10/14 10. Exponential Functions and Applications

Exponential Functions and Rates

Quantities that change by a constant multiplier over equal intervals are modeled exponentially. The general form is

Here, $a\neq 0$ is the initial value or vertical scale factor, $b$ is the base, and $x$ is the input or time variable. The variable's position in the exponent distinguishes an exponential function from a polynomial such as $x^3$.

Recognizing growth and decay

• If $b>1$, the function represents exponential growth.

• If $0<b<1$, the function represents exponential decay.

A percentage rate must be written as a decimal before it is used. A growth rate of $r$ uses the factor $1+r$, while a decay rate of $r$ uses the factor $1-r$. For example, growth of $6\%$ gives a factor of $1.06$, and decay of $6\%$ gives a factor of $0.94$.

For $P(t)=500(1.08)^t$, the initial amount is $500$, and the amount increases by $8\%$ each time period. For $V(t)=12{,}000(0.85)^t$, the value retains $85\%$ each period and therefore decreases by $15\%$.

Takeaway: Identify the initial value and the multiplier first. A multiplier greater than one means growth; a positive multiplier less than one means decay.

Graphs and Key Features

For the basic function $y=b^x$, the domain is $(-\infty,\infty)$ and the range is $(0,\infty)$. The graph has a $y$-intercept of $(0,1)$, because $b^0=1$, and its horizontal asymptote is $y=0$.

When the function is $f(x)=ab^x$, the initial value appears at the $y$-intercept:

Thus, the intercept is $(0,a)$. The graph increases when $b>1$ and decreases when $0<b<1$. Without a vertical shift, it approaches but does not reach $y=0$.

Reading a graph or equation

1. Evaluate the function at $x=0$ to find the initial value.

2. Examine the base to determine growth or decay.

3. Use the horizontal asymptote to describe the long-term behavior.

4. Check that the output remains positive when $a>0$ and no vertical shift is present.

Takeaway: The base controls the direction of change, while the coefficient controls the starting height.

Continuous Growth, Decay, and Half-Life

A continuous growth or decay model uses the natural exponential base $e$:

The initial amount is $A_0$, the amount at time $t$ is $A(t)$, and $k$ is the continuous growth or decay constant. Use matching time units for $t$ and $k$. When $k>0$, the quantity grows; when $k<0$, it decays.

Continuous growth example

A culture begins with $500$ cells and grows continuously at $3\%$ per hour. After $8$ hours:

So the culture contains approximately $636$ cells.

Half-life model

If a substance has initial mass $800$ grams and a half-life of $6$ years, a suitable model is

After $18$ years, three half-lives have passed:

Therefore, $100$ grams remain.

Finding a factor from two values

Suppose a population starts at $2{,}000$ and reaches $2{,}420$ after $3$ years. For $A(t)=A_0b^t$:

The population grows by approximately $6.5\%$ per year because the annual factor is about $1.065$.

Takeaway: Use $A_0e^{kt}$ when the rate is continuous, and use a half-life model when repeated halving describes the decay.

Compound Interest Models

Compound interest applies exponential growth to money. For principal $P$, annual rate $r$ written as a decimal, time $t$ in years, and $n$ compounding periods per year, use

Common values of $n$ include $1$ for annually, $2$ for semiannually, $4$ for quarterly, $12$ for monthly, and commonly $365$ for daily compounding.

For continuous compounding, use

Periodic compounding example

A deposit of $\$1{,}200$ earns $4\%$ annually, compounded quarterly, for $6$ years. Substitute $P=1200$, $r=0.04$, $n=4$, and $t=6$:

The balance is approximately $\$1{,}518.38$.

Continuous compounding example

With the same deposit, rate, and time:

The balance is approximately $\$1{,}525.97$.

Takeaway: Match the formula to how often interest is added, and keep the rate as a decimal.

Solving Exponential Equations

When the unknown appears in an exponent, first try to rewrite both sides with a common base. If that is not convenient, use logarithms.

Common-base method

The one-to-one property says that if $a^u=a^v$ for $a>0$ and $a\neq 1$, then $u=v$. For example:

Therefore,

Logarithm method

For $a^x=c$, where $a>0$, $a\neq 1$, and $c>0$, take the natural logarithm of both sides:

The power property gives $x\ln a=\ln c$, so

For example, if $2^x=7$, then

Linear expressions in the exponent

For $5e^{2x+1}=40$, isolate the exponential expression first:

Then take the natural logarithm:

and solve:

A dependable sequence is:

1. Isolate the exponential expression.

2. Take logarithms of both sides.

3. Use the power property to move the exponent in front of the logarithm.

4. Solve the resulting linear equation.

5. Check the result in the original equation.

Takeaway: Use common bases when possible; otherwise, isolate first and then apply logarithms.

Modeling and Interpreting Applications

Application problems become manageable when the situation is translated into a model before calculating.

General strategy

1. Identify the initial amount, usually the value at $t=0$.

2. Identify the growth or decay factor or rate, converting percentages to decimals.

3. Choose the appropriate model: $A=A_0b^t$, $A=A_0e^{kt}$, or a compound-interest formula.

4. Substitute the known values.

5. Use logarithms if the unknown is in an exponent.

6. Interpret the result with units and suitable rounding.

Finding time

Suppose an investment follows

To find when it reaches $\$4{,}000$, set the model equal to the target:

Divide by $2500$ and take logarithms:

The investment reaches the target after approximately $9.58$ years. If only complete years are counted, it exceeds the target during the tenth year.

Common checks

• Write $4\%$ as $0.04$, not $4$.

• Use $1+r$ for growth and $1-r$ for decay.

• Distinguish the number of compounding periods $n$ from the number of years $t$.

• Avoid rounding intermediate values too early.

• State units and explain what the numerical answer means.

• Do not treat exponential change as linear change: the amount added or removed varies because it is based on the current amount.

Takeaway: A correct model, consistent units, careful rounding, and a contextual interpretation are all part of a complete solution.

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