# Question 613e0

May 4, 2017

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

$\approx 3 \text{ years "69 1/4" days}$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble about the model to use}}$

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")$\textcolor{g r e e n}{\text{ and "-x%t" for decay}}$
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$\textcolor{b l u e}{\text{Determine a percentage that works for this question}}$

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 $t \to {P}_{t}$
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 $\ln \left(e\right) = 1$

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

For info only $\frac{x}{100} \approx 0.0572$ to 4 decimal places

->" approximately "5.7%
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$\textcolor{b l u e}{\text{Answering the question}}$

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

Divide both sides by 100

15=18/e^(x%t)

x%t=ln(18/15)#

$t = \ln \left(\frac{18}{15}\right) \div \ln \left(\frac{18}{17}\right) = 3.1897 \ldots$

$\approx 3 \text{ years "69 1/4" days}$