 In this video we present the solution to question number 13 from practice exam number one for math 1210 In which case we're asked to find all real solutions to the equation the natural log of x plus 1 minus the natural log of x equals 2 So when I see these two natural log on the left hand side, they both involve x's right in order to solve for x I'm gonna have to combine the x's together And so we can do that by condensing the logarithm using the second law of logarithms in fact So remember the second law of logarithms tells us that if you have a log base a of any number capital a and you subtract from it Log base a of some other value b this can combine together to become the log base a of a divided by b So subtracting logarithms brings them together as Division on the inside All right, it doesn't matter what the base is as long as the base is the same since they're both common law natural Excuse me. This will come together become the natural log of x plus 1 over x and this is equal to 2 The next thing after you condense the logs together, I would switch this from logarithmic form to exponential form Basically, we want to move the natural log from the left hand side to the right hand side, but we're not moving the natural log We're actually moving the base e so when we move the natural log the base e from the left hand side the right hand side Switches from a logarithm to an exponential so the corresponding Exponential form of this equation would be x plus 1 over x is equal to e squared or another way of thinking about it Is you're just going to exponentiate both sides of the equation e carrot here the right hand side becomes e squared e When you exponentiate it by a natural log they'll cancel out because they're inverse operations now e squared is just a number I don't want you to freak out too much about it It would be really no different if it was like the number 5 how we proceed forward from here We need to solve for x so I'm actually going to clear the denominators times both sides of the equation by an x In which case that gives me x plus 1 is equal to e squared x for which then we need to combine like terms I'm going to subtract x from both sides This then gives us 1 is equal to e squared x minus x which again the e squared is just a number right? We would just subtract the exponents, but if you don't like that We can think of it in terms of factoring right both these are divisible by x if you factor out the x you end up with e Squared minus 1 times x so that's what we have here. We have this e squared minus 1 Times x that's equal to 1 we need to divide by the coefficient which the coefficient is e squared minus 1 So those will cancel out and we get as our solution x equals 1 over e squared minus 1 Now one has to be cautious that whenever you solve a logarithmic equation We have to make sure that it actually fits inside the domain of the original functions, right? We have these natural locks here natural locks cannot accept zero if they cannot accept Negatives so when you look at the natural log of x right here You can't plug it a zero you can't plug it a negative nothing there is allowed when it comes to the natural log of x plus 1 This one actually allows you to plug anything up until negative one So the natural log of x has the more restrictive Number here are the more restrictive domain. We need to make sure that this is a positive number, right? Well 1 over e squared minus 1 that would be positive so long as e squared minus 1 is positive Which would imply e squared is greater than 1 which would imply e is greater than 1 The square root of 1 which is 1 which is true So this is in fact a positive number you could also check this with a scientific calculator very easily This does fit inside the domain of the problem So it is in fact the solution to this equation