If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Secant line with arbitrary point

## Video transcript

a curve has the equation y equals the natural log of X and passes through the points P equals e comma 1 and Q is equal to X natural log of X write an expression in X that gives the slope of the secant P and Q so I think I'm going to need my little scratch pad for this one right over here so this is the same question over again and let's just try to visualize this curve right over here so let me draw my let me draw my axes so let's say this is my this is my y axis this is my y axis and then this is my x axis this is my x axis right over here that's my x axis and natural log of X so let's think about a little bit the natural log of 0 e to what power is equal to 0 well that's going to be 1 so you're going to you're going to have the point 1 0 on this so that's 1 right over there and then the natural log of smaller and smaller numbers as we approach 0 is going to get more and more negative all the way going down to negative infinity so this curve is going to look it's going to look something like this just as we're it's going to look something it's going to look something like something like like that something like that and we know it also has the point e comma 1 on it so it has the point the natural log of e so if that's 1 see that will be about 2 that'll be about 3 so that is e right over here roughly e right over here and that's the point a comma 1 so that's the point a comma 1 right over there and I'll just mark that as P they already gave us a label for it so that is P and we want to find the secant line between P and an arbitrary Q for any given X the the Y value is going to be natural log of X so this could be so let's say that this is our Q right over here so let's say that this right over here is our actually I'll just make it easier I'll make the Q actually want to make it clear that it is a I'll do it out here I'm being a little indecisive so let's say that is our cue right over there and that's the point X natural log natural log of X and we want to find the slope of the secant line joining these two points the slope of the secant line joining these two points so this the slope of this line I want to try to make it so it doesn't look tangent so it's secant so you see it cuts through the curve so that's the secant line right over there to intersects the curve at P and at Q so I want to find the slope of that secant line well to find the slope of the secant line I just need to find the change in Y and the change in X between these two points so what's the change in so let's be clear here this is the point this is when X is equal to well it's just a kind of arbitrary X and this right over here is the point Ln X so what is our change in X our change in X this value right over here our change in X is just going to be X minus E is equal to X minus E and what is our change in Y our change in Y is going to be Ln X is going to be the natural log of X minus 1 minus 1 that's this distance right over here so the slope of this line the line that contains both of these points the slope and I could write M for slope is going to be our change in Y over our change in X which is equal to Ln X minus 1 over X minus E and now we can just input that and make sure that we got that right so let me try to remember it Ln X minus 1 over X minus e so let me input that so we have the natural log of X minus 1 over and it interprets it interprets it nicely for us right below so we know make sure that we know what we're doing over x minus E and now let us let us check our answer and we got it right