Question #613e0

1 Answer
May 4, 2017

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")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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"#