Question

Question #613e0

Answer 1

To have these counts the percentage has to be different.
Approximately `5.7%`

`~~3" years "69 1/4" days"`

Explanation:

`color(blue)("Preamble about the model to use")`

You have what I call 'cycle' calculation system of type `P(1+x%)^t`
You have what I call continuous calculation system `Pe^(x%xxt)`

I have come across the comment that banks do not use the second one to calculate interest they owe us as it would cost them more.

In other words `Pe^(x%xxt)>P(1+x%)^t`

Apparently it is normal practice when dealing with populations to use the `color(purple)(Pe^(+x% t)` model with `color(red)(+x%t" for growth")` `color(green)(" and "-x%t" for decay")`
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`color(blue)("Determine a percentage that works for this question")`

Important: The unit of measurement for percentage is % and % is worth `1/100=>color(red)(x%)->x xx % -> color(red)(x xx 1/100)`

Set time span as `t`
Set original condition population count as `P_o=1800`
Set population count at time `t ->P_t`
Set unknown percentage as `->color(red)(x%)`

Then initial condition is such that:

`P_(t=1)=1700 = 1800e^(-color(red)(x%)xx1)`

`(17cancel(00))/(18cancel(00))=1/e^(x%)`

`e^(x%)=18/17`

`x%ln(e)=ln(18/17)`

but `ln(e)=1`

`x%=ln(18/17) larr" kept as is because decimal is not precise"`

For info only `x/100~~0.0572` to 4 decimal places

`->" approximately "5.7%`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

`P_t=1500=1800e^(-x%t)`

Divide both sides by 100

`15=18/e^(x%t)`

`x%t=ln(18/15)`

`t=ln(18/15)-:ln(18/17) =3.1897...`

`~~3" years "69 1/4" days"`