Applications of Exponential Functions (y=a^x, a^-x only)

An exponential function is the appropriate model to use when a quantity is increasing or decreasing at a rate that depends on the quantity present.

For example, in the final rounds of a sporting competition, the number of competing teams is halved at every stage. Thus, if $16$16 teams reached the first semifinal, there would be $8$8 in the second semifinal, and so on. The number of teams playing drops from $16$16 to $8$8 and then to $4$4 and finally, $2$2 and we see that the reduction in the number of teams playing at each stage depends on the number in the previous round.

This is an example of exponential decay. The rate of decrease gets progressively smaller. Many processes show the opposite pattern and exhibit exponential growth. In this, the rate of increase increases progressively. 

 

Example 1

Exponential growth is seen in the compound interest process. If a bank deposit accumulates interest at, say, $3.1%$3.1% per annum, an amount of $\$100$$100 would grow to $\$103.10$$103.10 in $1$1 year, but in the next years it would grow to $\$106.30$$106.30, $\$109.60$$109.60, $\$113.00$$113.00, $\$116.50$$116.50, and so on.

We see that the amount of increase is getting larger at each stage because the interest is calculated on a slightly larger principle each year: $\$3.10$$3.10, $\$3.20$$3.20, $\$3.30$$3.30, ... At first, the amount of interest is small but eventually, if the money stays in the account, the interest payments will become very large.

Each year, the amount of the deposit is being multiplied by $1.031$1.031. So, the amount $A$A of the deposit after $n$n years should be $A=1.031^n\times100$A=1.031n×100. In $23$23 years, the initial deposit of $\$100$$100 will have grown to $A=1,031^{23}\times100=$A=1,03123×100= $\$201.00$$201.00.

 

 

Example 2

In a biological experiment, a sample of a microorganism was seen to double in size by cell division every $11$11 minutes.

We can work out how many cells there will be after several $11$11-minute periods by a step-by-step calculation if we know how many cells were present initially. Suppose there were $n$n cells present at the beginning of the experiment.

After $11$11 minutes there are $2n$2n cells. Then, after a further $11$11 minutes, there are $2\times2n$2×2n cells. After a third period of $11$11 minutes, there are $2\times2\times2n$2×2×2n cells. This pattern continues and we see that after $k$k groups of $11$11 minutes, there will be $2^kn$2kn cells.

The researcher observes that the increase occurs gradually and not in sudden bursts at the end of each $11$11 minutes. The number of minutes that have elapsed since the beginning of the experiment is $m=11k$m=11k. So, it seems reasonable to put $k=\frac{m}{11}$k=m11​ and write down a formula that will give the number of cells $N$N after $m$m minutes, for any number of minutes, not just multiples of $11$11. This is

$N=2^{\frac{m}{11}}n$N=2m11​n.

Suppose there were $5$5 cells present at the beginning of the experiment and the growth of the microorganism was observed for $100$100 minutes. How many cells would there be at the end of this time?

We have $n=5$n=5. The exponent $\frac{m}{11}$m11​ is $\frac{100}{11}\approx9$10011​≈9. So, on substituting these numbers into the formula, we calculate $N=2^9\times5=2560$N=29×5=2560 cells.

 

Example 3

A certain radioactive isotope decays in such a way that after $175$175 years only half of the original quantity of the isotope remains.

Suppose $10$10 kg of the substance existed initially. After $175$175 years, only $5$5 kg will be left and then after a further $175$175 years, only $2.5$2.5 kg will be left, and so on. After $k$k groups of $175$175 years, we would expect $A=10\times\left(\frac{1}{2}\right)^k$A=10×(12​)k of the substance to remain.

The number of years that have elapsed since the beginning of this experiment must be $y=175k$y=175k. So, we can put $k=\frac{y}{175}$k=y175​ into the formula so that we no longer have to convert the time elapsed into groups of $175$175 years but can use just years instead. 

Also, we can replace $\left(\frac{1}{2}\right)^k$(12​)k with the equivalent expression  $2^{-k}$2−k.

Thus, we arrive at a formula for the amount remaining after $y$y years:

$A=10\times2^{-\frac{y}{175}}$A=10×2−y175​

In $1000$1000 years there will be $A=10\times2^{-\frac{1000}{175}}=0.19$A=10×2−1000175​=0.19 kg of the isotope remaining.

 

Worked Examples

Question 1

Question 2

Question 3