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Oregon State University

Harvey Mudd College

Baylor University

00:33

Dungarsinh Puthvisinh

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

00:06

Jeffery Wang

0:00

01:09

Felicia Sanders

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So we want to find this on the solution to this equation. So we want to solve for X. Now we have that. The log, a rhythm of something involving X, is equal to three and two software X. We need toe isolate X. So recalled it. The natural law algorithm is log based. Eat. So it seems like we should take e to both sides. So this equation should be equivalent to e to the log of two X minus one equals e to the third. Okay, but e raised to the natural log, which is long base e. Remember, that natural log is really log based e cancels. And so we have two X minus one is equal to eat to the third. But now I could just use addition division, so I can add one. So two X is equal to e to the third plus one and then divide by two e to the third plus one, all divided by two, and this will be our final answer. So again, we're just using inverse relationship between logarithms and exponential is okay, so we have another equation and we want to software X. But this is actually going to illustrate something very interesting that can happen with equation solving. And we may see this from time to time. And it's something that we need to be very careful of when we're solving equations. But for now, let's just go in blind. Let's just say Okay, well, I want to solve this for X. So let me isolate X and I'm gonna be really clever, and I'm going to use properties of logarithms. I'm going to use the fact that if I'm adding to longer those with same base, I could just write them as a single longer them with arguments multiplied together. So I'm just writing this very clearly to say Okay, the log. I'm adding these two together, these air, all log based ease, I could just multiply the arguments. Well, that's great, because now I can just do my favorite trick and raised I could so take both sides and put him in the argument of e so e to this is equal to eat to this. That should be equivalent and then cancel the logs. I mean, if you if you think about it, you have long of something equals log of something else we're actually just using the fact that log is 1 to 1. So if the logs of something are the same than the actually what you're taking, the log of has to be the same. So in other words, X plus six times X minus three is equal to 10. But this is just a quadratic equation. And now be careful. Don't use the zero product property here. Don't say either X plus six is 10 or X minus three is 10 because it only works if a product of two things equal zero. So I actually need to expand this out. So let's do that. So this is X squared. We have minus three x plus six x that's plus three x minus 18. He goes 10. I can subtract 10 over whenever you're solving a quadratic equation are really in the equation. It's always a good idea to get everything on one side, because now, when I factor notice that this does indeed factor, so we'll have X plus seven X minus four. I have a product of things equal to zero, and now I can use the zero product property so either X plus seven zero or X minus 40 And so we have two solutions. X equals negative seven and X equals four. But hold on, hold the phone. We're missing something very important. Now pause for a second and see if you could tell me what it iss. Of course you can't actually tell me, but just think about it. So one of these solutions is actually invalid. So if I plug in X is equal to four, I have log of 10 plus log of one. That's gonna be a log of 10 plus zero. Log of +10 equals Log of five, plus log of two, which is long. 10. So X equals four works, but try to plug in negative seven. Log of negative seven plus six is log of negative one. But the domain of the longer than function, remember, is on Lee positive numbers. I can Onley plug in positive numbers into a longer than so something is going on here. It doesn't even make sense to plug in negative seven into this equation. And the moral of the story is this. Always check. Check your answers. So when you find answers, I mean, I'm guilty of not doing this all the time, but it's always a good idea to go and check back and make sure that your solutions air actually solutions. Because in this case, this guy is not a solution. We Onley have one solution X is equal to four. So this guy right here sexually has a special name. This is called an extraneous solution. And an extraneous solution basically comes from doing something to the problem that actually wasn't 100% of valid. So actually, when we applied these properties of the logarithms, there's actually something we needed to make sure off, and I'm not going to go into a lot of detail. But essentially, I'm doing this step going from here to here. We actually introduced this extraneous solution and that's OK. As long as we check our answer and realized that Okay, X equals negative. Seven isn't actually a solution. Onley X is equal to four is a solution. So it's just again all of this stuff from pre calculus is just something to keep in the back of your mind because we're going to rapid fire. When we get to calculus, we're just gonna be doing algebra left and right, and all of these things are going to come to play from time to time

Limits

Derivatives

Differentiation

Applications of the Derivative

Integrals