Lesson Explainer: Solving Exponential Equations Graphically | Nagwa Lesson Explainer: Solving Exponential Equations Graphically | Nagwa

Lesson Explainer: Solving Exponential Equations Graphically Mathematics • Second Year of Secondary School

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In this explainer, we will learn how to solve exponential equations using graphical methods.

Exponential functions appear throughout many different areas of sciences and their applications include radioactive decay, population modeling, compound interest, and the spread of viruses, just to name a few.

Since exponential functions appear in many different disciplines, the study of exponential functions is important. In particular, solving exponential equations is a skill that we can apply to many different problems. For example, to determine the length of time we would need to invest a given amount of money in a compound interest account, we would need to solve an exponential equation.

Before we demonstrate how to solve an exponential equation, let’s first recall the definition of an exponential function.

Definition: Exponential Functions

An exponential function is a function of the form 𝑓(π‘₯)=π‘Žβ‹…π‘ο—, where π‘Ž and 𝑏 are real constants, 𝑏 is positive, and 𝑏≠1.

There are numerous ways of finding solutions to exponential equations of the form π‘Žβ‹…π‘=𝑔(π‘₯), for some given function 𝑔(π‘₯). For example, we could use rearrangement. However, this is not always possible. Instead we will find solutions to these equations graphically.

Given the graph of the exponential function 𝑦=π‘Žβ‹…π‘ο—, any point on the curve has coordinates of the form (π‘₯,π‘Žβ‹…π‘). If we sketch 𝑦=𝑔(π‘₯) on the same axes, then any point of intersection lies on both curves. In particular, their 𝑦-coordinates will be equal, giving us π‘Žβ‹…π‘=𝑔(π‘₯).

Let’s see an example of how to use this technique to solve an exponential equation.

Suppose we are asked to solve 4=8, given the following graph of 𝑦=4.

We want to find the values of π‘₯ that satisfy this equation. We can do this by sketching 𝑦=8 on the graph and finding the coordinates of the points of intersection.

We can see that there is one point of intersection, where π‘₯=1.5. Since this lies on the curve 𝑦=4, we must have 4=8οŠ§οŽ–οŠ«. Hence, π‘₯=1.5 is a solution to this equation. In fact, since this is the only point of intersection, we can conclude there is only one solution to the equation. We can write the set of solutions to the equation as {1.5}; this is called the solution set.

It is worth noting that sometimes, the points of intersection will not have an exact value for their π‘₯-coordinate. In these instances, we are approximating the solutions graphically and we can use the minor grid lines to determine the estimate.

It is not always possible to solve these equations. For example, if we were instead asked to solve the equation 4=βˆ’1, then we would sketch 𝑦=βˆ’1 on the graph and look for the points of intersection.

We see that there are no points of intersection. In fact, we know that 4 is positive for any value of π‘₯, so it can never be equal to βˆ’1. We can say the equation has no solutions or, alternatively, we can say that the solution set of the equation is βˆ….

It is also possible to have multiple solutions. For example, consider the following diagram of 𝑦=4 and 𝑦=73π‘₯+53.

We can see that there are two points of intersection, one at π‘₯=βˆ’0.5 and the other at π‘₯=1. Therefore, these are the only two solutions to the equation 4=73π‘₯+53.

Hence, we can say that the solution set to this equation is {βˆ’0.5,1}. We can check that these are solutions to the equation by substitution. For example, we substitute π‘₯=1 into the equation to get 4=73(1)+534=1234=4.

Since this equation is true, we have confirmed that π‘₯=1 is a solution to the equation.

In our first few examples, we will use the given graphs of exponential functions to find the solution set of an exponential equation graphically.

Example 1: Determining the Solution Set of an Exponential Equation Graphically

Use the given graph of the function 𝑓(π‘₯)=2οŠ«οŠ±ο— to find the solution set of the equation 2=2οŠ«οŠ±ο—.

Answer

We recall that the solution set to an equation is the set of all values that satisfy that equation. In this question, we want to determine all of the values of π‘₯ that make the equation 2=2οŠ«οŠ±ο— true. We can do this by recalling that any point on a curve 𝑦=𝑓(π‘₯) will have coordinates (π‘₯,𝑓(π‘₯)).

We can start by rewriting the given equation as 2=2𝑓(π‘₯)=2.οŠ«οŠ±ο—

So, we need to find all values of π‘₯ such that 𝑓(π‘₯)=2. We can find these by locating all of the points on the curve with 𝑦-coordinate 2; we sketch 𝑦=2 on the diagram and find the coordinates of the point of intersection.

By reading the π‘₯-coordinate of the point of intersection, we can see that 𝑓(4)=2, and since this is the only point of intersection, this is the only value of π‘₯ such that 𝑓(π‘₯)=2.

Therefore, the only solution to the equation is π‘₯=4. We can verify this is a solution by substituting π‘₯=4 into the equation: 2=22=22=2.οŠͺ

Since this is true, this confirms that π‘₯=4 is a solution to the equation. Hence, the solution set of the equation 2=2οŠ«οŠ±ο— is {4}.

Example 2: Determining the Solution Set of an Exponential Equation Graphically

The diagram shows the graph of 𝑓(π‘₯)=2οŠ¨ο—. Use this graph to find the solution set of the equation 2=4οŠ¨ο—.

Answer

We recall that the solution set to an equation is the set of all values that satisfy that equation. We are given the graph of 𝑓(π‘₯)=2οŠ¨ο—, and so, any point on this graph will have coordinates of the form ο€Ήπ‘₯,2ο…οŠ¨ο—. Hence, we can find the values of 2οŠ¨ο— by using the 𝑦-coordinates of the curve. In particular, we can rewrite the given equation as 2=4𝑓(π‘₯)=4.οŠ¨ο—

Hence, to solve this equation, we need to find the values where 𝑓(π‘₯) outputs 4. We sketch 𝑦=4 on the same graph and find the coordinates of points of intersection.

The coordinates of the point of intersection are (1,4), so we know that 𝑓(1)=4. Since there is only one point of intersection, this is the only solution to the equation; hence, π‘₯=1 is the only solution to the equation and the solution set is {1}.

It is worth noting we can check that 1 is a solution by substituting π‘₯=1 into the equation: 2=42=44=4.()

The solution set of the equation is {1}.

In our next few examples, we will see how to apply this method to an exponential equation that we need to rearrange first.

Example 3: Finding the Solution Set of an Exponential Equation Graphically

The diagram shows the graph of 𝑓(π‘₯)=2οŠ¨ο—. Use this graph to find the solution set of the equation 2βˆ’12=4οŠ¨ο—.

Answer

We recall that the solution set to an equation is the set of all values that satisfy that equation. We can rewrite and rearrange the equation in terms of 𝑓 as 2βˆ’12=4𝑓(π‘₯)βˆ’12=4𝑓(π‘₯)=16.οŠ¨ο—

Therefore, the solutions to the equation are the values of π‘₯ such that 𝑓(π‘₯)=16. We can find these values by recalling that any point on the curve will have the form (π‘₯,𝑓(π‘₯)), so the solutions to the equation are the points on the curve with 𝑦-coordinate 16.

There is only one point on the curve with 𝑦-coordinate 16; this is the point (2,16). Hence, 𝑓(2)=16 and the only solution to the equation is π‘₯=2. We can confirm this is a solution to the equation by substituting π‘₯=2 into the original equation: 2βˆ’12=42βˆ’12=416βˆ’12=44=4.()οŠͺ

Since this equation is true, we have confirmed that π‘₯=2 is a solution to the equation.

The solution set of the equation 2βˆ’12=4οŠ¨ο— is {2}.

Example 4: Finding the Solution Set of an Exponential Equation Graphically

The diagram shows the graph of 𝑓(π‘₯)=2ο‘οŽ‘. Use this graph to find the solution set of the equation 2+5=9ο‘οŽ‘.

Answer

We recall that the solution set to an equation is the set of all values that satisfy that equation. Since we are given a graph of 𝑓(π‘₯), we will start by rewriting the equation in terms of 𝑓(π‘₯) and rearranging: 2+5=9𝑓(π‘₯)+5=9𝑓(π‘₯)=4.ο‘οŽ‘

Hence, we can solve the equation by finding the values of π‘₯ such that 𝑓 outputs 4. We can find these values by recalling that any point on the curve has coordinates of the form (π‘₯,𝑓(π‘₯)).

Since we want 𝑓(π‘₯)=4, we sketch the line 𝑦=4 and find the coordinates of any points of intersection.

We find there is only one point of intersection with coordinates (4,4). Hence, 𝑓(4)=4 and π‘₯=4 is the only solution to the equation. We can verify this is a solution to the equation by substituting π‘₯=4 into the given equation: 2+5=92+5=94+5=99=9.

Since this is true, we have verified that π‘₯=4 is a solution to the equation.

The solution set of the given equation is {4}.

In our final two examples, we will see how to apply this method when the exponential function is equal to a linear function instead of a constant.

Example 5: Solving an Exponential Equation given a Diagram

Use the following graph to find the solution set of the equation 2=π‘₯βˆ’2ο—οŠ±οŠ©.

Answer

We recall that the solution set to an equation is the set of all values that satisfy that equation. We are given the graphs of 𝑦=2ο—οŠ±οŠ© and 𝑦=π‘₯βˆ’2, and the 𝑦-coordinate of a point on the graph tells us the output of the function at that value of π‘₯. Hence, at the points of intersection between the two graphs, we will have 2=π‘₯βˆ’2ο—οŠ±οŠ©. We can therefore use the points of intersection to find the solutions of the equation.

The π‘₯-coordinates of the points of intersection are 3 and 4, so these are the two solutions to the equation. We can verify these are the solutions by substituting them back into the original equation.

Substituting for π‘₯=3, 2=3βˆ’22=11=1.

Substituting for π‘₯=4, 2=4βˆ’22=22=2.οŠͺ

Since both equations are true, we have confirmed that both π‘₯=3 and π‘₯=4 are solutions to the equation.

The solution set of the equation is {3,4}.

Example 6: Solving an Exponential Equation Graphically

The following graph shows the function 𝑓(π‘₯)=2οŠ§οŠ±ο—. Use this graph and plot the function 𝑓(π‘₯)=π‘₯+3 to find the solution set of the equation 2=π‘₯+3οŠ±ο—.

Answer

We recall that the solution set to an equation is the set of all values that satisfy that equation. Any point on the given curve will have coordinates of the form (π‘₯,2)οŠ±ο—. If we sketch 𝑓(π‘₯)=π‘₯+3, then the points of intersection between the two graphs will have equal π‘₯- and 𝑦-coordinates. Since the 𝑦-coordinates will be equal, we will have 2=π‘₯+3οŠ±ο— at these points, giving us the solutions to the equation.

There are a few ways to sketch 𝑓(π‘₯)=π‘₯+3; we can note that when π‘₯=0, 𝑓(0)=0+3=3.

So, the 𝑦-intercept is 3. We can solve 𝑓(π‘₯)=0 to find the π‘₯-intercept: 0=π‘₯+3π‘₯=βˆ’3.

So, the π‘₯-intercept is βˆ’3 and the line passes through these two points.

We can then read off the π‘₯-coordinate of the only intersection point between the line and curve as π‘₯=βˆ’1, which is the only solution to the given equation. We can verify this is a solution to the given equation by substituting π‘₯=βˆ’1 into the equation: 2=(βˆ’1)+32=22=2.()

Since the equation is true, we have confirmed that π‘₯=βˆ’1 is a solution to the equation.

The solution set of the equation is {βˆ’1}.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • The solution set of an equation is the set of all values that satisfy the equation.
  • We can solve equations of the form 𝑓(π‘₯)=𝑔(π‘₯) by finding the π‘₯-coordinates of the points of intersection between the curves 𝑦=𝑓(π‘₯) and 𝑦=𝑔(π‘₯).
  • Sometimes, we may need to rearrange the equation into a form that is easier to solve graphically.
  • There can be no solutions to an equation, in which case the solution set is βˆ….
  • We may not be able to determine the exact coordinates of the point of intersection from a graph. In these instances, we can use the minor grid lines to approximate the solutions to the equation.

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