WEBVTT
00:00:02.520 --> 00:00:04.770
Solving Exponential Equations Graphically
00:00:04.930 --> 00:00:16.620
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.
00:00:16.850 --> 00:00:24.220
Before we begin by trying to solve exponential equations graphically, letβs start by recalling what we mean by an exponential function.
00:00:24.530 --> 00:00:34.390
We recall that an exponential function is one of the form π of π₯ is equal to π multiplied by π to the power of π₯, where π is positive and π is not allowed to be equal to one.
00:00:34.730 --> 00:00:38.340
Well, itβs worth noting these are not the only example of exponential functions.
00:00:38.370 --> 00:00:46.280
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.
00:00:46.590 --> 00:00:50.130
However, for now, weβll just focus on this type of exponential function.
00:00:50.350 --> 00:00:55.050
This then means that an exponential equation is an equation involving an exponential function.
00:00:55.080 --> 00:01:00.450
So, thatβs one of the form π times π to the power of π₯ is equal to some function π of π₯.
00:01:00.760 --> 00:01:06.160
A solution to this equation would be a value of π₯ which makes both sides of the equation equal.
00:01:06.520 --> 00:01:10.000
Sometimes we can solve equations by using algebraic manipulation.
00:01:10.050 --> 00:01:13.420
However, for exponential equations, this is often very difficult.
00:01:13.420 --> 00:01:16.460
This is because π₯, our variable, appears in the exponent.
00:01:16.800 --> 00:01:20.020
So instead, weβll focus on finding these solutions graphically.
00:01:20.260 --> 00:01:24.270
To see how we might solve an exponential equation graphically, letβs start with an example.
00:01:24.310 --> 00:01:33.550
Letβs assume we were given the graph of the exponential function π¦ is equal to four to the power of π₯, and letβs say we were asked to solve the equation four to the power of π₯ is equal to one.
00:01:33.590 --> 00:01:39.880
This means we need to find the value of π₯ we substitute into our exponential function to output a value of one.
00:01:40.110 --> 00:01:41.870
We can do this by using our graph.
00:01:41.910 --> 00:01:48.960
Remember, the π¦-coordinate of every point on our curve tells us the output value for our function at that value of π₯.
00:01:49.150 --> 00:01:55.720
We want to know when our function outputs a value of one, it will output a value of one when its π¦-coordinate is equal to one.
00:01:55.750 --> 00:01:59.610
So, we sketch the line π¦ is equal to one onto our axes.
00:01:59.850 --> 00:02:02.660
We can then see when our curve has a π¦-value of one.
00:02:02.690 --> 00:02:06.520
It has a π¦-value of one when its π₯-value is equal to zero.
00:02:06.850 --> 00:02:14.540
In other words, weβve shown thereβs one point of intersection between the line π¦ is equal to one and the curve π¦ is equal to four to the power of π₯.
00:02:14.570 --> 00:02:17.060
Thatβs the point with coordinates zero, one.
00:02:17.400 --> 00:02:23.670
At this value of π₯, both our function four to the power of π₯ and the function one output the same value.
00:02:23.710 --> 00:02:25.290
They output the value of one.
00:02:25.640 --> 00:02:28.040
Therefore, this must be a solution to our equation.
00:02:28.110 --> 00:02:32.270
π₯ equals zero solves the equation four to the power of π₯ is equal to one.
00:02:32.580 --> 00:02:34.580
There are a few other things we can notice.
00:02:34.610 --> 00:02:39.000
For example, this is the only point of intersection between our line and curve.
00:02:39.430 --> 00:02:44.390
And every solution to our equation will be a point of intersection between the line and curve.
00:02:44.420 --> 00:02:51.070
So, because thereβs only one point of intersection between the line and the curve, we can conclude thereβs only one solution to this equation.
00:02:51.360 --> 00:03:01.320
Itβs 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 theyβre equal.
00:03:01.650 --> 00:03:05.820
Letβs start with substituting π₯ is equal to zero into the left-hand side of the equation.
00:03:05.970 --> 00:03:11.880
Substituting π₯ is equal to zero into the left-hand side of our equation, we get four to the power of zero.
00:03:12.130 --> 00:03:14.720
And we can evaluate this by using our laws of exponents.
00:03:14.750 --> 00:03:19.230
We know any nonzero number raised to the zeroth power is always equal to one.
00:03:19.600 --> 00:03:22.120
We can then do the same with the right-hand side of our equation.
00:03:22.410 --> 00:03:28.590
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.
00:03:28.900 --> 00:03:33.400
Therefore, when π₯ is equal to zero, both the left- and right-hand side of our equation are equal.
00:03:33.430 --> 00:03:36.930
This confirms that π₯ is equal to zero is a solution to our equation.
00:03:37.210 --> 00:03:40.730
We can use this exact same method to solve other exponential equations.
00:03:40.760 --> 00:03:45.490
For example, letβs solve the equation four to the power of π₯ is equal to five minus π₯.
00:03:45.720 --> 00:04:07.590
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 π¦ is equal to four to the power of π₯ and the line π¦ 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.
00:04:07.870 --> 00:04:10.960
We already have a graph of π¦ is equal to four to the power of π₯.
00:04:10.960 --> 00:04:15.150
So, on the same axis, we should sketch the line π¦ is equal to five minus π₯.
00:04:15.190 --> 00:04:17.460
First, we know that its π¦-intercept is five.
00:04:17.790 --> 00:04:20.080
We can also find the π₯-intercept of this line.
00:04:20.110 --> 00:04:23.200
We substitute π¦ is equal to zero and solve for our value of π₯.
00:04:23.230 --> 00:04:26.850
We see that the π₯-intercept of this line is when π₯ is equal to five.
00:04:27.240 --> 00:04:28.840
We can use this to plot the line.
00:04:28.880 --> 00:04:32.540
We know its π¦-intercept is at five and its π₯-intercept is also at five.
00:04:32.990 --> 00:04:37.860
Then the straight line connecting these two points is the line π¦ is equal to five minus π₯.
00:04:38.010 --> 00:04:45.600
Finally, we can see that thereβs 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.
00:04:45.940 --> 00:04:49.570
We can see that the π₯-coordinate of this point of intersection is one.
00:04:49.800 --> 00:04:52.870
So, we have that π₯ is equal to one is a solution to our equation.
00:04:52.900 --> 00:04:58.920
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.
00:04:59.160 --> 00:05:01.030
However, we do need to be careful.
00:05:01.060 --> 00:05:08.660
We sketch the line π¦ is equal to five minus π₯, and we use this to estimate the point of intersection between the line and the curve.
00:05:08.910 --> 00:05:15.450
So, we canβt be sure that π₯ is equal to one is the exact solution to our equation since we are approximating by using our sketch.
00:05:15.640 --> 00:05:24.700
To show that π₯ is equal to one is the solution to this equation, weβre going to need to substitute π₯ is equal to one into both the left- and right-hand side of our equation and check if theyβre equal.
00:05:25.060 --> 00:05:27.180
We can start with the left-hand side of our equation.
00:05:27.210 --> 00:05:30.570
Substituting π₯ is equal to one, we get four to the first power.
00:05:30.800 --> 00:05:35.630
And by using our laws of exponents, we know any number raised to the first power is equal to itself.
00:05:35.630 --> 00:05:37.910
So, four to the first power is equal to four.
00:05:38.180 --> 00:05:40.660
We can then do the same with the right-hand side of our equation.
00:05:40.690 --> 00:05:45.610
Substituting π₯ is equal to one, we get five minus one, which we can evaluate is equal to four.
00:05:45.980 --> 00:05:54.490
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.
00:05:54.820 --> 00:05:57.560
So far, all of our equations have had solutions.
00:05:57.590 --> 00:06:01.280
However, it is also possible that an equation has no solutions.
00:06:01.310 --> 00:06:08.990
For example, we can see that the line π¦ is equal to negative two and the curve π¦ is equal to four to the power of π₯ have no points of intersection.
00:06:09.310 --> 00:06:27.550
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 π¦ is equal to four to the power of π₯ and the line π¦ is equal to negative two.
00:06:27.850 --> 00:06:33.010
And instead of saying there are no solutions to this equation, we can introduce the idea of the solution set.
00:06:33.100 --> 00:06:37.630
The solution set of an equation is the set of all solutions to that equation.
00:06:37.920 --> 00:06:45.950
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.
00:06:46.350 --> 00:06:54.040
Letβs now go through an example where weβre given the graph of an exponential function and we need to use this to determine the solution set of an exponential equation.
00:06:56.990 --> 00:07:06.770
Use the given graph of the function π 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.
00:07:07.130 --> 00:07:13.930
In this question, weβre given the graph of an exponential function and this exponential function appears in the given exponential equation.
00:07:14.230 --> 00:07:17.750
We need to use this to determine the solution set of the equation.
00:07:17.820 --> 00:07:23.020
First, we recall the solution set of an equation is the set of all solutions to that equation.
00:07:23.350 --> 00:07:28.060
Therefore, weβre looking for the set of all values of π₯ which balance both sides of the equation.
00:07:28.360 --> 00:07:36.760
Another way of thinking about this is since two to the power of five minus π₯ is equal to the function π of π₯, we can substitute π of π₯ into our equation.
00:07:36.790 --> 00:07:39.280
This gives us the equation π of π₯ is equal to two.
00:07:39.310 --> 00:07:43.650
Weβre looking for the set of all values of π₯ such that π of π₯ is equal to two.
00:07:44.070 --> 00:07:52.520
To find these values of π₯, we can recall that every single point on the curve π¦ is equal to π of π₯ will have coordinates of the form π₯, π of π₯.
00:07:52.760 --> 00:07:59.660
In other words, the π¦-coordinates of the points on the curve tell us the outputs of our function for the given value of π₯.
00:07:59.940 --> 00:08:03.110
We want to determine the values of π₯ where our function outputs two.
00:08:03.410 --> 00:08:06.660
These will be the points on our curve with π¦-coordinate equal to two.
00:08:06.690 --> 00:08:11.490
So, we can find these by sketching the line π¦ is equal to two onto the same set of axes.
00:08:11.730 --> 00:08:15.730
We can see thereβs only one point on our curve of π¦-coordinate equal to two.
00:08:15.760 --> 00:08:22.670
It will be the point of intersection between the line π¦ is equal to two and the curve π¦ is equal to two to the power of five minus π₯.
00:08:22.890 --> 00:08:26.290
The π¦-coordinate of this point is two and its π₯-coordinate is four.
00:08:26.670 --> 00:08:29.890
In other words, when π₯ is equal to four, our function outputs two.
00:08:29.960 --> 00:08:31.850
π evaluated at four is two.
00:08:32.210 --> 00:08:35.360
Therefore, π₯ is equal to four is a solution to our equation.
00:08:35.390 --> 00:08:41.620
In fact, since this is the only point of intersection between the line and the curve, this is the only solution to our equation.
00:08:41.990 --> 00:08:45.680
This means the solution set to our equation is just the set containing four.
00:08:46.030 --> 00:08:52.480
Itβs also worth noting we can check our answer by substituting π₯ is equal to four into our equation or into our function.
00:08:52.690 --> 00:08:59.450
Substituting π₯ is equal to four into our function π of π₯, we get π evaluated at four is two to the power of five minus four.
00:08:59.770 --> 00:09:01.520
Five minus four is equal to one.
00:09:01.520 --> 00:09:03.930
So, this simplifies to give us two to the first power.
00:09:03.990 --> 00:09:07.390
And any number raised to the first power is just equal to itself.
00:09:07.390 --> 00:09:16.410
So, π 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.
00:09:16.810 --> 00:09:24.080
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.
00:09:26.420 --> 00:09:32.980
Letβs now go for another example where weβre given the graph of an exponential function, and we need to use this to solve an exponential equation.
00:09:36.040 --> 00:09:40.810
The diagram shows the graph of π of π₯ is equal to two to the power of two π₯.
00:09:40.840 --> 00:09:45.990
Use this graph to find the solution set of the equation two to the power of two π₯ is equal to four.
00:09:46.450 --> 00:09:52.700
In this question, weβre given the graph of an exponential function, and weβre asked to solve an exponential equation where this function appears.
00:09:52.850 --> 00:09:58.250
To do this, we start by recalling the solution set of an equation is the set of all solutions to that equation.
00:09:58.640 --> 00:10:04.200
This means weβre looking for all of the values of π₯ which solve the equation two to the power of two π₯ is equal to four.
00:10:04.650 --> 00:10:09.940
To help us do this, letβs start by replacing two to the power of two π₯ in our equation with π of π₯.
00:10:10.190 --> 00:10:14.620
This means the equation weβre asked to solve can be rewritten as π of π₯ is equal to four.
00:10:14.970 --> 00:10:20.140
In other words, weβre just looking for the values of π₯ such that our function π outputs a value of four.
00:10:20.560 --> 00:10:27.220
And we can recall that the π¦-coordinate of a point on our curve tells us the output value of a function for that value of π₯.
00:10:27.250 --> 00:10:33.880
So, we can find all of the values where our function outputs four by sketching the line π¦ is equal to four onto our diagram.
00:10:34.140 --> 00:10:39.270
And we can then see on our diagram thereβs only one point on our curve with π¦-coordinate equal to four.
00:10:39.660 --> 00:10:41.850
This is the point of coordinates one, four.
00:10:42.230 --> 00:10:46.730
And itβs worth reiterating this tells us that π evaluated at one is equal to four.
00:10:47.120 --> 00:10:49.950
And therefore, one is a solution to our equation.
00:10:50.260 --> 00:10:59.360
In fact, all solutions to our equation will be a point of intersection between the line π¦ is equal to four and the curve π¦ is equal to two to the power of two π₯.
00:10:59.550 --> 00:11:04.240
So, because we can see thereβs only one point of intersection, we know thereβs only one solution.
00:11:04.320 --> 00:11:09.530
Therefore, the solution set of the equation two to the power of two π₯ is just the set containing one.
00:11:11.970 --> 00:11:16.720
Letβs now see an example where we first need to rearrange the exponential equation that weβre given.
00:11:18.980 --> 00:11:23.720
The diagram shows the graph of π of π₯ is equal to two to the power of π₯ over two.
00:11:23.970 --> 00:11:30.420
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.
00:11:30.850 --> 00:11:34.430
In this question, weβre given the graph of an exponential function π of π₯.
00:11:34.470 --> 00:11:39.750
And weβre asked to use this to determine the solution set of an equation which contains our function π of π₯.
00:11:39.980 --> 00:11:46.200
To do this, we start by recalling the solution set of an equation is the set of all solutions to that equation.
00:11:46.510 --> 00:11:52.850
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.
00:11:53.240 --> 00:11:58.770
To answer this question, it can help us to rewrite our exponential equation in terms of the function π of π₯.
00:11:58.940 --> 00:12:06.710
Substituting two to the power of π₯ over two is equal to π of π₯ into our equation, we get π of π₯ plus five is equal to nine.
00:12:07.090 --> 00:12:11.270
We can simplify this equation further by subtracting five from both sides.
00:12:11.510 --> 00:12:16.320
We get π of π₯ is equal to nine minus five, which simplifies to give us π of π₯ is equal to four.
00:12:16.680 --> 00:12:20.780
So, we want to find the values of π₯ such that our function outputs a value of four.
00:12:21.160 --> 00:12:27.960
Remember that the π¦-coordinate of any point on our curve tells us the output value of our function at that value of π₯.
00:12:28.100 --> 00:12:31.530
So, we want to find all of the points on our curve with π¦-coordinate four.
00:12:31.530 --> 00:12:35.200
We do this by sketching the line π¦ is equal to four onto our diagram.
00:12:35.550 --> 00:12:39.250
We can see thereβs only one point of intersection between our line and our curve.
00:12:39.280 --> 00:12:41.790
And we can see that this point has π₯-coordinate four.
00:12:42.190 --> 00:12:47.640
Therefore, when we input a value of π₯ is equal to four into our function, the output value is four.
00:12:47.670 --> 00:12:49.430
π of four is equal to four.
00:12:49.910 --> 00:12:56.560
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.
00:12:56.940 --> 00:13:00.330
Therefore, the solution set of this equation is the set containing four.
00:13:00.660 --> 00:13:07.800
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.
00:13:08.090 --> 00:13:17.260
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.
00:13:17.260 --> 00:13:19.300
So, this is equal to two squared plus five.
00:13:19.670 --> 00:13:20.950
And then we can evaluate this.
00:13:20.950 --> 00:13:22.330
Two squared is equal to four.
00:13:22.330 --> 00:13:28.030
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.
00:13:28.400 --> 00:13:32.030
Therefore, four is a solution to our equation, and we know itβs the only solution.
00:13:32.110 --> 00:13:38.520
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.
00:13:40.820 --> 00:13:44.760
Letβs now see an example where our exponential equation involves a linear function.
00:13:47.090 --> 00:13:49.710
Use the graphs below to answer the following question.
00:13:49.800 --> 00:13:55.060
True or false: the equation two to the power of π₯ is equal to negative π₯ has no solution.
00:13:55.430 --> 00:13:57.780
In this question, weβre given the graph of two functions.
00:13:57.780 --> 00:14:01.060
Letβs start by determining which two functions these are the graphs of.
00:14:01.460 --> 00:14:05.850
First, we can see that our straight line passes through the origin, so its π¦-intercept is zero.
00:14:06.160 --> 00:14:10.510
Next, we can see for every one unit we go across, we travel one unit down.
00:14:10.510 --> 00:14:12.390
So, its slope is negative one.
00:14:12.660 --> 00:14:19.470
In slopeβintercept form, thatβs the line π¦ is equal to negative one π₯ plus zero, which is just π¦ is equal to negative π₯.
00:14:19.810 --> 00:14:26.230
Our other curve has the shape of an exponential function, and we can see it passes through the point with coordinates one, two.
00:14:26.600 --> 00:14:32.700
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.
00:14:32.990 --> 00:14:39.780
We could do this with other points on our curve to conclude that this is indeed a sketch of the curve π¦ is equal to two to the power of π₯.
00:14:40.070 --> 00:14:46.290
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.
00:14:46.620 --> 00:14:49.290
We might be tempted to try and solve this by using manipulation.
00:14:49.290 --> 00:14:53.590
However, this will be very difficult because π₯ appears in the exponent and not in the exponent.
00:14:53.870 --> 00:14:59.510
Instead, recall that a solution to this equation is a value of π₯ such that both sides of the equation are equal.
00:14:59.920 --> 00:15:08.380
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.
00:15:08.660 --> 00:15:10.410
We can do this directly from our graph.
00:15:10.440 --> 00:15:16.120
For the outputs of these two functions to be equal with the same π₯-input, they must have a point of intersection.
00:15:16.440 --> 00:15:21.030
This is because the π¦-coordinate tells us the outputs of this function for the given input.
00:15:21.360 --> 00:15:28.930
Hence, because thereβs 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.
00:15:29.290 --> 00:15:34.170
In fact, we can even approximate this value by trying to read off its π₯-coordinate from the graph.
00:15:34.360 --> 00:15:38.340
Doing this, we would get that π₯ is approximately equal to negative 0.6.
00:15:38.370 --> 00:15:45.040
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.
00:15:47.900 --> 00:15:54.530
In our final example, weβll solve an exponential equation graphically by also sketching a linear function on the same given graph.
00:15:57.150 --> 00:16:02.270
The following graph shows the function π sub one of π₯ is equal to two to the power of negative π₯.
00:16:02.330 --> 00:16:11.520
Use this graph and plot the function π 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.
00:16:11.940 --> 00:16:19.450
In this question, weβre given two functions π sub one of π₯ and π sub two of π₯, and weβre given a graph of the function π¦ is equal to π sub one of π₯.
00:16:19.600 --> 00:16:21.740
Weβre asked to find the solution set of an equation.
00:16:21.770 --> 00:16:31.660
And since π sub one of π₯ is equal to the left-hand side of this equation and π sub two of π₯ is equal to the right-hand side of this equation, the equation is π sub one of π₯ equals π sub two of π₯.
00:16:32.010 --> 00:16:33.930
We can solve this equation graphically.
00:16:33.990 --> 00:16:41.930
Any solution to this equation will be a point of intersection between the curve π¦ is equal to π sub one of π₯ and the line π¦ is equal to π sub two of to π₯.
00:16:42.280 --> 00:16:53.500
Because the point of intersection would have the same π¦-coordinate and the π¦- 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.
00:16:53.850 --> 00:16:56.360
We need to sketch the curve π¦ is equal to π₯ plus three.
00:16:56.400 --> 00:16:59.120
First, we note that its π¦-intercept will be at three.
00:16:59.470 --> 00:17:03.050
We can also find its π₯-intercept by substituting π¦ is equal to zero.
00:17:03.080 --> 00:17:05.670
Solving this, we get that π₯ is equal to negative three.
00:17:05.920 --> 00:17:07.190
We can then plot our line.
00:17:07.220 --> 00:17:10.610
Its π¦-intercept is at three, and its π₯-intercept is at negative three.
00:17:11.050 --> 00:17:12.680
This then allows us to plot our line.
00:17:12.710 --> 00:17:15.440
We just connect the π¦- and π₯-intercept with a straight line.
00:17:15.920 --> 00:17:20.970
Then, the only point of intersection between our line and our curve will be the only solution to our equation.
00:17:21.000 --> 00:17:24.120
We can read off its π₯-coordinate; its π₯-coordinate is negative one.
00:17:24.410 --> 00:17:29.360
Then, since the question ask us to write this as a solution set, weβll write this as the set containing negative one.
00:17:29.470 --> 00:17:37.230
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.
00:17:39.890 --> 00:17:42.510
Letβs now go over some of the key points of this video.
00:17:42.870 --> 00:17:51.440
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.
00:17:51.830 --> 00:17:56.810
And in particular, if an equation has no solutions, we can say the solution set is the empty set.
00:17:57.300 --> 00:18:08.810
Next, we saw that we can solve the equation π of π₯ is equal to π of π₯ by finding the π₯-coordinates of all of the points of intersection between the graphs π¦ is equal to π of π₯ and π¦ is equal to π of π₯.
00:18:09.260 --> 00:18:12.530
Every point of intersection is a solution to the equation.
00:18:12.640 --> 00:18:16.950
And if there are no points of intersection, then there are no solutions to the equation.
00:18:17.290 --> 00:18:21.340
Finally, we saw the graphical solutions to equations can be approximations.
00:18:21.370 --> 00:18:25.490
This is particularly true if we need to sketch one of the functions ourselves.
00:18:25.890 --> 00:18:31.490
In these cases, we should use the grid lines to try and make our approximation as accurate as possible.