning interest, a bacteria colony, a car losing value and a medicine leaving the blood all change by a percentage, not a fixed amount.
The graphs of y = a^x and y = a^−x
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|---|
| $y = 2^x$ | $0.25$ | $0.5$ | $1$ | $2$ | $4$ | $8$ |
| $y = 2^{-x}$ | $4$ | $2$ | $1$ | $0.5$ | $0.25$ | $0.125$ |
- Both curves pass through $(0, 1)$, because any number to the power $0$ is $1$.
- $y = 2^x$ rises more and more steeply: each step right doubles $y$.
- $y = 2^{-x}$ is its mirror image in the $y$-axis: each step right halves $y$.
- Neither curve touches the $x$-axis. Halving forever gets closer to $0$ but never reaches it. The $x$-axis is an asymptote.
Slide the base a. Both curves always pass through (0, 1).
With $a = 2$, check the readout against the table. Then raise $a$ to $3$: is the growth curve steeper or flatter for positive $x$? Bring $a$ down towards $1$ and describe what happens to both curves. Why can neither curve ever cross the $x$-axis?
Growth and decay models
$y = k \times a^x$
$k$ is the starting amount (the value when $x = 0$, the $y$-intercept).
$a$ is the multiplier per step: $a \gt 1$ is growth, $0 \lt a \lt 1$ is decay.
A growth rate of $r\%$ gives $a = 1 + \dfrac{r}{100}$. A decay rate of $r\%$ gives $a = 1 - \dfrac{r}{100}$.
Why $1 + r$? Growing by $6\%$ means you keep $100\%$ and add $6\%$, so you end with $106\%$ of what you had: multiply by $1.06$. Losing $15\%$ leaves $85\%$: multiply by $0.85$.
A bacteria colony starts with $50$ cells and doubles every hour. Write a model and find the number of cells after $5$ hours.
A new car costs $\$32{,}000$ and loses $15\%$ of its value each year. Find its value after $3$ years, and in which year it first drops below half its price.
On Bacteria, read the doubling time, then drag the rate to $50\%$ and see how much longer doubling takes. On Savings, compare the curve with the dashed straight line: in which year does the curve pull ahead? On Car value and Medicine the readout gives the time to halve: check Example 2 against it.
Interpreting and finding a model
A patient has $400$ mg of a medicine in the blood, and the body removes $20\%$ of what is left each hour. How much remains after $3$ hours, and when does it first fall below $100$ mg?
To find $k$ and $a$ from data, $k$ is the value when $x = 0$, and $a$ is any value divided by the one before it. If a population goes $1200$, $1236$, $1273.08$, then $a = 1236 \div 1200 = 1.03$: growth of $3\%$ per step.
Limitations
- Growth can't go on forever. Bacteria run out of food and space, so real growth slows down.
- Decay never quite reaches zero in the model, but in reality the medicine is gone, or the car is sold for scrap.
- The rate may change. Interest rates move, and a car's depreciation is fastest in the first years.
- Whole numbers. The model may say $1600.4$ bacteria: round sensibly.
Your turn
A town of $12{,}000$ people grows by $25\%$ every decade. Write a model and find the population after $3$ decades.
Show the working
$a = 1 + 0.25 = 1.25$, so $P = 12{,}000 \times 1.25^d$. After $3$ decades: $12{,}000 \times 1.953125 = 23{,}437.5$, about $23{,}438$ people.
What is the value of $y = 5^x$ when $x = 0$?
Any non-zero number to the power $0$ is $1$, so every graph $y = a^x$ passes through $(0, 1)$.
Find $y$ for $y = 2^{-x}$ when $x = 2$.
$2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4} = 0.25$.
Which model shows decay?
Decay needs a multiplier between $0$ and $1$: $0.9$. The value of $k$ does not decide growth or decay, and $10 + 2x$ is linear.
$N = 50 \times 2^t$. Find $N$ when $t = 5$.
$2^5 = 32$, and $50 \times 32 = 1600$.
An investment grows by $6\%$ a year. What is $a$ in $V = k \times a^t$?
You keep $100\%$ and add $6\%$: $1 + 0.06 = 1.06$.
$V = 32{,}000 \times 0.85^t$. Find the value, in dollars, after $3$ years.
$0.85^3 = 0.614125$, and $32{,}000 \times 0.614125 = 19{,}652$.
$M = 400 \times 0.8^t$ mg. How much remains after $3$ hours?
$0.8^3 = 0.512$, and $400 \times 0.512 = 204.8$ mg.
Which is a limitation of modelling a bacteria colony with $N = 50 \times 2^t$?
The first three are just features of the model. The real limit is that growth slows when resources run out.
Exam style: $\$5000$ is invested at $6\%$ per year, compounded yearly, so $V = 5000 \times 1.06^t$. Find the value after $10$ years, to the nearest cent.
$1.06^{10} \approx 1.790848$, so $V = 5000 \times 1.06^{10} \approx 8954.24$.
Exam style: Using the same model, after how many whole years does the investment first exceed $\$10{,}000$?
Double $5000$ is $10{,}000$. $t = 11$: $5000 \times 1.06^{11} \approx 9491.49$. $t = 12$: $5000 \times 1.06^{12} \approx 10{,}060.98$. So it first exceeds $\$10{,}000$ after $12$ years.