Get ready for AP® Calculus
You probably already solved a system of equations by graphing the equations and looking for intersection points. This method can actually be used to solve (or find an approximate solution to) any single equation, no matter what kind! This is a very exciting tool.
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- Are there tools in the future courses that will allow me to solve a problem like this?
That is without doing it using brute force and let the computer graph the functions for me.(6 votes)
- The tools needed to solve this equation algebraically aren't on Khan Academy. This equation can only be solved in terms of something called the Lambert W function, which is only presented in some college math courses.
However, Khan Academy will provide better tools to approximate the solution by hand in the calculus section.(10 votes)
- 1:12How would you solve that logarithm with algebra?(3 votes)
- You use exponentiation and rewrite it as an exponent.
For example, log base 2 (8)=x can be rewritten as 2^x=8 and we can solve x to get 3.(0 votes)
- [Instructor] Let's say you wanted to solve this equation, two to the x squared minus three power is equal to one over the cube root of x. Pause the video and see if you can solve this. Well you probably realize that this is not so easy to solve. The way that I would at least attempt to tackle it is, you would say this is two to the x squared minus three is equal to x to the, I could rewrite this, this is one over x to the 1/3, so this is x to the negative 1/3 power. Maybe I can simplify it by raising both sides to the negative three power. And so then I would get, if I raise something to an exponent, then raise that to an exponent, I could just multiply the exponents. So it would be two to the negative three x squared plus nine power. I just multiplied both of these terms times negative three, is equal to x to the negative 1/3 to the negative three. Negative 1/3 times negative three is just 1, so that's just going to be equal to x. So it looks a little bit simpler, but still not so easy. I could try to take log base two of both sides and I'd get negative three x squared plus nine is equal to log base two of x. But once again not havin' an easy time solve this. And the reason why I gave you this equation is to appreciate that some equations are not so easy to solve algebraically. But we have other tools, we have things like computers, we can graph things and they can at least get us really close to knowing what the solution is. And the way that we can do that is we can say hey, well what if I had one function or one equation, that was y is equal to two x, two to the x squared minus three I should say. And you had another that was y is equal to one over the cube root of x. And then you could graph each of these and then you could see where they intersect. Because where they intersect that means two to the x squared minus three is giving you the same y as one over the cube root of x. Or another way to think about it is they're going to intersect at an x value, where these two expressions are equal to each other. And so what we could do is we could go to a graphing calculator, or we could go to a site like Desmos and graph it and at least try to approximate what the point of intersection is, and so let's do that. So I graphed this ahead of time on Desmos. So you can see here, this is our two sides of our equation but now we've expressed each of them as a function. Right here in blue we have two, we have f of x, or I could even say this is y is equal to f of x which is equal to two to the x squared minus three. And then in this yellowish color I have y is equal to g of x which is equal to one over the cube root of x, and we can see where they intersect. They intersect right over there. And we're not going to get an exact answer, but even at this level of zoom and on a tool like Desmos you can keep zooming in in order to get to get a more and more precise answer. In fact you can even scroll over this and it'll even tell you where they intersect. But even if we're trying to approximate, just looking at the graph. We can see that the x value, right over here it looks like it is happening at around, let's see this is 1.5, and each of these is a tenth. So this is 1.6 and then it looks like it's about two thirds of the way to the next one. So this looks like it's about, I'll say this is approximately 1.66. And if you were to actually find the exact solution, you would actually find this awfully close to 1.66. So the whole point here is, is that, even when it's algebraically difficult to solve something you could set up or restate your problem or reframe your problem in a way that makes it easier to solve. You can set this up as, hey let's make two functions, and then let's graph them and see where they intersect And the x value where they intersect well that would be a solution to that equation, and that's exactly what we did right there. We're saying that hey, the x value, the x solution here is roughly 1.66.