 In this video, we provide the solution to question number 13 for the practice version of exam 4 for math 1050. In which case we're given a logarithmic equation, the natural log of x plus 1 minus the natural log of x is equal to 2. And we have to find all real solutions to this equation. Alright, you'll notice that there are two x's in this problem. There's the natural log of x plus 1 and the natural log of x. So our goal is we have to bring the x's together. And since we have a log minus a log, we can use the second law of logarithms. To bring this difference together, in which case this will turn into the natural log of x plus 1 over x. And this is then equal to 2. Like so. So remember, difference of logarithms becomes a logarithm of a quotient. Next, we want to move the log base e, aka the natural log to the other side. So a log base e on the left becomes exponential base e on the right. So we end up with x plus 1 over x is equal to e squared. Now e squared is just a number. Don't worry about approximating it because we just want to have an exact answer anyway. Just leave it as e squared for right now. Now we have to still solve for x. There's an x in the numerator, x in the denominator. So I'm going to clear the denominators times both sides by x. The x is on the left cancel leaving an x plus 1. And on the right hand side, we get e squared x. I know this can feel a little bit confusing because we have two letters, which mean two very different things. x is our unknown variable. e is a constant number. It's just a rational number like pi. So we're using the symbol there. So e squared is just a number even though we haven't computed what it is. Not a big deal. That's just the coefficient. If I subtract x from both sides, the left hand side becomes 1. The right hand side becomes e squared x minus x. We want to add together the coefficients or if you prefer factor out the x. In which case you get e squared minus 1x. That's now the coefficient of x. Divide both sides by e squared minus 1, e squared minus 1. And then we're going to get the answer x is equal to 1 over e squared minus 1. Do not approximate this. Leave this as an exact value. I should mention that this value is in fact going to be positive. How do I know that? Well, the numerator is 1. That's positive. The denominator for actions positive if the numerator and denominator have the same sign. So since the numerator is positive, what about the denominator? It's bigger than 2, right? It's 2.7 something, right? It's between 2 and 3 there. So e squared is bigger than 4. 4 minus 1 is 3. So e squared minus 1 is still positive. So we have a positive value. That's a good thing to know because one should check their solution. You should always check it with a logarithmic equation. Notice if I take a positive number and I stick it into the natural log of x there. We don't want to take the natural log of a negative number. That gives us something to imagine. Remember, we're only looking for real solutions here. So if we take the natural log of a negative, that would be problematic here. But as 1 over e squared minus 1 is positive, the natural log of that value is perfectly fine. And then at the other one, if you plug it in, you take a positive plus 1 that's still positive. We're within the domain of the problem. So we then can be rest assured that the correct answer is 1 over e squared minus 1. We don't bother approximating this. The exact value is expected.