NZ Level 7 (NZC) Level 2 (NCEA)
Applications of Exponential Functions
Lesson

## Application 1: Depreciation

#### Example 1

A particular sports convertible costs $\$48000$$48000. Its value depreciates ( which means to lose value) at a rate of 18%18% per year. 18%18% of 4800048000 is 86408640, so after 11 year the car will be worth \39360$$39360.

After the second year, the value of the car will drop by $18%$18% of $\$39360$$39360, or \7084.80$$7084.80, and so will be only be worth $\$32275.20$$32275.20. ##### Developing a formula Rather than continuing to work through each year in this way, we could approach the problem differently. If it is to drop in value by 18%18%, then it retains \left(1-\frac{18}{100}\right)=0.82=82%(118100)=0.82=82%. Thus, after the first year its value will be 48000\left(1-\frac{18}{100}\right)48000(118100). After the second year, we simply apply the same factor \left(1-\frac{18}{100}\right)(118100) to the new value. Its value would become: 48000\left(1-\frac{82}{100}\right)\left(1-\frac{18}{100}\right)=42000\times\left(1-\frac{18}{100}\right)^248000(182100)(118100)=42000×(118100)2 A pattern is emerging. After the third year the new value becomes: 48000\times\left(1-\frac{18}{100}\right)^348000×(118100)3 In fact the future value (often denoted FVFV) after nn years of the car is given by: FV_n=48000\times\left(1-\frac{18}{100}\right)^nFVn=48000×(118100)n ##### A general formula Note that we could simplify the formula to read FV_n=48000\left(\frac{82}{100}\right)^nFVn=48000(82100)n, but it is often left as it is to remind us of how it was constructed. In general, any item, with a current value of AA, and depreciation rate r%r%, will have a future value after nn years given by: FV_n=A\left(1-\frac{r}{100}\right)^nFVn=A(1r100)n The important principle here is that we now have a mechanism to find any future value quickly. For example, after 1010 years the car will be worth FV_{10}=48000\times\left(1-\frac{18}{100}\right)^{10}=6597.5FV10=48000×(118100)10=6597.5 In addition logarithms can be used to determine how many years it will take for the car to depreciate down to any particular value. For example, to determine when the car will be worth half its original value, simply solve the equation 24000=48000\times\left(1-\frac{18}{100}\right)^n24000=48000×(118100)n.  2400024000 == 48000\times\left(1-\frac{18}{100}\right)^n48000×(1−18100​)n 0.50.5 == \times\left(1-\frac{18}{100}\right)^n×(1−18100​)n 0.50.5 == 0.82^n0.82n \log0.5log0.5 == n\log0.82nlog0.82 nn \approx≈ 3.53.5 So the sports car will halve in value after 33 years and 66 months. ## Application 2: The Rule of 72 In a similar way, an amount of money invested with a bank will grow exponentially by earning bank interest. If an amount of \A$$A is invested at $r%$r% for $n$n years, then the future value becomes:

$FV_n=A\left(1+\frac{r}{100}\right)^n$FVn=A(1+r100)n

Note that the only difference between this formula and the previous one is the presence of a plus sign inside the brackets. The formula is developed in precisely the same way as the depreciation formula above.

Before the widespread availability of calculators, bankers used a 'rule of thumb' called the Rule of $72$72 to determine how long in years an amount invested with them took to double in value.

The rule of thumb was simply to divide the number $72$72 by the current interest rate. So for example if the bank offered $6%$6%, the number of years it took to double was approximately $\frac{72}{6}=12$726=12 years.

To check how accurate the rule is, using $r=6%$r=6% we find that, after $12$12 years an amount $A$A turns into:

$FV_{12}=A\left(1+\frac{6}{100}\right)^{12}=2.012A$FV12=A(1+6100)12=2.012A

which is slightly over the $2A$2A predicted.

At $4%$4%, the rule of thumb says $\frac{72}{4}=18$724=18 years to double, and the true value after $18$18 years turns out to be $2.025A$2.025A.

One way to investigate this remarkable observation is to solve for $n$n in the equation $2A=A\left(1+\frac{r}{100}\right)^n$2A=A(1+r100)n. Thus:

 $2A$2A $=$= $A\left(1+\frac{r}{100}\right)^n$A(1+r100​)n $2$2 $=$= $\left(1+\frac{r}{100}\right)^n$(1+r100​)n $\log2$log2 $=$= $n\log\left(1+\frac{r}{100}\right)$nlog(1+r100​) $n$n $=$= $\frac{\log2}{\log\left(1+\frac{r}{100}\right)}$log2log(1+r100​)​

Hence we need to compare the values of $\frac{\log2}{\log\left(1+\frac{r}{100}\right)}$log2log(1+r100) and $\frac{72}{r}$72r across a range of interest rates. The comparison is depicted in the graph below. Astonishingly, the rule of $72$72 is exceptionally close to the true doubling time.

##### A POSSIBLE EXPLANATION

The quantity $\frac{\log2}{\log\left(1+\frac{r}{100}\right)}$log2log(1+r100) can be written $\frac{\ln2}{\ln\left(1+\frac{r}{100}\right)}$ln2ln(1+r100) using the change of base formula. The denominator $\ln\left(1+\frac{r}{100}\right)$ln(1+r100) can be shown mathematically to to be very close to the fraction $\frac{r}{100}$r100 provided $r$r is reasonably small. Check this with your calculator.

Hence the quantity $\frac{\ln2}{\ln\left(1+\frac{r}{100}\right)}$ln2ln(1+r100) is very slightly less than $\frac{\ln2}{\frac{r}{100}}=\frac{100\ln2}{r}$ln2r100=100ln2r or approximately $\frac{69}{r}$69r. Then perhaps, after checking, the number $69$69 may have been lifted to $72$72 to compensate for the shortfall in the original approximation. The added advantage is that $72$72 has lots of factors!

What ever the reason, it is an excellent rule of thumb!

#### Worked Examples

##### Question 1

A car originally valued at $\$28000$$28000 is depreciated at the rate of 15%15% per year. The salvage value SS of the car after nn years is given by S=28000\left(1-\frac{15}{100}\right)^nS=28000(115100)n 1. Complete the following table of values for S=28000\left(1-\frac{15}{100}\right)^nS=28000(115100)n. Express your answers to 2 decimal places where necessary.  nn SS 22 44 66 88 1010 1212 \editable{} \editable{} \editable{} \editable{} \editable{} \editable{} 2. Choose the correct curve for S=28000\left(1-\frac{15}{100}\right)^nS=28000(115100)n Loading Graph... A Loading Graph... B Loading Graph... C Loading Graph... D Loading Graph... A Loading Graph... B Loading Graph... C Loading Graph... D 3. What is the value of the car after 33 years? Answer to two decimal places where necessary. 4. After how many years does the value of the car drop down to an amount of \6485.27$$6485.27?

##### Question 2

Consider the following questions.

1. Find the equation for the value, $A$A, of a $\$30003000 investment at $18$18% p.a., with interest compounded annually, after $n$n years.

2. Find the value of the investment after $6$6 years, rounding to the nearest cent.

 $n$n $A$A $1$1 $2$2 $3$3 $4$4 $5$5 $6$6 $7$7 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
4. Which of the following graphs correctly represents the function?

A

B

C

D

A

B

C

D

### Outcomes

#### M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

#### M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

#### 91257

Apply graphical methods in solving problems

#### 91261

Apply algebraic methods in solving problems