Exponential Functions

Follow us: Facebook
<
  Share this page: Facebooktwitterredditpinterest

This section covers:

Whether we like it or not, we need to revisit exponents and then start talking about logs, which will help us solve exponential and logarithmic equations. These types of equations are used in everyday life in the fields of Banking, Science, and Engineering, and Geology, as well as more fields.

We learned about the properties of exponents here in the Exponents and Radicals in Algebra Section, and did some solving with exponents here.

Also note that Factoring with Exponents and Solving Exponential Equations after Factoring can be found in the Advanced Factoring section here.

And when we study Geometric Sequences, we’ll see that they are a discrete form of an exponential function.

Introduction to Exponential Functions

Again, exponential functions are very useful in life, especially in the worlds of business and science. If you’ve ever earned interest in the bank (or even if you haven’t), you’ve probably heard of “compounding”, “appreciation”, or “depreciation”; these have to do with exponential functions.

Just remember when exponential functions are involved, functions are increasing or decreasing very quickly (multiplied by a fixed number). That’s why it’s really good to start saving your money early in life and let it grow with time.

Remember that exponential functions are named that because of the “\(x\)” in their exponents! Exponential functions are written \(y=a{{b}^{x}},\,\,b>0\). “\(b\)” is called the base of the exponential function, since it’s the number that is multiplied by itself “\(x\)” times (and it’s not an exponential function when \(b=1\)). \(b\) is also called the “growth” or “decay” factor.

When \(b>1\), we have exponential growth (the function is getting larger), and when \(0<b<1\), we have exponential decay (the function is getting smaller). This makes sense, since when you multiply a fraction (less than 1) many times by itself, it gets smaller, since the denominator gets larger.

Parent Graphs of Exponential Functions

Here are some examples of parent exponential graphs. I always remember that the “reference point” (or “anchor point”) of an exponential function (before any shifting of the graph) is \((0,1)\) (since the “\(e\)” in “exp” looks round like a “0”). (Soon we’ll learn that the “reference point” of a log function is \((1,0)\), since this looks like  the “lo” in “log”). We have to also remember that if the function shifts, this “reference point” will move. The graph of an exponential function (a parent function: one that isn’t shifted) has an asymptote of \(y=0\).

When the base is greater than 1 (a growth), the graph increases, and when the base is less than 1 (a decay), the graph decreases. But the domain and range are the same for both parent functions, and both graphs have an asymptote of \(y=0\).

Remember from Parent Graphs and Transformations that the critical or significant points of the parent exponential function \(y={{b}^{x}}\) are \(\displaystyle \left( {-1,\,\frac{1}{b}} \right),\left( {0,1} \right),\left( {1,b} \right)\).

<

“Parent” Exponential Graphs

Domain:  \(\left( {-\infty ,\infty } \right)\);   Range:  \(\left( {0,\infty } \right)\)

 

End Behavior:  \(\left\{ \begin{array}{l}x\to -\infty ,\,\,y\to 0\\x\to \infty ,\,\,\,\,\,\,y\to \infty \end{array} \right.\)

Domain:  \(\left( {-\infty ,\infty } \right)\);   Range:  \(\left( {0,\infty } \right)\)

 

End Behavior:   \(\left\{ \begin{array}{l}x\to -\infty ,\,\,y\to 0\\x\to \infty ,\,\,\,\,\,\,y\to \infty \end{array} \right.\)

Transformations of Exponential Functions

Remember again the generic equation for a transformation with vertical stretch \(a\), horizontal shift \(h\), and vertical shift \(k\) is \(f\left( x \right)=a\cdot {{b}^{{x-h}}}+k\) for exponential functions.

Remember these rules:

When functions are transformed on the outside of the \(f(x)\) part, you move the function up and down and do “regular” math, as we’ll see in the examples below. These are vertical transformations or translations.

When transformations are made on the inside of the \(f(x)\) part, you move the function back and forth (but do the opposite math – basically since if you were to isolate the \(x\), you’d move everything to the other side). These are horizontal transformations or translations.

When there is a negative sign outside the parentheses, the function is reflected (flipped) across the \(x\)-axis; when there is a negative sign inside the parentheses, the function is reflected across the \(y\)-axis.

For exponential functions, get the new asymptote by setting \(y=\) the vertical shift. The domain is always \(\left( {-\infty ,\infty } \right)\), and the range changes with the vertical shift.

Here are some examples, using t-charts::

Transformation

T-chart

Graph

\(y={{2}^{{x-4}}}+3\)

 

Parent function:

\(y={{2}^{x}}\)

 

For exponential functions, use –1, 0, and 1 for the \(x\) values of the parent function.

To get new asymptote, set \(y=\) to the vertical shift.

x + 4 x y y + 3
3 –1 .5 3.5
4 0 1 4
5 1 2 5

 

Asymptote:  \(y=3\)

Domain:  \(\left( {-\infty ,\,\,\infty } \right)\)

Range:  \(\left( {3,\,\,\infty } \right)\)

Translate 4 units right, 3 units up.

\(y=-2{{e}^{{x+1}}}+2\)

 

Parent function:

\(y={{e}^{x}}\)

 

Remember that \(e\approx 2.7\).

Since the coefficient is negative, the graph is reflected (flipped) across the \(x\)-axis.

x – 1 x y 2y + 2
 –2 –1 \(\frac{1}{e}\) 1.26
–1 0 1 0
 0 1 e –3.44

 

Asymptote:  \(y=2\)

Domain:  \(\left( {-\infty ,\,\,\infty } \right)\)

Range:  \(\left( {-\infty ,2} \right)\)

Vertical stretch by a factor of 2, reflect over the \(x\)-axis, translate up 2 units, left 1 unit.

Writing Exponential Equations from Points and Graphs

You may be asked to write exponential equations, such as the following:

<
    • Write an equation to describe the exponential function in form \(y=a{{b}^{x}}\), with a given base and a given point.
    • Write an exponential function in form \(y=a{{b}^{x}}\) whose graph passes through two given points. (You may be able to do this using Exponential Regression.)
    • For a certain graph, write the appropriate exponential function of the form \(f\left( x \right)=a\cdot {{b}^{{x-h}}}+k\), given a certain base (given a base and asymptote).
    • For a certain graph, write the appropriate exponential function of the form \(y=a{{b}^{x}}+k\) (an exponential function with a vertical shift).

Problem Solution
Write an equation to describe the exponential function in form \(y=a{{b}^{x}}\), with base 3 and passing through the point \(\left( {4,\,\,162} \right)\). The equation will be in the form \(y=a{{\left( 3 \right)}^{x}}\), since the base is 3. Plug in 4 for x and 162 for \(y\), and solve for \(a\):

\(\begin{align}y&=a{{\left( 3 \right)}^{x}}\\162&=a{{\left( 3 \right)}^{4}}\\a&=\frac{{162}}{{81}}=2\end{align}\)             The equation is \(y=2{{\left( 3 \right)}^{x}}\).

Write an exponential function in form \(y=a{{b}^{x}}\) whose graph passes through the two points \(\left( {3,\,\,10} \right)\) and \(\left( {5,\,\,40} \right)\). By plugging in the given points, the two equations we’ll have are \(10=a{{\left( b \right)}^{3}}\) and \(40=a{{\left( b \right)}^{5}}\). We need to find \(a\) and \(b\); this is a system of equations.

The trick is to solve the first equation for \({{b}^{3}}\), and then substitute this in the second equation by factoring the \({{b}^{5}}\) to make \({{b}^{3}}\cdot {{b}^{2}}\). We could have also just solved for \(b\), but this way is easier:

\(\require{cancel} \begin{array}{l}10=a{{b}^{3}};\,\,\,\,\,{{b}^{3}}=\frac{{10}}{a}\\40=a{{b}^{5}};\,\,\,\,40=a{{b}^{3}}{{b}^{2}}\\40=\cancel{a}\left( {\frac{{10}}{{\cancel{a}}}} \right){{b}^{2}};\,\,\,\,\,{{b}^{2}}=\frac{{40}}{{10}}=4\\b=2\,\,\,\,\text{(base can }\!\!’\!\!\text{ t be negative)}\end{array}\) \(\begin{array}{c}\text{Plug }b\text{ in either equation for }a:\\10=a{{\left( 2 \right)}^{3}}\\a=\frac{{10}}{8}=\frac{5}{4}\end{array}\)

The exponential function is: \(\displaystyle y=\frac{5}{4}{{\left( 2 \right)}^{x}}\). Check your points; they work!  √

 

Alternative Way:

We could also approach this problem like we do here in the Sequences and Series section and treat this exponential model as a geometric series.

 

To get the common ratio (the base of the exponential equation), we can subtract the \(x\)’s, and then take this root of the quotient of the \(y\)’s. The base is then is \(\displaystyle \sqrt[{5-3}]{{\frac{{40}}{{10}}}}=\sqrt{4}=2\).

Now we have \(y=a{{\left( 2 \right)}^{x}}\); plug in either point; for example, \(\displaystyle 10=a{{\left( 2 \right)}^{3}};\,\,a=\frac{5}{4}\). The exponential function is: \(\displaystyle y=\frac{5}{4}{{\left( 2 \right)}^{x}}\).

Find the equation of this graph with a base of .5 and horizontal shift of \(-1\):

 

We see that this exponential graph has a horizontal asymptote at \(y=-3\), and with the horizontal shift, we have \(y=a{{\left( {.5} \right)}^{{x+1}}}-3\) so far.

 

When you have a problem like this, first use any point that has a “0” in it if you can; it will be easiest to solve the system. Solve for \(a\) first using point \(\left( {0,-1} \right)\):

 

\(\begin{array}{c}y=a{{\left( {.5} \right)}^{{x+1}}}-3;\,\,\,-1=a{{\left( {.5} \right)}^{{0+1}}}-3;\,\,\,\,2=.5a;\,\,\,\,a=4\\y=4{{\left( {.5} \right)}^{{x+1}}}-3\end{array}\)

For the following graph, write the appropriate exponential function in the form \(y=a{{b}^{x}}+k\) (vertically shifted exponential function):

 

 

We see that this exponential graph has an asymptote at \(y=-3\), so it will have a vertical shift of –3, or \(k=-3\). Our equation will be in the form \(y=a{{b}^{x}}-3\).

 

We need to use a system of equations with the two points on the graph: \(\left( {0,1} \right)\) and \(\left( {1,\,-1} \right)\). When you have a problem like this, first use any point that has a “0” in it if you can; it will be easiest to solve the system.

 

Solve for \(a\) first using \(\left( {0,1} \right)\): \(\begin{array}{c}1=a{{b}^{0}}-3;\,\,\,\,\,a\left( 1 \right)=1+3;\,\,\,\,a=4\\y=4{{b}^{x}}-3\end{array}\)

Use this equation and plug in \(\left( {1,\,-1} \right)\) to solve for \(b\): \(-1=4{{b}^{1}}-3;\,\,\,\,\,4b=2;\,\,\,\,\,\,b=.5\)

The exponential function is \(y=4{{\left( {.5} \right)}^{x}}-3\). Graph this function and it works!   √

 

Easier way to do this: The distance between the horizontal asymptote and the \(y\)-intercept is 4, so we have a vertical stretch of 4. Thus, \(a=4\) (We can only do this when there’s no horizontal shift). Since the graph has a vertical shift of –3, we have \(y=4{{b}^{x}}-3\) so far.

Plug in the other point \(\left( {1,-1} \right)\) to get \(b\):

\(\begin{array}{c}-1=4{{b}^{1}}-3;\,\,\,\,\,2=4b;\,\,\,\,\,b=.5\\y=4{{\left( {.5} \right)}^{x}}-3\end{array}\)

 

Exponential Function Applications

Here are some compounding formulas that you’ll use in working with exponential applications. The second set of formulas are based on the first, but are a little bit more specific, since the interest is compounded multiply times during the year:

Rate Compounded Annually

Rate Compounded Multiple Times Per Year

Exponential Growth

\(A=P{{\left( {1+r} \right)}^{t}}\)

Exponential Decay

\(A=P{{\left( {1-r} \right)}^{t}}\)

Exponential Growth

\(A=P{{\left( {1+\frac{r}{n}} \right)}^{{nt}}}\)

Exponential Decay

\(A=P{{\left( {1-\frac{r}{n}} \right)}^{{nt}}}\)

\(A=\) ending amount

\(P=\) principle, or starting amount

\(r=\) growth rate (turn % to decimal) – per year

\(t=\) number of years that have passed

 

(if applicable)  \(n=\) number of times the interest is compounded per year; for example:

 

When Compounded \(n\) When Compounded \(n\)
Daily 365 Monthly 12
Weekly 52 Quarterly 4
Bi-Weekly 26 Semi-Annually 2

Note that the growth (or decay) rate is typically a percentage, but when it’s in the formula, it’s a decimal. The growth (or decay) factor is the actual factor after the rate is converted into a decimal and added or subtracted from 1 (they may ask you for the growth factor occasionally). When an amount triples, for example, we start with the original and add 200% to it, so the growth rate is 200% (in the formula, it’s 2.00, which will be added to the 1), but growth factor is 3 (1 + 2). Here are some examples:

Wording in Problem Rate Type

Factor

Appreciates at 5% .05 or 5% Growth  \(1+.05=1.05\)
Depreciate at 10% .10 or 10% Decay  \(1-.10=.90\)
Doubles 1 or 100% Growth  \(1+1=2\)
Triples 2 or 200% Growth  \(1+2=3\)

Halves, or Half-life

.5 or 50%

Decay

 \(1-.50=.50\)

Exponential Compounding

One thing that the early mathematicians found is that when the number of times the compounding takes place (“\(n\)” above) gets larger and larger, the expression \(\displaystyle A={{\left( 1+\frac{1}{n} \right)}^{n}}\) gets closer and closer to a mysterious irrational number called “\(e\)” (called “Euler’s number), and this number is about 2.718. (Think of this number sort of like “pi” – these numbers are “found in nature”.)  You can find  \({{e}^{x}}\)  on your graphing calculator using “2nd ln“, or if you just want “\(e\), you can use “2nd  ÷”.

So, if we could hypothetically compound interest every instant (which is theoretically impossible), we could just use “\(e\)” instead of \(\displaystyle {{\left( 1+\frac{1}{n} \right)}^{n}}\). This would be the highest amount of interest someone could earn at that interest rate, if it were possible to compound continuously.

Now we have two major formulas we can use. You’ll probably have to memorize these, but you’ll use them enough that it’s not that bad:

Rate Compounded Annually (\(n=1\)),

or Multiple Times per Year

Rate Compounded Continuously

Growth: \(r>0\);  Decay: \(r<0\)

Growth: \(\displaystyle A=P{{\left( {1+\frac{r}{n}} \right)}^{{nt}}}\);     Decay: \(\displaystyle A=P{{\left( {1-\frac{r}{n}} \right)}^{{nt}}}\)

 \(\displaystyle A=P{{e}^{{rt}}}\)

\(A=\) ending amount

\(P=\) principle, or starting amount

\(r=\) growth rate (turn % to decimal) – per year

\(t=\) number of years that have passed

(if applicable)  \(n=\) number of times the interest is compounded per year

I remember the \(A=P{{e}^{rt}}\) formula by thinking of the shampoo Pert”, and you can think of continuously washing your hair, using Pert. Thus you use the Pert formula with continuous compounding.

More Growth/Decay Equations

There are two more exponential equations that are a little more specific than some we’ve mentioned above:

  • \(\displaystyle y=a{{b}^{{\frac{t}{p}}}},\,\,\,\,b>0\), where \(a=\) the initial amount, \(b=\) the growth factor (or decay factor, if \(b>1\)), \(t=\) the time that has passed, \(p=\) the period for the growth or decay factor (the growth or decay interval), and \(y=\) the amount after the time that has passed.

For example, if a mice population triples every 4 years (the growth interval), and the population starts with 20 mice, and the problem asks how many mice will there be in 12 years, the formula will be \(\displaystyle y=20{{\left( 3 \right)}^{\frac{12}{4}}}\). (The \(\displaystyle \frac{t}{p}\) (3) makes sense, since if the population triples every 4 years, the mice population would triple 3 times in 12 years!)

Half-life problems are common in science,and, using this formula, the \(b\) is \(\displaystyle \frac{1}{2}\) or .5, and \(p\) is the number of years that it takes for something to halve (divide by 2). (Note that you can also solve half-life problems using the next formula). We’ll explore these half-life problems below here below, and in the Logarithmic Functions section here.

  • \(\displaystyle N\left( t \right)={{N}_{0}}{{e}^{{kt}}}\), which is called uninhibited growth or continuous growth. In this formula, \({{N}_{0}}=\) the initial amount, \(k=\) the growth rate (or decay rate, if \(k<0\)), \(t=\) the time that has passed, and \(N\left( t \right)\) is the amount after the specified time period.

Notice that this is the same formula as the continuous growth equation \(A=P{{e}^{rt}}\) above, just in a different format. Notice also that in this formula, the decay is when \(k<0\). This is because we are raising \(e\) to the \(k\), and when an exponent is negative, it’s the same as 1 over that base with a positive exponent. Thus, the “multiplier” will be less than 1, and we have a decay.

For example, for half-life problems, we’ll see that sometimes we have to solve first for \(k\) using logs, with \({{N}_{0}}\) as 2 and \(N\left( t \right)\) as 1. In these cases, \(k\) will be negative, since we have a decay.

We’ll show later that anytime we need to solve for a variable in the exponent, we’ll typically use logs. Here are all the exponential formulas we’ve learned:

<

Rate Compounded Annually

(\(n=1\)),

or Multiple Times per Year

Rate Compounded Continuously

General Exponential Equation

Uninhibited Growth Equation

Growth: \(\displaystyle A=P{{\left( {1+\frac{r}{n}} \right)}^{{nt}}}\)

Decay: \(\displaystyle A=P{{\left( {1-\frac{r}{n}} \right)}^{{nt}}}\)

 \(A=P{{e}^{{rt}}}\)

Growth:  \(r>0\)

Decay:\(r<0\)

\(\displaystyle y=a{{b}^{{\frac{t}{p}}}}\)

Growth:\(b>0\)

Decay:\(b<0\)

\(\displaystyle N\left( t \right)={{N}_{0}}{{e}^{{kt}}}\)

Growth:\(k>0\)

Decay:\(k<0\)

\(A=\) ending amount

\(P=\) principle, or starting amount

\(r=\) growth/decay rate (turn % to decimal) – per year

\(t=\) number of years that have passed

(if applicable)  \(n=\) number of times the interest is compounded per year

\(e=\) base (around 2.718)

\(y=\) ending amount

\(a=\) starting amount

\(b=\) growth/decay factor per time period

\(t=\) time that has passed

\(p=\) growth or decay period (interval)

\(\displaystyle N\left( t \right)=\) ending amount

\(\displaystyle {{N}_{0}}=\) starting amount

\(e=\) base (around 2.718)

\(k=\) growth/decay rate

\(t=\) the time that has passed

Remember when we put an exponent in the Graphing Calculator, we just use the “^” key!

Exponential Word Problems:

<

Here are some appreciation/depreciation problems:

Exponential Word Problem

Solution

You decide to buy a used car that costs $10,000. You’ve heard that the car may depreciate at a rate of 10% per year.

 

At that rate, what will the car be worth in 5 years?

Since the interest rate is compounded yearly, we can use the formula \(A=P{{\left( {1-r} \right)}^{t}}\) for exponential decay. The principle or starting amount, \(P\), is 10000. The rate \(r\) is .10, and the time \(t\) (in years) is 5.

\(\begin{align}A&=P{{\left( {1-r} \right)}^{t}}\\A&=10000{{\left( {1-.1} \right)}^{5}}\\&=\$5904.90\end{align}\)

 

The car will be worth $5904.90 in 5 years. It makes sense that it will be less, since we’re dealing with exponential decay.

The initial value of your car is $20,000. After 1 year, the value is $15,000

 

(a)  What is the percent decrease?  

(b)  Find the value of the car at this same rate after 5 years from the initial value.

(a)  The percent decrease is \(\frac{{\text{Old Price }-\text{ New Price}}}{{\text{Old Price}}}\times 100\), or \(\frac{{20000-15000}}{{20000}}=.25=25%\).

We could have also solved this by using our \(A=P{{\left( {1-r} \right)}^{t}}\) formula, as shown on the left; since we want the value after 1 year, \(t=1\):

 \(\begin{align}A&=P{{\left( {1-r} \right)}^{t}}\\15000&=20000{{\left( {1-r} \right)}^{1}}\\\frac{{15000}}{{20000}}&=1-r\\r&=.25=25\%\end{align}\)

 

(b)  Again, since the interest rate is compounded yearly, we can use the formula \(A=P{{\left( {1-r} \right)}^{t}}\) for exponential decay. The principle or starting amount, \(P\), is 20000. The rate \(r\) is .25 from above, and the time \(t\) (in years) is 5: 

  \(\begin{align}A&=P{{\left( {1-r} \right)}^{t}}\\A&=20000{{\left( {1-.25} \right)}^{5}}\\&=\$4746.09\end{align}\)

The car will be worth $4746.09 in 5 years. It makes sense that it will be less, since we’re dealing with exponential decay.

Megan has $20,000 to invest for 5 years and she found an interest rate of 5%. How much money will she have at the end of 5 years if:

 

  (a)  the interest rate compounds monthly?

  (b)  the interest rates compounds semi-annually?

Since the interest rate compounds at different intervals during the year, we need to use formula \(\displaystyle A=P{{\left( {1+\frac{r}{n}} \right)}^{{nt}}}\) for growth. The starting amount, or \(P\), is 20000. The rate \(r\) is .05, the time interval \(t\) (in years) is 5.

For part (a), the number of times the interest compounds per year (\(n\)) is 12, since it compounds monthly:

\(\displaystyle \begin{align}A&=P{{\left( {1+\frac{r}{n}} \right)}^{{nt}}}\\\,A&=20000{{\left( {1+\frac{{.05}}{{12}}} \right)}^{{\left( {5\times 12} \right)}}}\\&=\$25667.17\end{align}\)

Megan will have $25,667.17 at the end of 5 years, if the rate compounds monthly.

For (b), everything stays the same, but the number of times the interest compounds per year is 2, since it compounds semi-annually (twice per year):

\(\displaystyle \begin{align}A&=P{{\left( {1+\frac{r}{n}} \right)}^{{nt}}}\\A&=20000{{\left( {1+\frac{{.05}}{2}} \right)}^{{\left( {5\times 2} \right)}}}\\&=\$25601.69\end{align}\)

Megan will have $25,601.69 at the end of 5 years, if the rate compounds semi-annually.

Madison really wants to buy a car in 4 years and she wants to start saving for a down payment. 

 

If she deposits $3500 now with interest compounding continuously at 3%, what down payment will she have for her car? 

We’ll use the “shampoo” formula – PERT – since the interest is compounding continuously.  (Remember, you continuously wash your hair!)

 

Since we need to figure out how much Madison will have in 4 years, we are looking for \(A\). Since she has 3500 now, the starting value or \(P\) is 3500. She has 4 years to save, so \(t\) is 4, and since the interest rate is 3%, \(r\) is .03:

\(\begin{align}A&=P{{e}^{{rt}}}\\A&=3500{{e}^{{(.03)(4)}}}\\A&=3946.24\end{align}\)

Madison will have $3946.24 as a down payment for her new car.

Half-Life Problems:

Half-life problems deal with exponential decays that halve for every time period. For example, if we start out with 20 grams, after the next time period, we’d have 10, then 5, and so on. For these problems, the base (decay factor) of the exponential equation is .5.

The trick on half-life problems is to raise the .5 to \(\displaystyle \frac{{\text{time period we want}}}{{\text{time for one half-life}}}\), since this will give us the number of times the substance actually halves.

Here is our first example; note that we solve this same problem with logs here in the Logarithmic Functions section.

Half-Life Problem

Math

Notes

If a chemist has 40 grams of a substance that has a half-life of 6 hours, how much will there be after 18 hours? \(\begin{array}{c}y=a{{b}^{{\frac{t}{p}}}}\\y=40{{\left( {.5} \right)}^{{\frac{{18}}{6}}}}\\y=5\,\,\text{grams}\end{array}\) For this problem, we need to use the exponential equation \(\displaystyle y=a{{b}^{{\frac{t}{p}}}}\), since we have a time period for which the substance divides in two.

 

Since we’re dealing with a half-life problem, we know the decay factor is .5, since it halves every 6 hours (the decay interval, or time of a half-life).

 

Since the half-life is 6, but we want to know how much of the substance there will be after 18 hours, our “time” is actually \(\displaystyle \frac{{18}}{6}\) or \(\displaystyle \frac{{\text{time we are interested in}}}{{\text{time of a half-life}}}\), which is 3. Since we start with 40 grams, \(a=40\). Using the formula, we get \(y=\) 5 grams.

 

5 grams makes sense since in 6 hours, there will be 20 grams (half of 40), in 12 hours, there will be 10 grams (half of that), and in 18 hours, there will be 5 grams (half of that).     √

Here is another half-life problem:

Half-Life Problem

Math

Notes

A culture of bacteria triples every 8 hours.

 

 

(a)  If there are 100 bacteria now, how many will there be after 18 hours?

 

(b)  Find the bacteria population 2 hours earlier.

\(\begin{array}{c}y=a{{b}^{{\frac{t}{p}}}}\\\\y=100{{\left( 3 \right)}^{{\frac{{18}}{8}}}}\\y=1184\,\,\text{bacteria}\end{array}\) (a)  For this problem, we need to use the exponential equation \(\displaystyle y=a{{b}^{{\frac{t}{p}}}}\) again, since we have a time period for which the bacteria triples.

 

We know the growth factor is 3, since the bacteria triples. Since the bacteria triples every 8 hours, but we want to know how much of it there will be after 18 hours, our time is actually \(\displaystyle \frac{{18}}{8}\) or \(\displaystyle \frac{{\text{time we are interested in}}}{{\text{time bacteria triples}}}\). Since we start with 100 grams, \(a=100\).

 

Using the formula, we get \(y=1184\) bacteria. We have to round down since we can’t have part of a bacterium. This makes sense since the bacteria starts with 100 and triples about  times.   √

Solution 1:

\(\begin{array}{l}y=100{{\left( 3 \right)}^{{\frac{{-2}}{8}}}}\\y=76\,\,\text{bacteria}\end{array}\)

 

Solution 2:

\(\begin{array}{c}100=a{{\left( 3 \right)}^{{\frac{2}{8}}}}\\a=76\,\,\text{bacteria}\end{array}\)

(b)  Using the same formula, we can use –2 for the time, since we want the number of bacteria 2 hours earlier. We get 76 bacteria, which makes sense, since there would be less than 100. Round up since we are going backwards.

 

We could have also set up the formula the second way, since we want to find out the number of bacteria starting out, when we end up with  100 bacteria 2 hours later.

Here’s one more where we’ll see how important logs will be to solve these types of problems:

Half-Life Problem:

Suppose that a graduating class had 500 students graduating the first year, but after that, the number of students graduating declines by a certain percentage.

(a)   If the number of students graduating will be 400 in 2 years, what is the decay rate?

(b)  Using this same decay rate, in about how many years will there be less than 300 students?

Math

Notes

(a) \(\require{cancel} \displaystyle \begin{align}A&=P{{\left( {1-r} \right)}^{t}}\\400&=500{{\left( {1-r} \right)}^{2}}\\\left( {\frac{{400}}{{500}}} \right)&={{\left( {1-r} \right)}^{2}}\\{{\left( {\frac{{400}}{{500}}} \right)}^{{\frac{1}{2}}}}&={{\left( {{{{\left( {1-r} \right)}}^{{\cancel{2}}}}} \right)}^{{\cancel{{\frac{1}{2}}}}}}\,\\{{\left( {\frac{{400}}{{500}}} \right)}^{{\frac{1}{2}}}}&=1-r\\r&=1-\,{{\left( {\frac{{400}}{{500}}} \right)}^{{\frac{1}{2}}}}\approx .106\end{align}\) (a)  We’ll use the exponential decay formula, where \(P\) (starting number) is 500, \(A\) (ending number) is 400, and \(t\) (number of years) is 2. This type of problem is a little different, since we have to work “backwards” to get the rate.

 

Remember to raise both sides of the equation to the reciprocal of the exponent (exponent is 2, reciprocal is \(\displaystyle \frac{1}{2}\)) to get the variable \(r\) out from “within the exponent”.

 

We solve for \(r\) and get about .106 or about 10.6%.

(b)

\(\begin{align}A&=P{{\left( {1-r} \right)}^{t}}\\300&=500{{\left( {1-.106} \right)}^{t}}\end{align}\)

 

 

Help! How do we solve for \(t\)?

 

We don’t know how yet – we need logs, which we’ll learn later.

 

 

(b)  We’ll use the exponential decay formula, where \(P\) (starting number) is 500, the decay rate is .106, and we need to find \(t\) where A (ending number) is 300.

 

We’ll have to use “guess and check” to figure out what the \(t\) is, since we can’t really solve for it without using logarithms or logs. Let’s try \(t=10\) first, which is too little, 3 next, which is too large, and see if we can “zero in” on the correct value:

 

\(t\)

 \(500{{\left( {1-.106} \right)}^{t}}\)

Notes

\(t=10\)

 \(500{{\left( {1-.106} \right)}^{{10}}}\approx 163\) Too small: we need smaller number, since this is decay.

\(t=3\)

 \(500{{\left( {1-.106} \right)}^{3}}\approx 357\) Too large; we need larger number, since this is decay.

\(t=4\)

 \(500{{\left( {1-.106} \right)}^{4}}\approx 319\) Too large, but getting there.

\(t=4.6\)

 \(500{{\left( {1-.106} \right)}^{{4.6}}}\approx 299\) Pretty close!

In about 4.6 years, there will be less than 300 students. See how difficult this is?

Logs (which we’ll learn in the Logarithmic Functions section) will make it much easier!

 

(Note that we could solve this problem with a graphing calculator, for example, with \({{\text{Y}}_{1}}=300\) and \({{\text{Y}}_{2}}=500{{\left( {1-.106} \right)}^{t}}\). Then use CALC (2nd TRACE) 5 (intersect) to find the intersection, or use TABLE to see where \(x\) is when \(y=300\).)

Solving Exponential Functions by Matching Bases

In certain cases, we can solve an equation with a variable in the exponent by matching up the bases on each side, if we can. Remember that a base in an exponential equation is the number that has an exponent.

This method of matching bases to solve an exponential equation is also called the “One-to-One Property of Exponential Functions”.

As long as the bases are the same and we have just one base on each side of the equation, we can set the exponents equal to each other. This makes sense; if we had  \({{2}^{x}}=\,\,\,{{2}^{4}}\), we could see that \(x\) could only be 4, and nothing else. We can write this as the rule:

\({{b}^{x}}=\,\,\,{{b}^{y}}\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,x \,=\,\,\,y\)

Remember that when we multiply the same bases together with different exponents, we add the exponents. For example, \({{a}^{x}}\cdot {{a}^{y}}={{a}^{x+y}}\). Also remember that when we raise an exponent to another exponent, we multiply those exponents. For example, \({{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\). (Be careful, though, since technically \(\displaystyle {{a}^{{{{x}^{y}}}}}\) (without parentheses) is actually \(\displaystyle {{a}^{{\left( {{{x}^{y}}} \right)}}}\) – try examples on your calculator!)

The idea is to find the smallest base (it’s easier to use an integer and not a fraction) that can be used on both sides and match up the bases by raising the bases to get the original numbers.

Let’s solve the following equations and check our answers back after getting them (sometimes we have to use a calculator):

Changing Base to Solve Exponential Equations

 \(\displaystyle {{4}^{{x-3}}}=16\)

 

\(\displaystyle \begin{align}{{4}^{{x-3}}}&={{4}^{2}}\\x-3&=2\\x&=5\\\\\text{Check:}\\{{4}^{{5-3}}}&=16\\16&=16\end{align}\)

\({{10}^{{3x+2}}}=.1\)

 

\(\begin{align}{{10}^{{3x+2}}}&={{10}^{{-1}}}\\3x+2&=-1\\x&=-1\\\\\text{Check:}\\{{10}^{{3\left( {-1} \right)+2}}}&=.1\\.1&=.1\end{align}\) 

\({{9}^{{2x}}}={{27}^{{x+4}}}\)

 

\(\begin{align}{{\left( {{{3}^{2}}} \right)}^{{2x}}}&={{\left( {{{3}^{3}}} \right)}^{{x+4}}}\\{{3}^{{4x}}}&={{3}^{{3\left( {x+4} \right)}}}\\{{3}^{{4x}}}&={{3}^{{3x+12}}}\\4x&=3x+12\\x&=12\\\\\text{Check:}\\{{9}^{{2\left( {12} \right)}}}&={{27}^{{12+4}}}\\7.97..\times {{10}^{{22}}}&=7.97..\times {{10}^{{22}}}\end{align}\)

\(\displaystyle {{\left( {\frac{1}{4}} \right)}^{x}}={{8}^{{x-1}}}\)

 

\(\displaystyle \begin{align}{{\left( {{{2}^{{-2}}}} \right)}^{x}}&={{\left( {{{2}^{3}}} \right)}^{{x-1}}}\\{{2}^{{-2x}}}&={{2}^{{3x-3}}}\\-2x&=3x-3\\x&=\frac{3}{5}\\\\\text{Check:}\\{{\left( {\frac{1}{4}} \right)}^{{\frac{3}{5}}}}&={{8}^{{\frac{3}{5}-1}}}\\.435..&=.435..\end{align}\)

 \({{2}^{{4x}}}\cdot {{16}^{{x+3}}}={{4}^{{x-1}}}\)

 

\(\begin{align}{{2}^{{4x}}}\cdot {{\left( {{{2}^{4}}} \right)}^{{x+3}}}&={{\left( {{{2}^{2}}} \right)}^{{x-1}}}\\{{2}^{{4x}}}\cdot {{2}^{{4x+12}}}&={{2}^{{2x-2}}}\\{{2}^{{4x+4x+12}}}&={{2}^{{2x-2}}}\\8x+12&=2x-2\\x&=-\frac{7}{3}\\\\\text{Check:}\\{{2}^{{4\left( {-\frac{7}{3}} \right)}}}\cdot {{16}^{{-\frac{7}{3}+3}}}&={{4}^{{-\frac{7}{3}-1}}}\\.0098…&=.0098…\end{align}\)

Unfortunately, we can’t get common bases on both sides for most exponential equations. Soon we will get to the most common way to solve equations with variables in exponents – logarithms!

Exponential Regression

We learned about regression here in the Scatter Plots, Correlation, and Regression section, but didn’t really address Exponential Regression.

Let’s find an exponential regression equation to model the following data set using the graphing calculator:

\(x\) 1 4 6 10 11 13

20

\(y\) 3 5 7.5 16 18 27

96

As we did with linear and quadratic regressions, enter the data and perform regression in the calculator:

Keystroke Screens
Push STAT, then ENTER to go to EDIT.

 

To clear anything in the lists L1 or L2, move cursor to the top to cover the L1 or L2 and hit CLEAR (not DEL) and then hit ENTER.

 

To type in the data, start with right under the L1 and separate each entry by ENTER or by moving the cursor. The \(x\)’s go under L1 and the \(y\)’s go under L2.

 

If you need to edit the entries, you can access the lists the same way. You can type 2nd MODE (QUIT) to quit when you are done.

To get the best fit line for the \(x\) and \(y\) values that you’ve entered, push STAT, move cursor to CALC, and then push either 0, or move cursor down to ExpReg, and hit ENTER.

 

(With the newer calculators, there’s a setting to have STAT WIZARD: ON, which gives you the ability to enter more data when doing the STAT tests. Just keep hitting ENTER until you see the ExpReg screen.)

 

This will give us our \(a\) and \(b\) for the line \(y=a{{b}^{x}}\). Thus, the best exponential fit for this data is \(y=\left( {2.474} \right){{\left( {1.2} \right)}^{x}}\).

To put this best fit line into the calculator, click on Y1 =, and then VARS, 5 for Statistics (or move cursor to Statistics), move cursor to right to EQ, then push ENTER. (You must have first used the ExpReg function above.) You’ll see the Y1 filled with very long numbers (they don’t like to round on the calculator, I guess!) Now push GRAPH to graph over the points that you have from the Plot1.

 

Note:  You can also accomplish this by pushing STAT, over to CALC, scroll to ExpReg or hit 0, scroll down to Store RegEQ, then (before hitting ENTER), pushing ALPHA TRACE (F4) 1,  ENTER (for Y1), ENTER, or, after Store RegEQ, hit VARS, highlight (hit) Y-VARS, 1 (Function), (for Y1), ENTER, ENTER. Now the equation will be in Y1, and you can graph it by hitting GRAPH.

 

(For the older operating system, you push STAT, CALC, scroll to ExpReg or hit 0, then (before hitting ENTER), pushing VARS, scroll to Y-VARS, 1 (Function), 1 (for Y1), ENTER.)

 

We can do this same type of regression with natural logs (LnReg) when you learn about log functions.

Exponential Inequalities

You may have to solve inequality problems (either graphically or algebraically) with exponential functions. Remember that we learned about using the Sign Chart or Sign Pattern method for inequalities here in the Quadratic Inequalities section, and also we have the domain restriction that the argument of a log has to be \(>0\).

Let’s do a few inequality problems:

Exponential Inequality Notes/Graph
\({{81}^{x}}+9\ge 10\cdot {{9}^{x}}\)

 

\(\begin{align}{{81}^{x}}-10\cdot {{9}^{x}}+9&\ge 0\\{{\left( {{{9}^{2}}} \right)}^{x}}-10\cdot {{9}^{x}}+9&\ge 0\\{{9}^{{2x}}}-10\cdot {{9}^{x}}+9&\ge 0\\\left( {{{9}^{x}}-9} \right)\left( {{{9}^{x}}-1} \right)&\ge 0\\{{9}^{x}}-9&=0;\,\,\,x=1\\{{9}^{x}}-1&=0;\,\,\,x=0\end{align}\)

Usually these types of problems will either be a linear or quadratic inequality. Since we have a quadratic inequality, we have to set everything to 0, try to factor (we can!) and use a sign chart, since we don’t know if the factors are positive or negative.

 

The boundary points, or critical values, are the roots (setting the factors to 0) of the quadratic, as if it were an equality. To get the signs, we plug in a sample number in each interval to see if \(\left( {{{9}^{x}}-9} \right)\left( {{{9}^{x}}-1} \right)\) is positive or negative. For example, we can use \(\displaystyle \frac{1}{2}\) for the interval between 0 and 1:  \(\left( {{{9}^{{\tfrac{1}{2}}}}-9} \right)\left( {{{9}^{{\tfrac{1}{2}}}}-1} \right)=-12<0\)

The problem calls for \(\ge 0\), so we look for the plus sign(s), and our answers are inclusive (hard brackets). The answer is \(\left( {-\infty ,0} \right]\cup \left[ {1,\infty } \right)\).

Learn these rules, and practice, practice, practice!


For Practice: Use the Mathway widget below to try an Exponential Function problem. Click on Submit (the blue arrow to the right of the problem) to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Logarithmic Functions – you are ready!

<