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Course: Algebra 2   >   Unit 12

Exponential equation word problem

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Video transcript

How to Write an Exponential Function: Word Problems

Any function in the form \(y=ab^x\) is called an exponential function. Unlike linear functions where the growth rate is constant, exponential functions are characterized by the fact that the growth rate of the function is proportional to the function's current value. In this article, the method of writing an exponential function is explained step by step.

How to Write an Exponential Function: Word Problems

A step-by-step guide to Writing exponential function: word problems

Any function in the form \(y=ab^x\) is called an exponential function where \(a\) and \(b\) are fixed and \(x\) is an independent variable:

Variable \(y\): represents the output value

\(a\) represents the initial value

\(b\) represents the common ratio

\(x\) represents the number of times the original value has been multiplied by the common ratio

Unlike linear functions where the growth rate is constant, exponential functions are characterized by the fact that the growth rate of the function is proportional to the function’s current value and are used to model growth and decay processes, such as population growth, radioactive decay, and compound interest.

Follow the step-by-step procedure below to write an exponential function.

Step \(1\): Analyzing the problem and identifying the variables. At this stage, you should specify what the question is about and what the variables of the problem are.

Step \(2\): Using the variables, write the exponential function in the form of \(y=ab^x\).

Step \(3\): Determine the initial value (\(a\)) according to the information of the problem. \(a\) shows the value of the variable at the beginning of the time period.

Step \(4\): In this step, you must specify the common ratio (\(b\)). \(b\) is the coefficient by which the variable is multiplied in each time period.

Step \(5\): Specify the \(x\) variable that represents the number of time periods that have passed. For example, if in the problem it is said that a certain amount will be multiplied by \(4\) every month and after \(5\) months you will be asked for the amount, \(x=5\).

Step \(6\): Place the values of \(a\), \(b\) and \(x\) in the exponential function and simplify it.

Step \(7\): Check your work by inserting \(x\) values and seeing if they match the problem statement

Step \(8\): Write the final answer in one sentence

Writing exponential function: word problems-Example 1:

If a bank’s profits start at \($60,000\) and increase by \(25%\) each year, how much will the profits be after \(4\) years?

This problem is solved by using the exponential function. In the said problem, the profit of the company starts from \(60,000\) dollars and increases by \(25%\) every year, the profit is multiplied by \(1.2\) every year. With this information, we write the exponential function:

\(y = 60000 * 1.25^x\)

In this function, \(x\) is the number of past years and \(y\) represents the profit after \(x\) years.

The profit amount after \(4\) years is obtained by inserting \(x = 4\) in the function: \(y = 40000 * 1.25^4 = 60000 * 1.25 * 1.25 * 1.25 * 1.25 = 146484\)

by: Effortless Math Team about 5 months ago (category: Articles )

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how to solve exponential equations word problems

How Do You Solve a Word Problem Using an Exponential Function?

Word problems let you see math in the real world! This tutorial shows you how to create a table and identify a pattern from the word problem. Then you can see how to create an exponential function from the data and solve the function to get your answer!

Background Tutorials

Factors and greatest common factor.

What's a Factor?

What's a Factor?

Factors are a fundamental part of algebra, so it would be a great idea to know all about them. This tutorial can help! Take a look!

Function Definitions and Notation

What's a Function?

What's a Function?

You can't go through algebra without learning about functions. This tutorial shows you a great approach to thinking about functions! Learn the definition of a function and see the different ways functions can be represented. Take a look!

Evaluating Expressions

How Do You Evaluate an Algebraic Expression?

How Do You Evaluate an Algebraic Expression?

Plugging variables into an expression is essential for solving many algebra problems. See how to plug in variable values by watching this tutorial.

What is a Variable?

What is a Variable?

You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!

Rules of Exponents

What are Exponents?

What are Exponents?

Did you know that another word for 'exponent' is 'power'? To learn the meaning of these words and to see some special cases involving exponents, check out this tutorial!

Prime Factorization

How Do You Find the Prime Factorization of a Number Using a Tree?

How Do You Find the Prime Factorization of a Number Using a Tree?

To write the prime factorization for a number, it's often useful to use something called a factor tree. Follow along with this tutorial and see how to use a factor tree to find the prime factorization of a given number.

Further Exploration

Exponential functions.

How Do You Graph an Exponential Function Using a Table?

How Do You Graph an Exponential Function Using a Table?

Graphing an exponential function? No sweat! Create a table of values to give you ordered pairs. Then, plot those ordered pair on a coordinate plane and connect the points to make your graph! Follow along with this tutorial as it shows you all the steps.

Modeling with Exponential Functions

How Do You Identify Exponential Behavior from a Pattern in the Data?

How Do You Identify Exponential Behavior from a Pattern in the Data?

Take a look at how you identify exponential behavior from a pattern in your data. You'll also see how to figure out if that pattern represents exponential growth or exponential decay. Check it out!

Math Hints

Exponential Functions

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 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.

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 \(\displaystyle \left( {-1,\,\frac{1}{b}} \right),\left( {0,1} \right),\left( {1,b} \right)\).

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 :

Asymptote:   \(y=3\)

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

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

Translate 4 units right, 3 units up.

how to solve exponential equations word problems


Parent function:


Remember that \(e\approx 2.7\).

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

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 1 unit left, 2 units up.

how to solve exponential equations word problems

Writing Exponential Equations from Points and Graphs

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

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:

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  her e  below, and in the Logarithmic Functions section here .

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:

how to solve exponential equations word 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:

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: \(\begin{array}{c}300=500{{\left( {1-.106} \right)}^{t}};\,\,\frac{{300}}{{500}}={{\left( {1-.106} \right)}^{t}}\\\ln \left( {\frac{{300}}{{500}}} \right)=\ln {{\left( {1-.106} \right)}^{t}};\,\,\ln \left( {\frac{{300}}{{500}}} \right)=t\ln \left( {1-.106} \right)\\t=\frac{{\,\ln \left( {\frac{{300}}{{500}}} \right)}}{{\ln \left( {1-.106} \right)}}\approx 4.556\end{array}\)

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.)

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:

\(\displaystyle \begin{align}\frac{{\frac{5}{3}+\frac{{7.5}}{5}+\frac{{16}}{{7.5}}+\frac{{18}}{{16}}+\frac{{27}}{{18}}+\frac{{96}}{{27}}}}{6}&\approx 1.913\\y&\approx 3{{\left( {1.913} \right)}^{x}}\end{align}\)

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.

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!

how to solve exponential equations word problems

Select a Course Below

Exponential-Growth Word Problems

Log Probs Expo Growth Expo Decay

What is the formula for exponential growth?

Exponential growth word problems work off the exponential-growth formula, A  =  Pe rt , where A is the ending amount of whatever you're dealing with (for example, money sitting in an investment, or bacteria growing in a petri dish), P is the beginning amount of that same whatever, r is the growth constant, and t is time.

Content Continues Below

Exponential Growth and Decay on

Exponential Growth and Decay

The exponential-growth formula A  =  Pe rt is related to the compound-interest formula , and represents the case of the interest being compounded "continuously".

Note that the particular variables used in the equation may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:

A = P e r t A = P e k t Q = N e k t Q = Q 0 e k t

No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the growth constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

How do you solve exponential growth problems?

To solve exponential growth word problems, you may be plugging one value into the exponential-growth equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the doubling-time information to find the growth constant (probably by solving the exponential equation by using logarithms), and then using that value to find whatever the exercise requested.

What is an example of an exponential growth problem?

For this exercise, the units on time t will be hours, because the growth is being measured in terms of hours.

The beginning amount P is the amount at time t  = 0 , so, for this problem, P  = 100 .

The ending amount is A  = 450 at t  = 6 hours.

The only variable I don't have a value for is the growth constant k , which also happens to be what I'm looking for. So I'll plug all the known values into the exponential-growth formula, and then solve for the growth constant:

450 = 100 e 6 k

4.5 = e 6 k

ln(4.5) = 6 k

ln(4.5) / 6 = k = 0.250679566129...

They want me to round my decimal value (found by punching keys on my calculator) to two decimal places. So my answer is:

Many math classes, math books, and math instructors leave off the units for the growth and decay rates. However, if you see this topic again in chemistry or physics, you will probably be expected to use proper units ("growth-decay constant ÷ time"). If I had included this information in my solution above, my answer would have been " k  = 0.25 /hour". I doubt that your math class will even mention this, let alone require that you include this. Still, it's not a bad idea to get into the habit now of checking and reporting your units.

Note that the constant was positive, because it was a growth constant. If I had come up with a negative value for the growth constant, then I would have known to check my work to find my error(s).

In this problem, I know that time t will be in hours, because they gave me growth in terms of hours. As a result, I'll convert "a day and a half" to "thirty-six hours", so my units match.

I know that the starting population is P  = 100 , and I need to find A at time t  = 36 .


But what is the growth constant k ? And why do they tell me what the doubling time is?

They gave me the doubling time because I can use this to find the growth constant k . Then, once I have this constant, I can go on to answer the actual question.

So this exercise actually has two unknowns, the growth constant k and the ending amount A . I can use the doubling time to find the growth constant, at which point the only remaining value will be the ending amount, which is what they actually asked for. So first I'll find the constant.

If the initial population were, say, 100 , then, in 6.5 hours (being the specified doubling time), the population would be 200 . (It doesn't matter what starting value I pick for this part of my solution, as long as my ending value is twice as much. It's the "twice as much" that matters here, more than the number that it's twice of.) I'll set this up and solve for k :

200 = 100 e 6.5 k

2 = e 6.5 k

At this point, I need to use logs to solve:

ln(2) = 6.5 k

ln(2) / 6.5 = k

I could simplify this to a decimal approximation, but I won't, because I don't want to introduce round-off error if I can avoid it. So, for now, the growth constant will remain this "exact" value. (I might want to check this value quickly in my calculator, to make sure that this growth constant is positive, as it should be. If I have a negative value at this stage, I need to go back and check my work.)

Now that I have the growth constant, I can answer the actual question, which was "How many bacteria will there be in thirty-six hours?" This means using 100 for P , 36 for t , and the above expression for k . I plug these values into the formula, and then I simplify to find A :

A = 100 e 36(ln(2)/6.5) = 4647.75313957...

I will take the luxury of assuming that they don't want fractional bacteria (being the 0.7313957... part), so I'll round to the nearest whole number.

about 4648 bacteria

You can do a rough check of this answer, using the fact that exponential processes involve doubling times. The doubling time in this case is 6.5 hours, or between 6 and 7 hours.

If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, 800 in eighteen hours, 1600 in twenty-four hours, 3200 in thirty hours, and 6400 in thirty-six hours.

If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twenty-one hours, 1600 in twenty-eight hours, and 3200 in thirty-five hours.

The answer I got above, 4678 bacteria in thirty-six hours, fits nicely between these two estimates.

Algebra Tutors

Note: When you are given a nice, neat doubling time, another method for solving the exercise is to use a base of 2 . First, figure out how many doubling-times that you've been given. In the above case, this would start by noting that "a day and a half" is 36 hours, so we have:

36 ÷ 6.5 = 72/13

Use this as the power on 2 :

100 × 2 (72/13) = 4647.75314...

Not all algebra classes cover this method. If you're required to use the first method for every exercise of this type, then do so (in order to get the full points). Otherwise, this base- 2 trick can be a time-saver. And, yes, you'd use a base of 3 if you'd been given a tripling-time, a base of 4 for a quadrupling-time, etc.


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