Finding Exponential Equations in Context

Certain growth and decay situations arise that naturally lead to the formulation of a rule that can be used generally.

Example 1 

An amount of $$5000$5000 is invested at $6$6%pa until it triples in value. When will that be?

Increasing an amount A by $6$6% is equivalent to applying the factor $1.06$1.06, simply because $A\left(1.06\right)=A\left(1+0.06\right)=A+\frac{6}{100}A$A(1.06)=A(1+0.06)=A+6100​A.

The amount $6$6% is referred to the rate of growth of the investment.

In our problem, the factor must be applied each successive year as illustrated in this simple table:

Year	Amount	Pattern
$0$0	$5000$5000	$5000$5000
$1$1	$5000\left(1.06\right)$5000(1.06)	$5000\left(1.06\right)^1$5000(1.06)1
$2$2	$5000\left(1.06\right)\left(1.06\right)$5000(1.06)(1.06)	$5000\left(1.06\right)^2$5000(1.06)2
$3$3	$5000\left(1.06\right)\left(1.06\right)\left(1.06\right)$5000(1.06)(1.06)(1.06)	$5000\left(1.06\right)^3$5000(1.06)3
$4$4	$5000\left(1.06\right)\left(1.06\right)\left(1.06\right)\left(1.06\right)$5000(1.06)(1.06)(1.06)(1.06)	$5000\left(1.06\right)^4$5000(1.06)4

The pattern becomes clear. After $n$n years the principal  amount of $5000 will have the future value given by $A_n=5000\left(1.06\right)^n$An​=5000(1.06)n. 

The future value will be triple the principal when $A_n=5000\left(1.06\right)^n>15000$An​=5000(1.06)n>15000. We can solve this using logarithms:

$5000\left(1.06\right)^n$5000(1.06)n	$>$>	$15000$15000
$\left(1.06\right)^n$(1.06)n	$>$>	$3$3
$n\log1.06$nlog1.06	$>$>	$\log3$log3
$n$n	$>$>	$\frac{\log3}{\log1.06}$log3log1.06​
$n$n	$>$>	$18.8542$18.8542
$\therefore$∴  $n$n	$=$=	$19$19

Thus it will take $19$19 years to triple the principal.

Example 2 

A problem gambler is expected to lose $12$12% of his stake each night he plays the poker machines at his local club. He starts with $$1000$1000 on Monday night, puts the entire amount through the machine, and then leaves with his takings. He uses his takings from the previous night to repeat the pattern on the next night, putting the entire amount through the machine and leaving once again with his takings. Based on his expected losses, when do we expect him to have $10$10% of his original amount left?

An expected loss of $12$12% is equivalent to saying that he is expected to leave with $88$88% of his money each night.

So after night $n$n, the amount remaining $T$T will be given by $T_n=1000\left(1-0.12\right)^n=1000\left(0.88\right)^n$Tn​=1000(1−0.12)n=1000(0.88)n. 

For example, the amount left after $5$5 nights becomes $T_5=1000\left(0.88\right)^5=527.73$T5​=1000(0.88)5=527.73.

Of course this is only an expected position, and he may well be slightly worse off or better off than this. We do know with some certainty that, on average, people who gamble in this way will inevitably lose their money.

The answer to the specific question asked can be determined by solving $1000\left(0.88\right)^n=0.1\left(1000\right)$1000(0.88)n=0.1(1000) as follows:

$1000\left(0.88\right)^n$1000(0.88)n	$=$=	$0.1\left(1000\right)$0.1(1000)
$\left(0.88\right)^n$(0.88)n	$=$=	$0.1$0.1
$n\log\left(0.88\right)$nlog(0.88)	$=$=	$\log\left(0.1\right)$log(0.1)
$n$n	$=$=	$\frac{\log\left(0.1\right)}{\log\left(0.88\right)}$log(0.1)log(0.88)​
$n$n	$\approx$≈	$18$18

Hence, after $18$18 days the gambler expects to have lost $$900$900 of his original stake. 

Other comments

The same strategy applies to similar forms of growth and decay. For example, the appreciation (meaning gain in value) of artworks and antiques, the growth rate of populations, the decay rates of radioactive material, the slow decline to extinction of certain species of animals.

Whenever there is a constant percentage periodic rate of growth or decay from a base population the same equation development applies. We generally refer to it as an exponential rate of growth or decay. 

 

Worked Examples

Question 1

Question 2

Question 3