### Video Transcript

In this video, we will learn how to
solve real-world problems involving exponential functions. To do that, weβll remember the
general exponential function forms. Weβll start with π of π₯ equals π
to the π₯ power, where the base π is a positive number other than one. However, when we want to model
real-world data in an exponential form, we need to modify this slightly. We use π of π₯ equals π΄ times π
to the π₯ power. Itβs still true that π is a
positive number other than one. The independent variable, the
power, usually represents some unit of time. And the variable π΄ represents the
initial value of what the function is measuring.

We can easily see what the initial
value is when π₯ is zero, when no time has passed. Weβve already said that the base π
must be a positive number other than one. This value for the base π tells us
something about the rate at which the quantity changes over time. It tells us how our initial value
is changing. And these changes fall in one of
two categories. Weβll either have a function that
increases over time or a function that decreases over time. An increase represents exponential
growth, while a decrease represents exponential decay. When weβre dealing with exponential
growth, the π-value is greater than one. And when weβre dealing with
exponential decay, the π-value will be less than one.

But remember, weβve already said
that the π-value must be positive. And that means for exponential
decay the π-value must fall between zero and one. Before we look at any examples,
thereβs one more thing we need to note about this rate, this π-value. In exponential modeling, very often
weβll be dealing with percent increase or decrease. And we need to think carefully
about how we translate a percent increase or a percent decrease into a base
exponential value.

Letβs say we want to model a three
percent decrease over time. Since we know that this is a
decrease, we know that weβre looking for a base exponential value between zero and
one. We also know that three percent is
three out of 100 or, written as a decimal, 0.03. However, this function models the
change in the initial quantity over time. And that means weβre not modeling
how much we lost; weβre modeling how much is remaining over every unit of time. And if three percent is the
decrease, the 97 percent remains. And that means as a function, we
could call this π of π₯ equals π times 0.97 to the π₯ power. Itβs also possible to model this as
the fraction 97 over 100 to the π₯ power, though itβs more common in this type of
modeling to use the decimal form.

Now letβs think about how we would
model a three percent increase. That percentage written as a
decimal is 0.03. This is telling us that every unit
of time, weβre gaining three percent of the initial amount. And when weβre working with
exponential increase, that π-value will be greater than one. And whatβs happening here is we
have 100 percent of what we started with plus the three percent increase every unit
of time. And we model that as π of π₯
equals π times 1.03 to the π₯ power. In modeling, there are a few common
ratios we see again and again. If the value is doubling every unit
of time, then the π-value equals two. And if the changes over time is
half, the π-value is equal to one-half. Now weβre ready to consider some
examples.

The number of people visiting a
museum is decreasing by three percent a year. This year there were 50,000
visitors. Assuming the decline continues,
write an equation that can be used to find π, the number of visitors there will be
in π‘ yearsβ time.

When weβre dealing with the percent
decrease, weβre not dealing with a linear function, so we know weβll need an
exponential function to model this. This means weβll use the general
form π of π₯ equals π΄ times π to the π₯ power. Our π-variable is the rate of
change. The π₯-variable represents the unit
of time weβre measuring. And π΄ represents the initial
value. We need to be clear here that we
want to model the number of visitors there will be at the museum.

A decrease in three percent of the
visitors means that 97 percent of the visitors are maintained. Since weβre modeling the number of
visitors, weβll use 97 percent. Weβll write this percent in decimal
form 0.97. We know that our π₯-value is being
measured in π‘ years. And our starting value, our initial
value, are the 50,000 visitors from this year. Our equation is modeled with a
capital π so that we have the equation π equals 50,000 times 0.97 to the π‘
power.

In our next example, weβre given a
model, and we need to interpret data from that model.

A population of bacteria decreases
as a result of a chemical treatment. The population π‘ hours after the
treatment was applied can be modeled by the function π of π‘, where π of π‘ equals
6000 times 0.4 to the π‘ power. What was the population when the
chemical was first applied? And what is the rate of population
decrease?

Our function, thatβs the general
form π of π₯ equals π΄ times π to the π₯ power. In this form, the π΄ represents the
initial value, which means we identify 6000 as the initial value. One way to check that this is true
would be to plug in π‘ equals zero. When zero time has passed, we know
that the bacteria has its original population. 0.4 to the zero power equals one,
and 6000 times one equals 6000, which confirms the value for the initial
population. So we turn our attention to the
rate of population decrease.

For this model, we know that the
π-value tells us something about the rate. The function is written so that it
tells us how many of the bacteria remain after π‘ hours. If 0.4 remain, then 0.6 is the
amount that has been decreased. If we start with one whole amount,
some rate of decrease will give us 0.4 of the population that remain. And that value here is 0.6. We usually want to write this in
percent form, so we say that there was a 60 percent decrease. This model is showing us that after
the chemical treatment, thereβs a 60 percent decrease per hour of the population of
bacteria.

In our next example, weβll write a
model and then use it to solve for a quantity after a certain amount of time.

A microorganism reproduces by
binary fission, where every hour each cell divides into two cells. Given that there were 15,141 cells
to begin with, determine how many cells there were after five hours.

Since this microorganism is
reproducing, we will expect more cells and not less, which means weβre expecting
exponential growth. Our unit of time is every hour. That means we can let π‘ be equal
to the hours after the initial count. If every hour one cell divides into
two cells, one cell becomes two cells in an hour. After an additional hour, the two
cells become four. This represents a doubling of the
cells every hour.

So weβll need to take our
exponential form π of π₯ equals π΄ times π to the π₯ power, where π΄ is our
initial value, 15,141. π is the rate. Since our rate is doubling, π is
equal to two. And our variable will be π‘. Itβll be units of time. We now wanna take this function and
use it to solve for how many cells there were after five hours, which means we need
to calculate 15,141 times two to the fifth power, which is 484,512. After five hours, we can expect
that this microorganism will have 484,512 cells.

In our final example, weβll take
some data weβre given and use it to create a population model.

The US census is taken every 10
years. The population of Texas was 3.05
million in 1900 and 20.9 million in 2000. By modeling the growth as
exponential, answer the following questions. Write an exponential function in
the form π of π equals π nought times π to the π power to model the population
of Texas, in millions, π decades after 1900. Round your value of π to three
decimal places if necessary. According to the model, what was
the population of Texas in 1950? Give your answer to three
significant figures. And finally, rewrite your function
in the form π of π¦ is equal to π nought times π to the π¦ power, where π¦ is the
time in years after 1900. Round your value of π to four
decimal places.

Letβs start with what we know. In 1900, the population was 3.05
million. And by 2000, that value was 20.9
million. We know the general form of the
exponential function π of π₯ equals π΄ times π to the π₯ power. Weβre following this general form
with the function π of π equals π nought times π to the π power, where our π
represents time in decades after 1900. And that means our initial value π
nought should be the population in 1900. Since weβre working in millions, we
can leave this as 3.05. π is our unknown value. To solve for π, we can use our
other data point.

We know that in 2000 there was a
population of 20.9 million. We also know that 2000 is 10
decades after 1900. Plugging all of this in, we can use
that information to solve for π. To get π by itself, we divide both
sides of the equation by 3.05, which gives us 6.8524 continuing is equal to π to
the 10th power. Instead of rounding this, weβll
just leave it in our calculator as is. To isolate π, weβll raise both
sides of this equation to the one-tenth power. π to the 10th power to the
one-tenth power equals π, and 6.8524 continuing to the one-tenth power equals
1.212228 continuing. We want to round this π-value to
three decimal places.

The fourth decimal place has a
two. So we say that π equals 1.212. This π-value thatβs greater than
one tells us weβre dealing with population growth. And if we think of the decimal
0.212 as a percent, we can say that the population is growing at a rate of about
21.2 percent every decade. And weβve created a model to
calculate what the population would be π decades after 1900. π of π equals 3.05 times 1.212 to
the π power. Using this model, we want to
estimate what the population was in 1950. 1950 is 50 years after 1900, which
is five decades. To calculate this, we wanna take π
of five, 3.05 times 1.212 to the fifth power, which is equal to 7.9765
continuing. Three significant figures in this
case would be to the second decimal place. If we round to the second decimal
place, we get 7.98. Based on our model, we can expect
that the population of Texas in 1950 would have been 7.98 million.

For part three of this question, we
want to rewrite our exponential model, where our unit of time is years instead of
decades. It will be a really similar process
to what we did in the first part. Weβll still have the same initial
value of 3.05. But to solve for π¦, weβll use our
second data point. In 2000, the population was 20.9
million. And that was 100 years after our
initial value. So weβll plug in 100 for π¦, and
then we can solve for π. By dividing both sides of the
equation by 3.05, we get 6.8524 continuing equals π to the one hundredth power. Again, we donβt wanna round this
6.852 continuing yet. Weβll just leave it in our
calculator so that we can raise both sides of this equation to the one over 100
power.

π to the one hundredth power to
the one over 100 power equals π. And 6.8524 continuing to the one
over 100 power equals 1.01943 continuing. Weβre rounding to four decimal
places this time, which means weβll use π equals 1.0194. Our π-value is less than our
π-value. According to the π-value, the
population growth per year was 1.94 percent versus a 21.2 percent population growth
decade over decade. So for our yearly model, we have π
of π¦ equals 3.05 times 1.0194 to the π¦ power.

From here, weβre ready to review
our key points. The exponential form for modeling
real-world problems is π of π₯ equals π΄ times π to the π₯ power, where π is the
initial quantity, π is how the quantity changes over time β when π is greater than
one, we have exponential growth, when π is between zero and one, we have
exponential decay β and π₯ is the unit of time.