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Solving equations by graphing: graphing calculator

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- [Instructor] We are told we want to solve the following equation, the negative natural log of two x is equal to two times the absolute value of x minus four, all of that minus seven. One of the solutions is x is equal to 0.5. Find the other solution. They say hint, use a graphing calculator and round your answer to the nearest tenth. So pause this video and have a go at this if you like, and then we'll work on this together. And I encourage you to have a go on it, (laughing) a go at it. All right now let's work on this together. Now the key here is to realize that we might be able to solve this by graphing, or at least approximate the solutions to this by graphing. And the way we do that is, if we have an equation, especially a hairy equation like this, in one variable, we can set y equal to the left and then set y equal to the right and then graph each of those functions and then think about where they intersect. Because they'll intersect at an x-value that gives us the same y-value, and that means that the two sides are the same. So what do I mean by that? Well we could set y is equal to the negative natural log of two x. So we could have one equation or one function like that. And then we could have another equation or function that y is equal to two times the absolute value of x minus four minus seven. And let's see where they intersect. And the x-values where they intersect, those are going to be solutions to that. So I'm going to use Desmos as my graphing calculator. So let's type in the two sides. So first I'll do the left side. So if y is equal to the negative natural log of two x. And actually, let me make my color to be the same or as close as I can, so maybe closer to that bluer color. Okay and then the net one, I want y is equal to two times, two times the absolute value, actually I don't know whether Desmos prefers, I'll use that, actually that works, okay, x minus four, and then I will close my absolute value, and then I have minus seven. And I will do this in the red color so that we can keep track of things. Okay. So those are my two graphs. And now, I just need to think about where they intersect. And one of the solutions is x equal to 0.5. That's not the one they want. They want the other solution so to speak. So let's see. So let's see, we have one solution, actually let me zoom in a little bit. So when x is equal to 0.5, that's this solution, that's this solution right over here. It looks like y is equal to zero there. But then the other point of intersection seems to be right over here, and actually Desmos has a nice little feature where it'll tell us that point right over there. But you could even approximate it. You can see that x is over six and that each of these, let's see one, two, three, four, five, each of those is .2. So it's going to be 6.2 something is what I would do, and they want us to round to the nearest tenth anyway. So you don't even need to use that feature, but you can see very clearly that when x is equal to approximately 6.238, that we get a y is equal to negative 2.54. Or another way to think about it is, when x is approximately equal to, when x is approximately equal to 6.2, that the two sides of this equation are going to be equal to each other, are approximately equal to each other. And we're done, we've just solved using a, or at least approximated a solution using graphing in a graphing calculator.