Question

Precalculus help?

Answer 1

I tried this:

Explanation:

I tried using the suggestion (in the question) of geometric sequence considering a sequence of the type:

`a_n=("first value")*("common ratio")^(n-1)`

representing the population of the herd after `n` years.

I considered:

`"first value"=47`

`"common ratio"=51/47=56/51=61/56~~1.09`

Giving the explicit formula:

`color(red)(a_n=47*(1.09)^(n-1))`

at `n=7` we get:

`a_7=47*(1.09)^(7-1)=78.8~~79`

To get a number bigger than `85` we need:

`47*(1.09)^(n-1)>85`

rearrange:

`(1.09)^(n-1)>85/47`

`(1.09)^(n-1)>1.8`

let us take the natural log of both dsides:

`ln[(1.09)^(n-1)]>ln(1.8)`

we can use a pocket calculator and some properties of logs to solve it:

`(n-1)ln(1.09)>ln(1.8)`
`n-1>ln(1.8)/ln(1.09)`
`n>ln(1.8)/ln(1.09)+1`
`n>7.82~~8` years