# A population of 1200 badgers grow by 6% every 8 years. How long does it take the population of...

## Question:

A population of 1200 badgers grow by 6% every 8 years. How long does it take the population of badgers to grow 2000?

## Exponential Growth:

Consider a population of people in a city. Say that there are like 500 people living in the city and all of them are married. These are 250 couples. So if these couples have more than one child, then these children eventually grow old and let's say they start families of their own. These children thus have multiple children of their own. Now the growth of the population here would be more than just linear and actually takes a more exponential behavior. This is essentially exponential growth.

Given:

• {eq}\displaystyle N_0 = 1200 {/eq} is the initial badger population
• {eq}\displaystyle N = 2000 {/eq} is the current badger population

If the badger population grows by 6% every 8 years, then let us assume that in one year, we have 6/8% or 0.75% growth annually. So in eight years, we have {eq}\displaystyle N = 1.06N_0 {/eq}. And eight years after that, we have another 6% growth but this time it reflects the current population. We can thus multiply another 1.06. We can do this n times to set up an equation of the form:

{eq}\displaystyle N = N_0 (1.06)^{t/8} {/eq}

We want t here so we isolate it by first isolating the exponent:

{eq}\displaystyle \frac{N}{N_0} = (1.06)^{t/8} {/eq}

We now take the natural logarithm of both sides:

{eq}\displaystyle ln \Big[\frac{N}{N_0}\Big] = ln(1.06)^{t/8} {/eq}

We then take the exponent down:

{eq}\displaystyle ln \Big[\frac{N}{N_0}\Big] = \frac{t}{8} ln(1.06) {/eq}

And we isolate t:

{eq}\displaystyle t = \frac{8}{ ln(1.06)} \left(ln \Big[\frac{N}{N_0}\Big] \right) {/eq}

We now substitute:

{eq}\displaystyle t = \frac{8}{ ln(1.06)} \left(ln \Big[\frac{2000}{1200}\Big] \right) {/eq}

Evaluating this expression gives us:

{eq}\displaystyle \boxed{t = 70.13\ y } {/eq} 