# Thread: Exponential Models and Application help

1. ## Exponential Models and Application help

I am provided with an example for an Exponential mode, but I can't understand parts of it, I'd appreciate if some one could explain them a bit more.

Question:

There are apporximately ten times as many red kangaroos as grey kangatoos in a certain area. If the population of grey kangaroos increases at a rate of 11% per annum while that of the red kangaroos decreases at 5% per annum, find how many years must elapse before the proportions are reversed, assuming the same rates continue to apply.

Solution

Let $P$ = population of grey kangaroos at the start.

Therefore the number of grey kangaroos after $n$ years = $P(1.11)^n$

The number of red kangaroos after $n$ years = $10P(0.95)^n$

Question 1 : Why is the percentage raised to the power of $n$ (+n years later)?

When Proportions are reversed:

$P(1.11)^n = 10* [10P(0.95^n)]$
$(1.11)^n = 100(0.95)^n$

Question 2 : Why is $[10P(0.95^n)]$ multiplied by $10$?

2. Originally Posted by 22upon7
I am provided with an example for an Exponential mode, but I can't understand parts of it, I'd appreciate if some one could explain them a bit more.

Question:

There are apporximately ten times as many red kangaroos as grey kangatoos <<<<< see question 2 in a certain area. If the population of grey kangaroos increases at a rate of 11% per annum while that of the red kangaroos decreases at 5% per annum, find how many years must elapse before the proportions are reversed, assuming the same rates continue to apply.

Solution

Let $P$ = population of grey kangaroos at the start.

Therefore the number of grey kangaroos after $n$ years = $P(1.11)^n$

The number of red kangaroos after $n$ years = $10P(0.95)^n$

Question 1 : Why is the percentage raised to the power of $n$ (+n years later)?

When Proportions are reversed:

$P(1.11)^n = 10* [10P(0.95^n)]$
$(1.11)^n = 100(0.95)^n$

Question 2 : Why is $[10P(0.95^n)]$ multiplied by $10$?
I'll take the 2nd line of your calculation:

$(1.11)^n = 100(0.95)^n~\implies~\dfrac{(1.11)^n}{(0.95)^n}=10 0~\implies~\left(\dfrac{1.11}{0.95}\right)^n=100$

Use logarithms to determine n:

$n = \log_{\frac{111}{95}}(100)=\dfrac{\ln(100)}{\ln(11 1)-\ln(95)} \approx 29.6\ years$

3. Thanks, I understand the final Log part of the question.

What I don't understand is why the percentage is raised to the power of n/number of years?

Also doesn't the equation for red kangaroos [10P(0.95^n)] already reflect the fact that it is 10 times more than the number of Grey Kangaroos?

Thanks again,

Dru

4. Originally Posted by 22upon7
Thanks, I understand the final Log part of the question.

What I don't understand is why the percentage is raised to the power of n/number of years?

Also doesn't the equation for red kangaroos [10P(0.95^n)] already reflect the fact that it is 10 times more than the number of Grey Kangaroos? <<<<< Yes

Thanks again,

Dru
Let's start at the beginning:

At the beginning you have a population of P animals and the population increases by 11% per year:

$
year\ 0 : P$

$year\ 1: P + P*11\% = P(1+11\%) = P(1+0.11)$
$year\ 2: P(1+0.11) + P(1+0.11) * 0.11 = P(1+0.11)(1 + 0.11) = P(1+0.11)^2$
$year\ 3: P(1+0.11)^2 + P(1+0.11)^2 * 0.11 = P(1+0.11)^2(1 + 0.11) = P(1+0.11)^3$

....

$year\ n: P(1+0.11)^{n-1} + P(1+0.11)^{n-1} * 0.11 = P(1+0.11)^{n-1}(1 + 0.11)$ $= P(1+0.11)^n$