Whether we like it or not, we need to revisit exponents and then start talking about Logarithms, 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 and many more fields.
Note that we learned about the properties of exponents here in the Exponents and Radicals in Algebra section, and did some solving with exponents here. Factoring with Exponents and Solving Exponential Equations after Factoring can be found in the Advanced Factoring section here. Inverses of Exponential Functions can be found in the Inverses of Functions 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 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. When exponential functions are involved, functions are increasing or decreasing very quickly (multiplied by a fixed number each time period). That’s why it’s really good to start saving your money early in life and let it grow with time.
I like to think that exponential functions are named that because of the “$ x$” in their exponents; they are written as $ 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$, we have exponential decay (the function is getting smaller). This makes sense, since when you multiply a fraction (less than 1) many times by itself, it gets smaller, since the denominator gets larger. The “$ a$” in the functions is the stretch (or compression, if $ a<1$), and it is the $ \boldsymbol {y}$-intercept when there are no horizontal or vertical shifts.
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”). If the exponential function shifts right or left, or up and down, this “anchor point” will move. An exponential that isn’t shifted up or down has an asymptote of $ y=0$, meaning the graph gets closer and closer to the $ x$-axis without ever touching it.
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. 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 \infty \\x\to \infty ,\,\,\,\,\,y\to 0\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{{b}^{{x-h}}}+k$ ($ y=a{{b}^{{x-h}}}+k$) for exponential functions.
As you’ll see in the transformations below, it turns out that the $ \left( {0,1} \right)$ parent-function reference point of a transformed exponential function is $ \left( {h,a+k} \right)$. But, if you use t-charts, you won’t need to memorize this!
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, as we saw in the Parent Graphs and Transformations section
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 $ y=a{{b}^{x}}+k$, given an asymptote.
- For a certain graph, write the appropriate exponential function of the form $ y=a{{b}^{{x-h}}}+k$, given a certain base and asymptote.
- For a certain exponential function $ f\left( x \right)=a{{b}^{{x-h}}}+k$, the new $ (0, 1)$ reference point is $ \left( {h,a+k} \right)$, if you’re only asked for this point. Also, to find the “$ a$”, given an exponential graph and a transformed reference point, you can subtract the $ y$-value of the asymptote ($ k$) from the $ y$-value of the new reference point ($ a=\left( {a+k} \right)-k$).
Exponential Function Applications and Word Problems
Here are some compounding formulas that you’ll use in working with exponential applications, such as in appreciation (growth) and depreciation (decay). The second set of formulas are based on the first, but are a little bit more specific, since the interest is compounded multiply times during the year:
Note that the growth (or decay) rate is typically a percentage, but when it’s in the formula, it’s a decimal. The growth (or decay) factor is the actual factor after the rate is converted into a decimal and added or subtracted from 1 (they may ask you for the growth factor occasionally). When an amount triples, for example, we start with the original and add 200% to it, so the growth rate is 200% (in the formula, it’s 2.00, which will be added to the 1), but growth factor is 3 (1 + 2). Here are some examples:
Continuous Compounding
One thing that the early mathematicians found is that when the number of times the compounding takes place (“$ 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 “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 $ \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:
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.
Note: Sometimes you’ll see a problem that calls for simple interest, which is linear and not exponential. This equation for simple interest is $ I=Prt$, where $ P=$ principal, or starting amount, $ r=$ annual interest rate, and $ t=$ number of years. To get the full amount in an account after the $ t$ years, the equation is $ A=P+Prt$, or $ A=P\left( {1+rt} \right)$. There is a simple example here of using simple interest.
Here are some problems using compound interest formulas. Note that if we need to solve for a variable in an exponent, such as $ t$, we’ll need to use logs, as in in the problems here.
More Growth/Decay Equations
There are two more exponential equations that are a little more specific:
- $ \displaystyle y=a{{b}^{{\frac{t}{p}}}},\,\,\,\,b>0$, where $ a=$ the initial amount, $ b=$ the growth/decay factor, $ t=$ the time that has passed, $ p=$ the period for the growth or decay factor (the interval in which the growth/decay happens), and $ y=$ the amount after the time that has passed.
For example, if a mice population triples every 4 years (the growth interval), the population starts with 20 mice, and the problem asks how many mice will there be in 12 years, the formula is $ \displaystyle y=20{{\left( 3 \right)}^{{\frac{{12}}{4}}}}=540$ mice. (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, $ 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, when the “multiplier” is less than 1, the amount is decreasing, and we have a decay. Makes sense!
(We’ll see later that we typically have to solve for $ k$ first, using logarithms. For example, for half-life problems, we need to solve for $ k$ when $ {{N}_{0}=2}$ and $ N\left( t \right)=1$; $ k$ will be negative in this case).
Here are all the exponential formulas we’ve learned:
Remember when we put an exponent in the Graphing Calculator, we just use the “^” key!
Here are more exponential word problems. we’ll see more of these types of problems in the Logarithmic Functions section:
As an example of how we can also do these types of problems using logarithms, we could also use the uninhibited growth formula:
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.
For half-life problems, .5 base is raised to $ \displaystyle \frac{{\text{time period we want}}}{{\text{time for one half-life}}}$, since this is 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.
Why Use Logs?
Here’s one more exponential problem where we’ll see how important logs will be to solve these types of problems:
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 positive integer base 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; we can always solve these using a graphing calculator, like this:
(We will also see another 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. (I’m using the TI-84 Plus CE calculator.)
$ x$ | 1 | 4 | 6 | 10 | 11 | 13 | 20 |
$ y$ | 3 | 5 | 7.5 | 16 | 18 | 27 | 96 |
Note: Without using regression on the calculator, we could find an approximate model by averaging successive ratios. Then use the first “$ y$” to get the “$ a$”. This method isn’t as accurate as the calculator method below, but it’s a nice trick:
$ \displaystyle \begin{align} \frac{{\displaystyle \frac{5}{3}+\displaystyle \frac{{7.5}}{5}+\displaystyle \frac{{16}}{{7.5}}+\displaystyle\frac{{18}}{{16}}+\displaystyle\frac{{27}}{{18}}+\displaystyle \frac{{96}}{{27}}}}{6}&\approx 1.913\\y&\approx 3{{\left( {1.913} \right)}^{x}}\end{align}$
As we did with linear and quadratic regressions, here’s how to 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 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. We’ll do Logarithmic Inequalities here.
Let’s do an inequality problem:
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.
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On to Logarithmic Functions – you are ready!