Solving equations by graphing: graphing calculator | Algebra 2 | Khan Academy

We are told we want to solve the following equation: that the negative natural log of 2x is equal to 2 times the absolute value of x minus 4, all of that minus 7. 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. I encourage you to have a go at it. 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. 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. 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 equal to the negative natural log of 2x, so we could have one equation or one function like that. Then, we could have another equation or a function that y is equal to 2 times the absolute value of x minus 4 minus 7. And let's see where they intersect. The x values where they intersect are going to be solutions to that.

I'm going to use Desmos as my graphing calculator, so let's type in the two sides. First, I'll do the left side. So if y is equal to the negative natural log of 2x, 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 next one I want is y equal to 2 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. 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. 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, 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.

Actually, Desmos has a nice little feature; it'll tell us that point right over there. But you could even approximate it. You can see that x is over 6, 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. 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, we get y is equal to negative 2.54. Another way to think about it is when x is approximately equal to 6.2, that the two sides of this equation are going to be approximately equal to each other, and we're done. We’ve just solved using, or at least approximated, a solution using graphing in a graphing calculator.