a growing city had a population of
500,000 in 2005 in 2010 the population
was
760,000 we want to assume exponential
growth and then Express the population
after T years as a function of T predict
the population in
2025 then determine in which year the
population will reach 1
million so we want to start by finding
the exponential function that's going to
model this population so we'll be using
this exponential function here where P
of T would be the population after a
certain number of years P Sub 0 or P not
would be the initial
population K would be the exponential
growth rate and T would be the time and
years so let's start by finding the
exponential function for this population
so when we read these first two
sentences the starting population would
be 500,000 so P sub Z or P not is equal
to 500,000
this is in 2005 then in 2010 the
population was 760,000 so P of
T would be equal to
760,000 and then from 2005 to 2010
that's a span of five years so T is
going to be equal to five so now we'll
perform substitution into our
exponential function and then solve for
K our exponential growth rate so we want
to solve solve the equation
760,000 =
500,000 * e ra the power of K * T but
since T is 5 we'll have
5K now we want to solve this exponential
equation for K so we'll first isolate
the exponential part so we'll divide
both sides by
500,000 this simplifies to
one and then on the left side of 760,000
ID
500,000 which is equal to
1.52 so we have 1.52 would equal e to
the power of
5K and now since we have base e here
instead of taking the common log of both
sides we'll take the natural log of both
sides and on the right side we can apply
the power property of log to move this
5K to the
front so now we'd have natural log
1.52
equals 5K * natural log e but natural
log e is equal to 1 if we have natural
log e this is log base e and since e
raised to the first power is equal to
e this is equal to
one so this simplifies to one so to
solve this for K we just need to divide
both sides by
five of course we could divide both
sides by natural log e but again that's
just equal to one so it's not going to
change
anything so we have K is equal to this
quotient which we'll have to get a
decimal approximation
for so we have natural log of
1.52 / 5
and the more decimal places that we use
for K the more accurate our answer is
going to be let's go ahead and take this
out to six decimal
places so we'll have
0.083
742 and now we have our exponential
function that's going to model this
population we'll have P of T is equal to
the initial population of
5,000 * e ra the power of 0 8374298849
P of 20 to approximate the population in
the year
2025 so now we'll go back to the
calculator and approximate this
value second natural log brings up e the
power of 8374298849
I so we want to substitute 1 million for
p of T and solve for T so we'll have 1
million equals
500,000 e raised the power of 0.0
8374
2T so we're going to isolate the
exponential part so we'll divide both
sides by
500,000 simplifies to one this would be
2 so we have 2 equals
e raised to this
exponent just as we did before we'll now
take the natural log of both sides of
the equation and then apply the power
property of logarithms so we'll move
this exponent to the front of the
logarithm so we'll have natural log 2
equals
0.083 472 T times natural log e but as
we showed before natural log e
simplifies to 1 so to solve this for T
we just need to divide by this decimal
coefficient again this simplifies to one
so let's go back to the calculator this
quotient would be the value of T which
will be the number of years after 2005
so I can see that t is approximately
8.28
years so this would be 8.28 years after
the Year
2005 and since the base year is the Year
2005 2005 plus
8.28 means that the population would
reach 1 million
during the
[Music]
year
2013 okay I hope you found this helpful