Growth and decay

Exponential and logarithmic functions, grade 12 (4 units) Lesson 9 of 20Learning

Growth and decay

A process of growth or decay

In a process of growth or decay the amount is multiplied by the same number $\displaystyle q$ in every unit of time – every year, every month or every hour. The number $\displaystyle q$ is called the growth factor (when $\displaystyle q>1$ ) or the decay factor (when $\displaystyle 0<q<1$ ).

An amount that grows by $\displaystyle p$ percent in every unit of time is multiplied by $\displaystyle {q}={1+\frac{p}{100}}$ : a growth of 4% – $\displaystyle {q}={1.04}$ .
An amount that falls by $\displaystyle p$ percent is multiplied by $\displaystyle {q}={1-\frac{p}{100}}$ : a fall of 10% – $\displaystyle {q}={0.9}$ .

After $\displaystyle t$ units of time the amount has been multiplied by $\displaystyle q$ $\displaystyle t$ times:

\displaystyle f(t)=f(0)\cdot q^{t}

$\displaystyle f(0)$ – the amount at the start.
$\displaystyle f(t)$ – the amount after $\displaystyle t$ units of time.

The amount after a given time

Dana deposited 20,000 ₪ in a savings plan, at an interest of 4% a year.
The example below calculates the sum after 5 years, and then after more money is added.

The growth factor: $\displaystyle {q}={1.04}$ , and the unit of time is a year.
After 5 years: $\displaystyle {f(5)}={20000\cdot1.04^5\approx24333.06}$ ₪.
After those 5 years Dana added 6,000 ₪. From here the process starts again, from a starting amount of $\displaystyle {24333.06+6000}={30333.06}$ ₪.
3 years later: $\displaystyle 30333.06\cdot1.04^3\approx34120.57$ ₪.

The graph of growth and decay

The graph of $\displaystyle {f(t)}={f(0)\cdot q^t}$ is the graph of an exponential function. In the drawing: Dana’s savings, $\displaystyle {f(x)}={20000\cdot1.04^x}$ , and beside it an amount that falls by 10% every year, $\displaystyle {g(x)}={20000\cdot0.9^x}$ .

\displaystyle f(x)=20000\cdot1.04^{x}

\displaystyle g(x)=20000\cdot0.9^{x}

\displaystyle y=0

Both graphs cut the $\displaystyle y$ -axis at the starting amount, $\displaystyle (0,20000)$ .
When $\displaystyle q>1$ the graph increases, and when $\displaystyle 0<q<1$ it decreases.
In decay the amount comes closer and closer to 0 and never reaches it: the horizontal asymptote is $\displaystyle {y}={0}$ .

A uniform unit of time

In $\displaystyle {f(t)}={f(0)\cdot q^t}$ the time $\displaystyle t$ is measured in the same unit as $\displaystyle q$ . When the time is given in another unit, we change it to the unit of $\displaystyle q$ .

Dana’s interest is 4% a year. After 18 months, $\displaystyle {t}={1.5}$ years have passed: $\displaystyle 20000\cdot1.04^{1.5}\approx21211.92$ ₪.
The growth factor for a month is the number that, multiplied by itself 12 times, gives $\displaystyle 1.04$ : $\displaystyle 1.04^{\frac{1}{12}}\approx1.00327$ .
So the sum grows by about 0.33% a month, not by $\displaystyle \frac{4}{12}\%$ .

Finding the decay factor from two amounts

Car A is worth 60,000 ₪ today, and its value falls every year by 8% of its value the year before.
Car B is worth 79,000 ₪ today, and its value falls every year by a fixed percent.
In 10 years the two cars will be worth the same.

The example below finds by what percent the value of car B falls every year.

The value of car A in 10 years: $\displaystyle 60000\cdot0.92^{10}\approx26063.31$ ₪.
For car B: $\displaystyle {79000\cdot q^{10}}={26063.31}$ , so $\displaystyle q^{10}\approx0.32991$ .
$\displaystyle {q}={0.32991^{\frac{1}{10}}\approx0.895}$ .
$\displaystyle {1-0.895}={0.105}$ : the value of car B falls by about 10.5% every year.

Exercise 1

50,000 people live in a town.
The number of residents grows by 3% every year.

What is the growth factor?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power