Question

A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 years? A species of beetles grows 32% every year. Suppose 100 beetles are released into a field.

How many beetles will there be in 20 years? A species of beetles grows 32% every year. Suppose 100 beetles are released into a field.
About when will there be 100,000 beetles?

Answer 1

`P_(10) = 1605` beetles

`P_(20) = 25791` beetles

In 24.88 years there will be 100 000 beetles

Explanation:

Let's call
`P_t` the beetle population that is in year t. If t starts in year 0, with 100 beetles, and the population grows 32% each year, then the population of beetles that will occur the following year is:

`P_1 = P_0 + 0.32P_0`

If we write this equation for a year t, then
`P_t` will have the following form:

`P_t = P_0 (1 + 0.32) ^ t`

Now we find
`P_t = 10`

`P_(10) = 100 (1 + 0.32) ^ {10}`

`P_(10) = 1605` beetles

In 20 years there will be:

`P_(20) = 100 (1 + 0.32) ^ {20}`

`P_(20) = 25791` beetles

To know when there will be 100 000 beetles we equal Pt to 100 000 and we clear t.

`100000 = 100 (1 + 0.32) ^ t\\\\ ln (1000) = t * ln (1 + 0.32)\\\\ t = (ln (1000))/(ln (1 + 0.32))`

t = 24.88 years

In 24.88 years there will be 100 000 beetles

Answer 2

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32

Given that 100 beetles are released into a field.

So that means initial number of beetles P=100

Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

`A=P(1+r)^n` where A is number of beetles at any year n.

Plug the given values into above formula we get:

`A=100(1+0.32)^n`

`A=100(1.32)^n`

now plug n=10 years

`A=100(1.32)^(10)=100(16.0597696605)=1605.97696605`

Hence answer is approx 1606 beetles will be there after 10 years.

To find answer for 20 years plug n=20 years

`A=100(1.32)^(20)=100(257.916201549)=25791.6201549`

Hence answer is approx 25791 beetles will be there after 20 years.

Now we have to find time for 100000 beetles so plug A=100000

`A=100(1.32)^n`

`100000=100(1.32)^n`

`1000=(1.32)^n`

`log(1000)=n*log(1.32)`

`(\log\left(1000\right))/(\log\left(1.32\right))=n`24.8810001465=n

Hence answer is approx 25 years.