This section covers:
- Exponential Functions
- Parent Graphs of Exponential Functions
- Exponential Regression
- Exponential Function Applications
- Exponential Word Problems
- Solving Exponential Functions by Matching Bases
- Factoring with Exponents
- Factoring to Solve Exponential Equations
- More Practice
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, Engineering, and Geology, as well as many 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. Note that Transformations, Inverses, Compositions, and Inequalities of Exponents can be found here.
Exponential Functions
Exponential functions are very useful in life, especially in the world of finance. If you’ve ever earned interest in the bank (or even if you haven’t), you’ve probably heard of “interest rate”, “compounding”, “appreciation”, or “depreciation”.
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.
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 , 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 (where the a in \(y=a{{b}^{x}}\) is 1). I always remember that the “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”), and the “anchor point” of a log function, which we’ll review soon, is (1, 0) (since this looks like the “lo” in “log”). We have to also remember that if the function shifts, this “anchor point” will move.
Notice also that 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.
These parent graphs can be transformed like the other parent graphs in the Parent Functions and Transformations section, and in the Transformations, Inverses, Compositions, and Inequalities of Exponents/Logs section.
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:
As we did with linear and quadratic regressions, enter the data and perform regression in the calculator:
We can do this same type of regression with natural logs (LnReg) when you learn about log functions.
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:
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:
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 \(A={{\left( 1+\frac{1}{n} \right)}^{n}}\) 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 “2^{nd} ln“, or if you just want “e”, you can use “2^{nd} ÷”.
So, if we could hypothetically compound interest every instant (which is theoretically impossible), we could just use “e” instead of \({{\left( 1+\frac{1}{n} \right)}^{n}}\). This would be the highest amount of interest someone could earn at that interest rate, if it were possible to compound continuously.
So 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:
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:
- , 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 \(y=20{{\left( 3 \right)}^{\frac{12}{4}}}\). (The \(\frac{t}{p}\) (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 \(\frac{1}{2}\) 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.
- , which is called uninhibited growth or continuous growth. In this formula, = the initial amount, k = the growth rate (or decay rate, if k < 0), t = the time that has passed, and 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 as 2 and as 1. In these cases, k will be negative, since we have a decay. We’ll explore these half-life problems below here.
We’ll show later that anytime we need to solve for a variable in the exponent, we’ll typically use logs. So here are all the exponential formulas we’ve learned:
Remember when we put an exponent in the Graphing Calculator, we just use the “^” key!
Exponential Word Problems:
Problem:
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?
Solution:
Problem:
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.
Solution:
Problem:
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 rate compounds semi-annually?
Solution:
Half-Life Problem:
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?
Solution:
Note that we solve this same problem with logs here in the Logarithmic Functions section.
Problem:
A culture of 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.
Solution:
(Note that if the bacteria tripled every hour, we could just use the formula \(y=a{{b}^{t}}\).)
Problem:
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?
Solution:
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?
Solution:
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.
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:
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):
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!
Factoring with Exponents
Factoring and Solving with exponents can be a bit trickier. Note that learned about the properties of exponents here in the Exponents and Radicals in Algebra Section, and did some solving with exponents here. Note also that we’ll discuss Exponential Functions here. In your Pre-Calculus and Calculus classes, you may see algebraic exponential expressions that need factoring and possibly solving, either by taking out a Greatest Common Factor (GCF) or by “unfoiling”. These really aren’t that bad, if you remember a few hints:
- To take out a GFC with exponents, take out the factor with the smallest exponent, whether it’s positive or negative. (Remember that “larger” negative numbers are actually smaller). This is even true for fractional exponents. Then, to get what’s left in the parentheses after you take out the GCF, subtract the exponents from the one you took out. For example, for the expression \({{x}^{-5}}+{{x}^{-2}}+x\), we’d take out \({{x}^{-5}}\) for the GCF to get \({{x}^{-5}}\left( {{x}^{-5-\,-5}}+{{x}^{-2-\,-5}}+{{x}^{1-\,-5}} \right)={{x}^{-5}}\left( {{x}^{0}}+{{x}^{3}}+{{x}^{6}} \right)={{x}^{-5}}\left( 1+{{x}^{3}}+{{x}^{6}} \right)\). (Remember that “ – – ” is the same as “+ +” or “+”). Multiply back to make sure you’ve done it correctly.
- For fractional coefficients, find the common denominator, and take out the fraction that goes into all the other fractions. So for the fraction you take out, the denominator is the least common denominator of all the fractions, and the numerator is the Greatest Common Factor (GCF) of the numerators. For example, for the expression \(\frac{3}{4}{{x}^{2}}-\frac{2}{4}x+\frac{16}{4}=\frac{1}{4}\left( 3{{x}^{2}}-2x+16 \right)\) (since nothing except for 1 goes into 3 and 2 and 16). Multiply back to make sure you’ve done it correctly.
- For a trinomial with a constant, if the largest exponent is twice that of the middle exponent, then use substitution like u, for the middle exponent, “unfoil”, and then put the “real” expression back in. For example, for \({{x}^{\frac{2}{3}}}-{{x}^{\frac{1}{3}}}-2\) , let \(u={{x}^{\frac{1}{3}}}\) , and we have \({{u}^{2}}-u-2\) , which factors to \(\left( u-2 \right)\left( u+1 \right)\). We can then translate back to \(\left( {{x}^{\frac{1}{3}}}-2 \right)\left( {{x}^{\frac{1}{3}}}+1 \right)\) and solve from there (set each to 0 and solve). Always foil back to make sure you’ve done it correctly. We call this method u-substitution or simply u-sub.
Let’s do some factoring. Learning to factor these will actually help you a lot when you get to Calculus:
Factoring to Solve Exponential Equations
After factoring, you may be asked to solve the exponential equation. Here are some examples, some using u-substitution.
Note that the third problem uses log solving that we will learn here in the Logarithmic Functions section.
Math | Notes |
\(\displaystyle 2{{x}^{{\frac{1}{2}}}}+{{x}^{{\frac{1}{4}}}}-15=0\)
Let \(\displaystyle u={{x}^{{\frac{1}{4}}}}\) \(\displaystyle \begin{array}{c}2{{u}^{2}}+u-15=0\,\,\,\,\,\,\,\,\,\,\left( {2u-5} \right)\left( {u+3} \right)=0\\u=\frac{5}{2},-3\end{array}\)
Substitute \(\displaystyle {{x}^{{\frac{1}{4}}}}\) back in for \(u\) for both factors: \(\displaystyle {{x}^{{\frac{1}{4}}}}=\frac{5}{2};\,\,\,{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{4}}={{\left( {\frac{5}{2}} \right)}^{4}};\,\,\,x=\frac{{625}}{{16}}\) \(\require{cancel} \displaystyle \cancel{{{{x}^{{\frac{1}{4}}}}=-3}}\); No Solution
Check: | Let’s use \(u\)-substitution here, since we see that the largest exponent \(\left( {\frac{1}{2}} \right)\) is twice that of the middle exponent \(\left( {\frac{1}{4}} \right)\). Use the trick to let \(u\) equal to the variable in the middle: \(u={{x}^{{\frac{1}{4}}}}\), and then \(\displaystyle 2{{x}^{{\frac{1}{2}}}}=2{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{2}}=2{{u}^{2}}\). (Remember that when we raise an exponent to another exponent, we multiply exponents).
Now we can factor more easily. We can factor this one; if you can’t, you can always use the Quadratic Formula.
Then we’ll put the \(u\)’s back in. Remember that we want to end up with \(\displaystyle {{x}^{1}}\) by itself, so we have to raise both sides to the reciprocal of the exponent. For the second factor, we have no solution, since we have an even exponent in the denominator and we need to end up with a negative number – can’t be done with real numbers.
Try the answer back in for \(x\) to make sure it works; \(\frac{{625}}{{16}}\) works! You can use the STO> X function in your graphing calculator to check (see left). |
\(\displaystyle {{9}^{x}}+{{3}^{{x+1}}}=4\) \(\displaystyle \begin{array}{l}{{\left( {{{3}^{2}}} \right)}^{x}}+\left( {{{3}^{x}}} \right)\left( {{{3}^{1}}} \right)=4\\{{3}^{{2x}}}+\left( 3 \right){{3}^{x}}=4\\{{\left( {{{3}^{x}}} \right)}^{2}}+\left( 3 \right){{3}^{x}}=4\end{array}\)
Let \(\displaystyle u={{3}^{x}}\) \(\begin{array}{c}{{u}^{2}}+3u-4=0\,\,\,\,\,\,\,\,\,\,\left( {u-1} \right)\left( {u+4} \right)=0\\u=1,\,-4\end{array}\)
Substitute \({{3}^{x}}\) back in for \(u\) for both factors: \({{3}^{x}}=1;\,\,\,\,\,x=0\) \(\cancel{{{{3}^{x}}=-4}}\); No Solution | It looks like we need the “3”’s to match, so let’s break up the second term and change the base of the first term – it works! Remember that when we multiply the same base, we add exponents. Also remember that when we raise an exponent to another exponent, we multiply exponents.
We’ll then let \(u={{3}^{x}}\), and then \({{9}^{x}}={{\left( {{{3}^{2}}} \right)}^{x}}={{3}^{{2x}}}={{3}^{{x\cdot 2}}}={{\left( {{{3}^{x}}} \right)}^{2}}={{u}^{2}}\).
For the first factor, we just know that anything raised to 0 = 1, so the solution is 0. For the second factor, we get no solution, since we can’t raise 3 to anything to get a negative number.
Try the 0 back in the original and see if it works – it does: \({{9}^{0}}+\left( 3 \right){{3}^{0}}=4;\,\,\,\,1+3\left( 1 \right)=4\) |
\({{\left( {{{3}^{x}}-27} \right)}^{3}}+18{{\left( {{{3}^{x}}-27} \right)}^{2}}=0\) \(\displaystyle \begin{array}{c}{{\left( {{{3}^{x}}-27} \right)}^{2}}\left[ {{{{\left( {{{3}^{x}}-27} \right)}}^{1}}+18} \right]=0\\{{\left( {{{3}^{x}}-27} \right)}^{2}}\left( {{{3}^{x}}-9} \right)=0\\\sqrt{{{{{\left( {{{3}^{x}}-27} \right)}}^{2}}}}=\pm 0=0;\,\,\,\,\,\,\,\left( {{{3}^{x}}-9} \right)=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}^{x}}=27;\,\,\,\,\,\,\,\,\,{{3}^{x}}=9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=3;\,\,\,\,\,\,\,\,\,x=2\end{array}\) | For this one, we don’t really need \(u\)-sub, but since we can just take out the expression with the smallest exponent, like we did earlier.
After factoring, we set each factor to 0. We have to think about what exponents will work when 3 is raised to that exponent – to get 27 and 9 (or we could have used logs).
We get that both 3 and 2 work; plug them back in the original to make sure they are correct. |
Here’s one where we have to use logs to solve:
\(2{{e}^{{5x}}}-5{{e}^{{2x}}}-12{{e}^{{-x}}}=0\) \(\displaystyle \begin{array}{c}{{e}^{{-x}}}\left( {2{{e}^{{6x}}}-5{{e}^{{3x}}}-12} \right)=0\\\\\text{Let }u={{e}^{{3x}}}\\{{e}^{{-x}}}\left( {2{{u}^{2}}-5u-12} \right)=0\\{{e}^{{-x}}}\left( {2u+3} \right)\left( {u-4} \right)=0\\{{e}^{{-x}}}=0\,\,\,\,\,\,\,\,\,u=-\frac{3}{2}\,\,\,\,\,\,\,\,\,\,u=4\\{{e}^{{-x}}}=0\,\,\,\,\,\,\,\,\,{{e}^{{3x}}}=-\frac{3}{2}\,\,\,\,\,\,\,\,\,\,{{e}^{{3x}}}=4\\\,\,\,\,\cancel{{{{e}^{{-x}}}=0}}\,\,\,\,\,\,\,\,\,\cancel{{{{e}^{{3x}}}=-\frac{3}{2}}}\,\,\,\,\,\,\,\,\ln {{e}^{{3x}}}=\ln 4\\3x=\ln 4;\,\,\,\,\,\,\,\,x=\frac{{\ln 4}}{3}\end{array}\) | First, factor out the GCF, which is \({{3}^{{-x}}}\) (smallest exponent).
Then, set \(u={{e}^{{3x}}}\), since \(\displaystyle {{e}^{{6x}}}={{\left( {{{e}^{{3x}}}} \right)}^{2}}\) (note that \(u\) is in the middle term).
Factor, and set each factor to 0 (including the GCF).
Note that we had to “throw away” the extraneous solutions, since (positive) \(e\) raised to anything can’t be 0 or negative. |
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!