Lesson Video: Solving Exponential Equations Graphically Mathematics

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

18:33

Video Transcript

Solving Exponential Equations Graphically

In this video, we will learn how to determine the number of solutions to an exponential equation given graphically, and we will also see how to apply this to solve exponential equations using graphical methods. Before we begin by trying to solve exponential equations graphically, let’s start by recalling what we mean by an exponential function.

We recall that an exponential function is one of the form 𝑓 of π‘₯ is equal to π‘Ž multiplied by 𝑏 to the power of π‘₯, where 𝑏 is positive and 𝑏 is not allowed to be equal to one. Well, it’s worth noting these are not the only example of exponential functions. For example, we could multiply our value of π‘₯ by a constant, or we could add a constant onto this value of π‘₯, and this would still be an exponential function. However, for now, we’ll just focus on this type of exponential function. This then means that an exponential equation is an equation involving an exponential function. So, that’s one of the form π‘Ž times 𝑏 to the power of π‘₯ is equal to some function 𝑔 of π‘₯. A solution to this equation would be a value of π‘₯ which makes both sides of the equation equal.

Sometimes we can solve equations by using algebraic manipulation. However, for exponential equations, this is often very difficult. This is because π‘₯, our variable, appears in the exponent. So instead, we’ll focus on finding these solutions graphically. To see how we might solve an exponential equation graphically, let’s start with an example. Let’s assume we were given the graph of the exponential function 𝑦 is equal to four to the power of π‘₯, and let’s say we were asked to solve the equation four to the power of π‘₯ is equal to one. This means we need to find the value of π‘₯ we substitute into our exponential function to output a value of one. We can do this by using our graph. Remember, the 𝑦-coordinate of every point on our curve tells us the output value for our function at that value of π‘₯.

We want to know when our function outputs a value of one, it will output a value of one when its 𝑦-coordinate is equal to one. So, we sketch the line 𝑦 is equal to one onto our axes. We can then see when our curve has a 𝑦-value of one. It has a 𝑦-value of one when its π‘₯-value is equal to zero. In other words, we’ve shown there’s one point of intersection between the line 𝑦 is equal to one and the curve 𝑦 is equal to four to the power of π‘₯. That’s the point with coordinates zero, one. At this value of π‘₯, both our function four to the power of π‘₯ and the function one output the same value. They output the value of one. Therefore, this must be a solution to our equation. π‘₯ equals zero solves the equation four to the power of π‘₯ is equal to one.

There are a few other things we can notice. For example, this is the only point of intersection between our line and curve. And every solution to our equation will be a point of intersection between the line and curve. So, because there’s only one point of intersection between the line and the curve, we can conclude there’s only one solution to this equation. It’s also worth noting we can check that π‘₯ is equal to zero is a solution to our equation by substituting π‘₯ is equal to zero into both sides of the equation and making sure they’re equal. Let’s start with substituting π‘₯ is equal to zero into the left-hand side of the equation.

Substituting π‘₯ is equal to zero into the left-hand side of our equation, we get four to the power of zero. And we can evaluate this by using our laws of exponents. We know any nonzero number raised to the zeroth power is always equal to one. We can then do the same with the right-hand side of our equation. However, the right-hand side of this equation is just a constant value of one, so the value of π‘₯ does not affect this value. Therefore, when π‘₯ is equal to zero, both the left- and right-hand side of our equation are equal. This confirms that π‘₯ is equal to zero is a solution to our equation.

We can use this exact same method to solve other exponential equations. For example, let’s solve the equation four to the power of π‘₯ is equal to five minus π‘₯. Once again, since a solution to our equation is a value of π‘₯ such that both sides of our equation are equal, we can find the solutions to our equation by looking for points of intersection between the curve 𝑦 is equal to four to the power of π‘₯ and the line 𝑦 is equal to five minus π‘₯ because the points of intersection will have the same output for both of our functions, in other words, they will be solutions to our equation.

We already have a graph of 𝑦 is equal to four to the power of π‘₯. So, on the same axis, we should sketch the line 𝑦 is equal to five minus π‘₯. First, we know that its 𝑦-intercept is five. We can also find the π‘₯-intercept of this line. We substitute 𝑦 is equal to zero and solve for our value of π‘₯. We see that the π‘₯-intercept of this line is when π‘₯ is equal to five. We can use this to plot the line. We know its 𝑦-intercept is at five and its π‘₯-intercept is also at five. Then the straight line connecting these two points is the line 𝑦 is equal to five minus π‘₯.

Finally, we can see that there’s one point of intersection between our curve and our line, and this will be the point where the outputs of both of these functions are equal. We can see that the π‘₯-coordinate of this point of intersection is one. So, we have that π‘₯ is equal to one is a solution to our equation. And in fact, since this is the only point of intersection between the line and the curve, this is the only solution to our equation. However, we do need to be careful. We sketch the line 𝑦 is equal to five minus π‘₯, and we use this to estimate the point of intersection between the line and the curve. So, we can’t be sure that π‘₯ is equal to one is the exact solution to our equation since we are approximating by using our sketch.

To show that π‘₯ is equal to one is the solution to this equation, we’re going to need to substitute π‘₯ is equal to one into both the left- and right-hand side of our equation and check if they’re equal. We can start with the left-hand side of our equation. Substituting π‘₯ is equal to one, we get four to the first power. And by using our laws of exponents, we know any number raised to the first power is equal to itself. So, four to the first power is equal to four. We can then do the same with the right-hand side of our equation. Substituting π‘₯ is equal to one, we get five minus one, which we can evaluate is equal to four. Therefore, since both the left- and right-hand side of our equation are equal when π‘₯ is equal to one, we can conclude that π‘₯ is equal to one is a solution to our exponential equation.

So far, all of our equations have had solutions. However, it is also possible that an equation has no solutions. For example, we can see that the line 𝑦 is equal to negative two and the curve 𝑦 is equal to four to the power of π‘₯ have no points of intersection. This means if we were asked to solve the equation four to the power of π‘₯ is equal to negative two by using the given diagram, we would be able to conclude that there are no solutions to this equation because any solution to this equation would be a point of intersection between the curve 𝑦 is equal to four to the power of π‘₯ and the line 𝑦 is equal to negative two. And instead of saying there are no solutions to this equation, we can introduce the idea of the solution set.

The solution set of an equation is the set of all solutions to that equation. So, instead of saying the equation four to the power of π‘₯ is equal to negative two has no solutions, we can say that its solution set is the empty set. Let’s now go through an example where we’re given the graph of an exponential function and we need to use this to determine the solution set of an exponential equation.

Use the given graph of the function 𝑓 of π‘₯ is equal to two to the power of five minus π‘₯ to find the solution set of the equation two to the power of five minus π‘₯ is equal to two.

In this question, we’re given the graph of an exponential function and this exponential function appears in the given exponential equation. We need to use this to determine the solution set of the equation. First, we recall the solution set of an equation is the set of all solutions to that equation. Therefore, we’re looking for the set of all values of π‘₯ which balance both sides of the equation. Another way of thinking about this is since two to the power of five minus π‘₯ is equal to the function 𝑓 of π‘₯, we can substitute 𝑓 of π‘₯ into our equation. This gives us the equation 𝑓 of π‘₯ is equal to two. We’re looking for the set of all values of π‘₯ such that 𝑓 of π‘₯ is equal to two.

To find these values of π‘₯, we can recall that every single point on the curve 𝑦 is equal to 𝑓 of π‘₯ will have coordinates of the form π‘₯, 𝑓 of π‘₯. In other words, the 𝑦-coordinates of the points on the curve tell us the outputs of our function for the given value of π‘₯. We want to determine the values of π‘₯ where our function outputs two. These will be the points on our curve with 𝑦-coordinate equal to two. So, we can find these by sketching the line 𝑦 is equal to two onto the same set of axes. We can see there’s only one point on our curve of 𝑦-coordinate equal to two. It will be the point of intersection between the line 𝑦 is equal to two and the curve 𝑦 is equal to two to the power of five minus π‘₯. The 𝑦-coordinate of this point is two and its π‘₯-coordinate is four. In other words, when π‘₯ is equal to four, our function outputs two. 𝑓 evaluated at four is two.

Therefore, π‘₯ is equal to four is a solution to our equation. In fact, since this is the only point of intersection between the line and the curve, this is the only solution to our equation. This means the solution set to our equation is just the set containing four.

It’s also worth noting we can check our answer by substituting π‘₯ is equal to four into our equation or into our function. Substituting π‘₯ is equal to four into our function 𝑓 of π‘₯, we get 𝑓 evaluated at four is two to the power of five minus four. Five minus four is equal to one. So, this simplifies to give us two to the first power. And any number raised to the first power is just equal to itself. So, 𝑓 evaluated at four is equal to two, which is exactly the same as the right-hand side of our equation, confirming that π‘₯ is equal to four is a solution to our equation. Therefore, we were able to show the solution set of the equation two to the power of five minus π‘₯ is equal to two is just the set containing four.

Let’s now go for another example where we’re given the graph of an exponential function, and we need to use this to solve an exponential equation.

The diagram shows the graph of 𝑓 of π‘₯ is equal to two to the power of two π‘₯. Use this graph to find the solution set of the equation two to the power of two π‘₯ is equal to four.

In this question, we’re given the graph of an exponential function, and we’re asked to solve an exponential equation where this function appears. To do this, we start by recalling the solution set of an equation is the set of all solutions to that equation. This means we’re looking for all of the values of π‘₯ which solve the equation two to the power of two π‘₯ is equal to four. To help us do this, let’s start by replacing two to the power of two π‘₯ in our equation with 𝑓 of π‘₯. This means the equation we’re asked to solve can be rewritten as 𝑓 of π‘₯ is equal to four.

In other words, we’re just looking for the values of π‘₯ such that our function 𝑓 outputs a value of four. And we can recall that the 𝑦-coordinate of a point on our curve tells us the output value of a function for that value of π‘₯. So, we can find all of the values where our function outputs four by sketching the line 𝑦 is equal to four onto our diagram. And we can then see on our diagram there’s only one point on our curve with 𝑦-coordinate equal to four. This is the point of coordinates one, four.

And it’s worth reiterating this tells us that 𝑓 evaluated at one is equal to four. And therefore, one is a solution to our equation. In fact, all solutions to our equation will be a point of intersection between the line 𝑦 is equal to four and the curve 𝑦 is equal to two to the power of two π‘₯. So, because we can see there’s only one point of intersection, we know there’s only one solution. Therefore, the solution set of the equation two to the power of two π‘₯ is just the set containing one.

Let’s now see an example where we first need to rearrange the exponential equation that we’re given.

The diagram shows the graph of 𝑓 of π‘₯ is equal to two to the power of π‘₯ over two. Use this graph to find the solution set of the equation two to the power of π‘₯ over two plus five is equal to nine.

In this question, we’re given the graph of an exponential function 𝑓 of π‘₯. And we’re asked to use this to determine the solution set of an equation which contains our function 𝑓 of π‘₯. To do this, we start by recalling the solution set of an equation is the set of all solutions to that equation. In this case, it will be the set of all values of π‘₯, such that two to the power of π‘₯ over two plus five is equal to nine. To answer this question, it can help us to rewrite our exponential equation in terms of the function 𝑓 of π‘₯. Substituting two to the power of π‘₯ over two is equal to 𝑓 of π‘₯ into our equation, we get 𝑓 of π‘₯ plus five is equal to nine. We can simplify this equation further by subtracting five from both sides. We get 𝑓 of π‘₯ is equal to nine minus five, which simplifies to give us 𝑓 of π‘₯ is equal to four. So, we want to find the values of π‘₯ such that our function outputs a value of four.

Remember that the 𝑦-coordinate of any point on our curve tells us the output value of our function at that value of π‘₯. So, we want to find all of the points on our curve with 𝑦-coordinate four. We do this by sketching the line 𝑦 is equal to four onto our diagram. We can see there’s only one point of intersection between our line and our curve. And we can see that this point has π‘₯-coordinate four. Therefore, when we input a value of π‘₯ is equal to four into our function, the output value is four. 𝑓 of four is equal to four. And in fact, since this is the only point of intersection between our line and our curve, this is the only solution to our equation. Therefore, the solution set of this equation is the set containing four.

We can check that π‘₯ is equal to four is a solution to our equation by substituting π‘₯ is equal to four into the left-hand side of our equation. Substituting π‘₯ is equal to four into the left-hand side of our equation, we get two to the power four over two plus five, which we can simplify four over two is equal to two. So, this is equal to two squared plus five. And then we can evaluate this. Two squared is equal to four. So, we get four plus five, which is equal to nine, which we can see is exactly equal to the right-hand side of this equation. Therefore, four is a solution to our equation, and we know it’s the only solution. Therefore, the solution set of the equation two to the power of π‘₯ over two plus five is equal to nine is the set containing four.

Let’s now see an example where our exponential equation involves a linear function.

Use the graphs below to answer the following question. True or false: the equation two to the power of π‘₯ is equal to negative π‘₯ has no solution.

In this question, we’re given the graph of two functions. Let’s start by determining which two functions these are the graphs of. First, we can see that our straight line passes through the origin, so its 𝑦-intercept is zero. Next, we can see for every one unit we go across, we travel one unit down. So, its slope is negative one. In slope–intercept form, that’s the line 𝑦 is equal to negative one π‘₯ plus zero, which is just 𝑦 is equal to negative π‘₯. Our other curve has the shape of an exponential function, and we can see it passes through the point with coordinates one, two. If we substitute π‘₯ is equal to one into the function two to the power of π‘₯, we can see this outputs a value of two. We could do this with other points on our curve to conclude that this is indeed a sketch of the curve 𝑦 is equal to two to the power of π‘₯.

We need to use these graphs to determine whether or not the equation two to the power of π‘₯ is equal to negative π‘₯ has a solution. We might be tempted to try and solve this by using manipulation. However, this will be very difficult because π‘₯ appears in the exponent and not in the exponent. Instead, recall that a solution to this equation is a value of π‘₯ such that both sides of the equation are equal. In other words, we need to input a value of π‘₯ into the function two to the power of π‘₯ and then put the same value into the function negative π‘₯ to get the same output. We can do this directly from our graph. For the outputs of these two functions to be equal with the same π‘₯-input, they must have a point of intersection. This is because the 𝑦-coordinate tells us the outputs of this function for the given input.

Hence, because there’s one point of intersection between the line and the curve, we can conclude that two to the power of π‘₯ is equal to negative π‘₯ has one solution. In fact, we can even approximate this value by trying to read off its π‘₯-coordinate from the graph. Doing this, we would get that π‘₯ is approximately equal to negative 0.6. Therefore, we were able to show that it is false that the equation two to the power of π‘₯ is equal to negative π‘₯ has no solutions.

In our final example, we’ll solve an exponential equation graphically by also sketching a linear function on the same given graph.

The following graph shows the function 𝑓 sub one of π‘₯ is equal to two to the power of negative π‘₯. Use this graph and plot the function 𝑓 sub two of π‘₯ is equal to π‘₯ plus three to find the solution set of the equation two to the power of negative π‘₯ is equal to π‘₯ plus three.

In this question, we’re given two functions 𝑓 sub one of π‘₯ and 𝑓 sub two of π‘₯, and we’re given a graph of the function 𝑦 is equal to 𝑓 sub one of π‘₯. We’re asked to find the solution set of an equation. And since 𝑓 sub one of π‘₯ is equal to the left-hand side of this equation and 𝑓 sub two of π‘₯ is equal to the right-hand side of this equation, the equation is 𝑓 sub one of π‘₯ equals 𝑓 sub two of π‘₯. We can solve this equation graphically. Any solution to this equation will be a point of intersection between the curve 𝑦 is equal to 𝑓 sub one of π‘₯ and the line 𝑦 is equal to 𝑓 sub two of to π‘₯. Because the point of intersection would have the same 𝑦-coordinate and the 𝑦- coordinate is the output of the function for the given π‘₯ coordinator, which means the outputs of the function would be the same, so our equation would be solved.

We need to sketch the curve 𝑦 is equal to π‘₯ plus three. First, we note that its 𝑦-intercept will be at three. We can also find its π‘₯-intercept by substituting 𝑦 is equal to zero. Solving this, we get that π‘₯ is equal to negative three. We can then plot our line. Its 𝑦-intercept is at three, and its π‘₯-intercept is at negative three. This then allows us to plot our line. We just connect the 𝑦- and π‘₯-intercept with a straight line. Then, the only point of intersection between our line and our curve will be the only solution to our equation. We can read off its π‘₯-coordinate; its π‘₯-coordinate is negative one.

Then, since the question ask us to write this as a solution set, we’ll write this as the set containing negative one. Therefore, we were able to show the solution set of the equation two to the power of negative π‘₯ is equal to π‘₯ plus three is just the set containing negative one.

Let’s now go over some of the key points of this video. First, the solution set of an equation is the set of all solutions to that equation, in other words, is the set of all values which satisfy the equation. And in particular, if an equation has no solutions, we can say the solution set is the empty set. Next, we saw that we can solve the equation 𝑓 of π‘₯ is equal to 𝑔 of π‘₯ by finding the π‘₯-coordinates of all of the points of intersection between the graphs 𝑦 is equal to 𝑓 of π‘₯ and 𝑦 is equal to 𝑔 of π‘₯.

Every point of intersection is a solution to the equation. And if there are no points of intersection, then there are no solutions to the equation. Finally, we saw the graphical solutions to equations can be approximations. This is particularly true if we need to sketch one of the functions ourselves. In these cases, we should use the grid lines to try and make our approximation as accurate as possible.

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