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

Current time:0:00Total duration:4:19

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 joining 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. Now let's just try to visualize
this curve right over here. So let me draw my axes. So let's say 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 it 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 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 something like this. It's going to look
something like that. And we know it also has
the point e comma 1 on it. So it has a point,
the natural log of e. So if that's 1,
that'll 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 e comma 1. So that's the point e
comma 1 right over there. And I'll just mark
that as P there to give 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 y value is going
to be natural log of x. So let's say that this
is our Q right over here. So let's say this right
over here is our-- actually, I want to make it clear that
it is a, I'll do it our here. I'm being a little indecisive. So let's say that is
our Q 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. 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. It 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 lnx. 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 lnx 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-- I could write m for slope-- is
going to be our change in y over our change in x,
which is equal to lnx 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--
lnx 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
nicely for us right below, so we make sure that we're
doing over x minus e. And now let us check our answer. And we got it right.