NZ Level 7 (NZC) Level 2 (NCEA)
Transforming Exponential Expressions
Lesson

In consumer arithmetics we learnt about compound interest and its corresponding formula $A=P\left(1+r\right)^n$A=P(1+r)n. This is in fact an example of exponential growth, where an amount increases by the same rate, but an increasing amount, at regular time intervals. The opposite of exponential growth is exponential decay, where the same thing happens except the amount decreases instead of increases. We've also seen an example of this in reducing balance depreciation, where the formula used is $A=P\left(1-r\right)^n$A=P(1r)n, and can be seen as the same as the compound interest formula where the rate is negative.

This same formula can be applied to other types of exponential growth and decay questions, whether it's talking about decreasing prices, bacteria multiplying, or something entirely different. Let's recap what it means:

Exponential growth and decay formula

$A=P\left(1+r\right)^t$A=P(1+r)t

where:

• $A$A is the resulting amount after the growth/decay
• $P$P is the principal amount, or initial amount
• $r$r is the rate of change, and indicates growth if $r>0$r>0, and decay if $r<0$r<0
• $t$t is the number of time periods that has passed from the start (same as $n$n in consumer arithmetic)

For example, if the population of a town is represented by $10000\times1.02^t$10000×1.02t and $t$t is the number of years, then that means it currently has a population of $10000$10000, it grows by $0.02=2%$0.02=2% every year, and that in $3$3 years the population would be $10000\times1.02^3$10000×1.023 ≈ $10612$10612.

## Rates at different time periods

Sometimes we want to see what the equivalent rate is for a different time period in the same situation, For example using the same case as before, we know that the annual growth rate is $2%$2%, but what would be the monthly growth rate? It is not as simple as dividing $2%$2% by $12$12, as $10000\left(1+\frac{2%}{12}\right)^{36}\approx10618$10000(1+2%12)3610618 (note that $36$36 months$=$=$3$3 years), which is not the answer we worked out before.

Instead let's try and restructure the formula into the form $A=P\left(1+r\right)^T$A=P(1+r)T, where $T$T is the number of months.

 $A$A $=$= $10000\times1.02^t$10000×1.02t $=$= $10000\times1.02^{\left(\frac{1}{12}\times12\times t\right)}$10000×1.02(112​×12×t) $=$= $10000\left(1.02^{\frac{1}{12}}\right)^{\left(12\times t\right)}$10000(1.02112​)(12×t) $\approx$≈ $10000\times1.0017^{12t}$10000×1.001712t $=$= $10000\left(1+0.0017\right)^{12t}$10000(1+0.0017)12t $=$= $10000\left(1+0.0017\right)^T$10000(1+0.0017)T as $12t$12t calculates how many months have passed

Wow look at that! With some simple exponential manipulation we managed to figure out that equivalently, the population grows at approximately $0.17%$0.17% each month.

#### Worked example

Justine bought an oil painting for $\$16001600 whose value is given by the formula $A=1600\times0.94^t$A=1600×0.94t, and $t$t is the number of years passed.

(a) What's the annual depreciation rate?

Think: What's $r$r in this case?

Do: $1600\times0.94^t=1600\left(1-0.06\right)^t$1600×0.94t=1600(10.06)t,

and so the annual depreciation rate is $0.06=6%$0.06=6%.

(b) Rearrange the formula into the form $A=P\left(1-r\right)^T$A=P(1r)T, where $T$T is the number of quarters passed, rounding numbers to $4$4 d.p., and find the quarterly depreciation rate.

Think: How many quarters are there in a year? What should $T$T be in terms of $t$t?

Do:

There are $4$4 quarters in a year, so $T=4t$T=4t.

 $1600\times0.94^t$1600×0.94t $=$= $1600\times0.94^{\left(\frac{1}{4}\times4\times t\right)}$1600×0.94(14​×4×t) $=$= $1600\left(0.94^{\frac{1}{4}}\right)^{\left(4\times t\right)}$1600(0.9414​)(4×t) $=$= $1600\times0.9847^{4t}$1600×0.98474t $=$= $1600\left(1-0.0153\right)^{4t}$1600(1−0.0153)4t

So then the quarterly depreciation rate is $1.53%$1.53%.

#### Practice questions

##### Question 1

The population of a particular mining town increased $160%$160% in $9$9 years, from $5100$5100 in 2004 to $13260$13260 in 2013. Assuming that the population increased at a constant annual rate, answer the following.

1. Find an expression for $A$A, the size of the population $y$y years after 2004. Write the expression such that it includes the annual rate of growth, correct to four decimal places.

2. Hence state the annual rate of growth. Give the rate as a percentage correct to two decimal places.

##### question 2

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.

1. What is the initial value of the function?

2. Express the function in the form $f\left(t\right)=\frac{8}{7}\left(1-r\right)^t$f(t)=87(1r)t, where $r$r is a decimal.

3. Does the function represent growth or decay of an amount over time?

decay

A

growth

B

decay

A

growth

B
4. What is the rate of decay per time period? Give the rate as a percentage.

### Outcomes

#### M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

#### 91261

Apply algebraic methods in solving problems