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In real-life settings, there exists phenomena where quantities either grow or decay exponentially. Typically, these are situations like uninhibited population growth, radioactive decay, or compound interest. Studying these in terms of growth and decay helps understand their behavior in the long run.

The constant r can then be interpreted as the *rate of growth*, in decimal form. A value of 0.06, for instance, means that the quantity increases by $6%$ over every unit of time. As is the case with all exponential functions, a is the y-coordinate of the y-intercept.

The constant r can then be interpreted as the *rate of decay*, in decimal form. A value of 0.12, for instance, would mean that the quantity decreases by $12%$ over every unit of time.

In an ideal environment, bacteria populations grow exponentially and can be modeled with an exponential growth function. The bacteria Lactobacillus acidophilus duplicates about once every hour. A single bacteria cell is placed in an ideal environment. State and interpret the constants a and r for the growth that will occur. Then, write a function rule describing this growth.

Show Solution *expand_more*

The constant a is the initial value of the quantity, in this case the number of bacteria. There was only one bacteria placed in the environment, so a is 1. The constant r is the rate of growth. Since the bacteria duplicate every hour, the amount of bacteria doubles every hour. This corresponds to an increase by $100%.$ Thus, r is 1. Substituting this into the rule of an exponential growth function gives
which can be simplified as
Since this is an exponential growth function, population will grow faster and faster, without bound. In a real environment, this would not happen, since the available space and nutrition would have to be infinite. At some point, the environment would no longer be ideal, so the growth would slow down or stop.

For an exponential growth function written as
the quantity (1+r) is called the *growth factor.*
Likewise, for an exponential decay function,
(1−r) is called the *decay factor.* In context, it is often the rate of growth or decay, r, that is given. Using r, it is possible to evaluate the corresponding growth or decay factor. As an example, a growth of $10%$ every unit of time gives the growth factor

(1+r)=(1+0.1)=1.1.

Similarly, it can be necessary to use the growth or decay factor to find the rate of growth or decay. For instance, a factor of 0.85 indicates a decay. Thus, r can be found by equating 0.85 with the decay factor (1−r), and solving the equation. Show Solution *expand_more*

To begin, let's analyze the given function rule. It's written in the form $y=a⋅b_{x},$ where a is the initial value and b is constant multiplier/growth factor. The constant multiplier, 0.88, is less than 1, so it is a decay factor. Therefore, the function shows decay. Since the decay factor is always equal to 1−r, we can write the equation.

0.88=1−r,

which can be solved for r.
Thus, the rate of decay is 0.12, or $12%,$ per year. The initial value is 800, and the constant multiplier is 0.88. Using this information, we can graph the exponential decay function by plotting some points that lie on H and connecting them with a smooth curve.

When money is deposited to a savings account, interest is accrued, often yearly — different types of interest work in different ways. Compound interest is when the interest earned is added to the original amount and future interest accrues into a larger amount. To find the balance of an account at any given time, an exponential growth function can be used. When the interest is compounded yearly, the balance can be modeled by a function.

In this context, P stands for the *principal*, which is the initial amount of money, and r is the interest rate in decimal form. If the interest is not compounded yearly, the function looks a little different.

The constant n is the number of times the interest is compounded per year, while r is still the annual interest rate. For an account with the principal $100 and an annual interest of $15%$ compounded twice a year, the growth function is:

Notice that this function grows continuously, whereas, in reality, the account balance only increases at the times of compound. Graphing the function together with the actual balance of the account will highlight how it can be used in practice.

Every time the interest is compounded, in this case every half year, the value of B is equal to the account balance. However, at all other times, it is not. To find, for instance, the account balance after 1.75 years, B(1.5) should then be evaluated, since that was the last time interest compounded.

One savings account, with a principal of $100, offers an annual interest rate of $15%$ compounded twice a year. Find the balance in the account after 5 years. Another savings account with the same principal will have the same balance after 5 years. However, the interest is compounded monthly. Find the interest rate of the second account.

Show Solution *expand_more*

First, we'll find the function rule describing the growth of the first account. It is given that P=100,r=0.15, and n=2. Substituting these values in the compound interest formula gives
Let's simplify this function before continuing.
Since the interest will accrute for 5 years, t=5. Therefore, we can find the account balance by evaluating B(5).
The account balance is $206.10 after 5 years. Now we can consider the second account. We know the balance of both accounts is equal, at least when both just had their interest compounded. This means that B(t) also describes the growth in the second account. However, since the interest in the second account accrues monthly, or 12 times a year, n=12.
Thus, the exponent in the rule should be 12t. By using the equality
and the power of a power property, we can rewrite B(t) so that it's possible to find the monthly interest rate. Having the exponent 12t means that the base of the power is equal to the monthly growth factor, which we can then use to find the monthly interest rate.

$B(t)=100(1+20.15 )_{2t}$

CalcQuot

Calculate quotient

$B(t)=100(1+0.075)_{2t}$

AddTerms

Add terms

$B(t)=100⋅1.075_{2t}$

$B(t)=100⋅1.075_{2t}$

Substitute

t=5

$B(5)=100⋅1.075_{2⋅5}$

Multiply

Multiply

$B(5)=100⋅1.075_{10}$

UseCalc

Use a calculator

$B(5)=206.10315…$

RoundDec

Round to

$B(5)≈206.10$

$B(t)=100⋅1.075_{2t}$

Rewrite

Rewrite 2t as $1/6⋅12t$

$B(t)=100⋅1.075_{1/6⋅12t}$

ProdInExponent

$a_{m⋅n}=(a_{m})_{n}$

$B(t)=100⋅(1.075_{1/6})_{12t}$

UseCalc

Use a calculator

$B(t)=100⋅(1.01212…)_{12t}$

RoundDec

Round to

$B(t)≈100⋅1.012_{12t}$

We find an approximate monthly growth factor 1.012, which corresponds to a rate of growth that is 0.012. Thus, the monthly interest rate is roughly $1.2%.$

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