ANNOUNCER: The following content is provided under a
Creative Commons license.
Your support will help MIT OpenCourseWare continue to
offer high quality educational resources for free.
To make a donation, or to view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu.
PROFESSOR: Hi.
Our lesson today is called the 'Inverse Logarithm'.
And what it will do, among other things, is give us a
very nice chance to revisit inverse functions in general,
only now applied specifically to the natural logarithm
function that we talked about last time.
So I call today's lesson, as I say, 'Inverse Logarithms'.
And to see what's coming up here, simply recall that last
time, we invented, so to speak, a new function called
the 'natural log of x'.
This had nothing to do with exponents.
It was what?
This was the function whose derivative with respect to 'x'
was '1/x' and passed through the point (1 , 0).
That uniquely defined this particular function.
The point being what?
That this particular function, since the domain is for
positive 'x', '1/x' is positive.
The curve is always rising, which means that the function
itself must be one to one.
And because the function is one to one, it means that the
inverse function exists.
This is no different from any other example of forming f
inverse, given a one to one function called 'f'.
So what we do is now, and remember how
you invert this function.
If you want to do this thing in slow motion, you first
rotate this through 90 degrees, then flip it over.
If you want to do it faster, the inverse graph of this is
the reflection of this curve with respect to the line 'y'
equals 'x'.
In any event, if we do this, we find that the graph of the
function 'y' equals 'inverse log x' is
this particular curve.
And notice the correspondence.
If this is the point (1 , 0) on the log curve, the inverse
point is (0 , 1).
In other words, the inverse log of 0 is 1.
OK.
Now, the idea is something like this.
Just to have a review, again, of how--
see, the punchline I want to make for today's lesson, right
at the beginning, is that the nice thing about studying
inverse functions is that once you know the function of which
you're taking the inverse, you automatically get all the
mileage you want out of the inverse function.
Well, by way of illustration, if the function natural log
had the usual logarithmic property, we would suspect
that its inverse should have the
usual exponential property.
Now what is the usual exponential property?
The exponential function is characterized by what?
That if you take 'f of 'x1 plus x2'', that's 'f of x1'
times 'f of x2'.
So you think of it in terms of the
usual exponential notation.
If you have, say, 10 raised to the 'x1 plus x2' power, that's
the same as 10 to the 'x1' power times
10 to the 'x2' power.
In other words, if our ln function is genuinely a
logarithmic function, we would expect the inverse of that
function to be a genuine exponential function.
And just to work with inverses again, let's see if this is
indeed true.
Let's see if it's really true that the 'inverse log of 'x1
plus x2'' is the 'inverse log of x1' times the
'inverse log of x2'.
You see, the whole idea is what?
Let's give these things names.
Let 'y1' be the 'inverse log of x1' and 'y2' be the
'inverse log of x2'.
Now, since we've already studied the natural log
function, the most natural thing to do here is to invert
these, namely 'y1' equals the inverse 'natural log of x1' is
the same as saying 'x1' is the 'natural log of y1'.
And similarly, this is the same as saying that 'x2' is
the 'natural log of y2'.
And now we add equals to equals, and we get that 'x1
plus x2' is 'natural log y1' plus 'natural log y2'.
But allegedly, we understand the properties of the natural
log, because if we didn't understand those, it would be
kind of futile to be studying the inverse function.
What do we know about the natural log?
We know that the 'natural log of 'y1 times y2', by
definition of a logarithmic function, is 'log
y1' plus 'log y2'.
In other words, 'natural log y1' plus 'natural log y2', by
the property of being logarithmic, is just the
'natural log of 'y1 times y2''.
And now, taking this relationship and inverting it
to say this is the same as saying what?
The 'inverse natural log of 'x1 plus x2'' is equal to 'y1'
times 'y2'.
And if we now observe that 'y1' was the 'inverse natural
log of x1', and 'y2' is the 'inverse natural log of x2',
we have the result that we claimed we would have at the
beginning here.
Now again, here's one example where it's not the result that
I'm not interested in.
But I'm interested in showing, again, what we meant when we
said that once we learn a new concept, we can bring to it
all of our old knowledge.
You see, all of these things come from our basic
definition.
Well, let's carry on one step further.
You see, this was an arithmetic property.
Let's see if we can get some calculus properties about the
inverse natural log based on our knowledge of the natural
log itself.
Well, for example, in a course of this type, the most natural
thing to do, I guess, is given any function, we would always
like to be able to talk about its derivative, provided, of
course, the function is differentiable.
So for example, a typical question that one might ask is
find 'dy dx', if 'y' is 'inverse natural log x'.
Again, this works the same way as it did for the inverse trig
functions, for the inverse of everything that we've done so
far in this course.
We start with this particular relationship and right away
translate it into that which we're more familiar with,
namely we translate 'y' equals the 'inverse natural log of x'
into 'x' equals 'natural log y'.
Now you see again, since I know how to differentiate 'log
y' with respect to 'y', the derivative of 'natural log y'
with respect to 'y' was by definition '1/y'.
From this relationship here, I can find
that 'dx dy' is '1/y'.
And knowing that 'dx dy' is '1/y', using my inverse
function general theory that tells me that to find the
derivative of 'y' with respect to 'x', all I have to do is
invert the derivative of 'x' with respect to 'y', I wind up
with the fact that 'dy dx' is equal to 'y' itself.
In other words, the 'inverse log x' function is
characterized by the fact that it's its own derivative.
In other words, it's a non-destructible type of
function with respect to differentiation.
This is a rather powerful property.
You see what this thing says?
It says that the derivative of 'y' with respect
to 'x' is 'y' itself.
By the way, just as an aside, we can do the converse of this
particular problem.
You see, we started with 'y' equals 'inverse log x' and
showed that this particular equation was satisfied.
Notice that if we start with this particular equation,
starting with 'dy dx' equals 'y', separate the variables,
and lo and behold, we wind up with 'dy over y' equals 'dx'.
And without carrying out the details, I leave this to you,
because it is straightforward.
Notice that as we look at the left hand side here, we
hopefully will think immediately
of the natural logarithm.
Namely, what do we want here?
The function whose derivative with respect to 'y' is '1/y'.
Again, you see what I'm saying is notice that if we had
started with this particular differential equation, we
could have shown that a logarithm, and hence solving
for 'y' explicitly, would have led to the
inverse logarithm also.
But at any rate, notice then that we can find the
derivative of the inverse log just by knowing how to find
the derivative of the natural log itself, which is as we
expect things should be.
Again, and this is a notational thing, we do not
use the language in general "inverse log x."
In other words, it's quite conventional to talk about
'inverse sine x'.
You may notice that sometimes in the text, sometimes in my
lectures I may have used the word 'arcsin x'.
But the general notation is sine to the minus 1.
However, when it comes to the natural logarithm function,
the inverse is usually not written as
'inverse natural log x'.
It's usually abbreviated, and let me say this, by the symbol
'e to the x'.
Now the reason I say by the symbol is this.
If I wanted to, I can say, look, If I've never heard of
exponents before, let this symbol be an abbreviation for
'inverse log x'.
Does 'inverse log x' exists?
Yes, I even drew the graph of it at the beginning of this
particular lecture.
However, again, if you are tempted to use exponents, the
notation 'e to the x' is in keeping in line with the idea
of our previous lecture.
I say what?
This matches the identification of 'natural log
x' with the 'log of x' to the base 'e'.
Remember, in our previous lecture, we mentioned that
this particular function exists without having to talk
about a base.
But if you wanted to identify it with a traditional
logarithm, the base you would have to pick is the base 'e',
where we showed that 'e' was some number between 2 and 4.
In other words, again, it's just like our 'dy dx' versus
'dy' divided by 'dx' and other symbols that we
invented this way.
Namely, if I look at 'e to the x' as being the inverse of the
natural log, or whether I look at it as being 'e' raised to
the 'x' power, where 'e' is that number that's someplace
between 2 and 4, I don't get into any trouble either way
because of the natural identification of choosing e
to be the base of my system.
But I just mention that in passing.
Now you see, the rest of today's lecture
will go fairly quickly.
And the reason that it will go fairly quickly is that once
we've established what the function is that we're talking
about, every other property of that function is going to
follow from the principles that we've already learned.
For example, in terms of our new notation, since the
derivative of 'e to the x' with respect to 'x' is 'e to
the x' itself, remember, 'e to the x' now can
be viewed as what?
'e to the x' power, or, to be on safer grounds, if you want
to be consistent, it's just an abbreviation for 'inverse
natural log x'.
But at any rate, if 'u' is now any differentiable function of
'x', notice that to find the derivative of 'e to the u'
with respect to 'x', by the chain rule, I can say what?
It's the derivative of 'e to the u' with respect to 'u'
times 'du dx'.
We've just shown that the derivative of 'e to the u'
with respect to 'u' is 'e' to the 'u' again.
And therefore, the derivative of 'e to the u' with respect
to 'x' is 'e to the u' times 'du dx'.
By the way, this again leads to another interesting thing
that makes integrals tough to handle, in a way.
We've mentioned this before, but here's another nice,
natural environment to bring this problem up in again.
When we discussed the second fundamental theorem of
integral calculus to show how one actually would have to
understand areas to be able to find a function whose
derivative was 'e to the 'minus x squared', we
mentioned, we didn't prove it, we mentioned that there was no
familiar function whose derivative with respect to 'x'
was 'e to the 'minus x squared'.
So this would be a very difficult problem to handle.
Now, let's look at this one instead.
Let's look at integral '2x 'e to the minus x squared' dx'.
To the untrained eye, it would appear that this is even
messier than this.
In other words, this is just 'e to the 'minus x squared''.
This one has 'e to the 'minus x squared'' with a '2x'
dangling in front of it, and that looks even tougher.
But notice that if we look at our exponent, 'minus x
squared', the derivative, or the differential of 'minus x
squared' is what?
It's 'minus '2x dx''.
In other words, the '2x dx' is precisely what you need to
reduce this to the form integral ''e to the u' du'.
In other words, working this thing out in more specific
detail, if I let 'u' equal 'minus x squared', 'du'
becomes 'minus '2x dx'', and therefore integral '2x 'e to
the 'minus x squared'' dx' just becomes what?
Well, the 'e to the 'minus x squared'' just becomes 'e to
the minus u'.
And '2x dx' is just 'minus du'.
OK?
I'm sorry, 'u' is 'minus x squared'.
'u' is 'minus x squared'.
So 'minus x squared' is 'u'.
So this becomes 'minus e to the 'u du''.
And that, of course, is just minus.
See, the minus will come outside.
The integral of 'e to the u' with respect to 'u' is
just 'e to the u'.
And since 'u' is equal to 'minus x squared', all we're
saying is that if you differentiate minus 'e' to the
'minus x squared', you wind up with what?
'2x 'e to the 'minus x squared'', the reason being
that you must multiply this by the derivative of the exponent
with respect to 'x'.
The root of the exponent is 'minus 2x'.
'Minus 2x' times minus 1 is '2x'.
If you want to see this in more concise differential
notation, you see, what we're saying is that the integral of
'e' to some power with respect to that same power is 'e' to
that power plus a constant.
In other words, the function that one would have to
integrate 'e to the 'minus x squared'' with respect to to
wind up with 'e to the 'minus x squared'' would be 'minus x
squared' itself.
And you see, now, in the next step, since this is a true
statement, the differential of 'minus x squared'
is minus '2x dx'.
And now you see, factoring out the minus sign and multiplying
through by minus 1, we arrive at the same result that we
wound up with before.
But again, this is the kind of material that we can drill on
extensively in the exercises in this particular unit.
The technique is what?
That the basic building block of the so-called exponential
function, the inverse natural log, is that when you
differentiate it, you essentially do not destroy it.
In other words, the derivative of 'e to the u' with respect
to 'x' is 'e to the u' times 'du dx'.
And I thought that in closing today's lesson, I might as
well show you a very powerful application of this particular
result in a non-obvious situation.
And it's something that we call second-order
differential equations.
I've picked out here a differential equation.
Let me just show you what I have in mind over here.
Suppose I tell you that 'y' is a twice-differentiable
function of 'x' that satisfies the following identity, that
the second derivative of 'y' with respect to 'x' minus 5
times the first derivative of 'y' with respect to 'x' plus 6
times 'y' is identically 0.
And the question is, if this equation is to be obeyed, what
must 'y' be?
And a very interesting technique for solving this
kind of a problem, you see, this is called a second-order
differential equation because the highest derivative that
appears is the second derivative.
OK, a rather powerful technique using the
exponential is available to us for problems of this sort.
And the idea hinges on this.
As a trial solution, which I'll call 'y sub t', let's try
'e to the rx', where 'r' happens to be a constant.
You see, the whole idea is if I differentiate 'e to the rx'
with respect to 'x', I get 'e to the rx' back again, only
with a factor of 'r', namely the derivative of my exponent
in this case.
See, 'r' is a constant.
The derivative of 'rx' with respect to 'x' is just 'r'.
So notice that the first derivative of 'y sub t' with
respect to 'x' is 'r 'e to the rx''.
The second derivative of 'y sub t' with
respect to 'x' is what?
'r' is a constant.
I differentiate 'e to the rx'.
That brings down another factor of 'r' and leaves me
with 'e to the rx'.
You see, the whole idea being, notice that 'e to the rx' is a
common factor of 'y sub t', 'y sub 't prime'', 'y sub 't
double prime''.
If I now substitute these results back into my original
equation, look what I wind up with. 'y double prime' becomes
'r squared 'e to the rx''.
Minus '5y prime', that's minus '5r 'e to the rx'', plus '6y',
'6e to the rx'.
And that must equal zero.
And here's the key point.
'e to the rx' is now a common factor.
I factor that out.
If the product of two numbers is 0, one of the
factors must be 0.
But notice from our graph of the exponential, the inverse
logarithm, 'e to the x', 'e to the rx' can never be negative
and can never be 0, in fact. 'e to the
rx' is always positive.
Therefore, if 'e to the rx' is always positive, it must be
the other factor which is 0.
But 'r squared' minus '5r' plus 6 is not a second-order
differential equation.
It's a second-degree polynomial equation.
It's a familiar quadratic equation, which I can solve
quite easily.
In other words, I find what?
That 'r' must be either 2, or 'r' must equal 3.
In other words, my claim is that either 'e to the 2x' or
'e to the 3x' must be a solution of
this particular equation.
Of course, we can do more with that, which we will in a later
part of calculus, not in this package's work.
But we're not going to study differential equations in
great detail here.
But for our present purposes, I think this illustrates how
one can use the fact that it's a rather powerful structural
property when the derivative of a function with respect to
'x' is the function itself.
By the way, as a quick check, notice that if 'y' equals 'e
to the 2x', 'y prime' is twice 'e to the 2x'.
'y double prime' is '4 e to the 2x'.
And therefore 'y double prime' minus '5y prime'
plus '6y' is what?
It's '4e to the 2x' minus '10 e to the 2x', '6 e to the 2x'.
And that, in fact, is genuinely, identically zero.
In a similar way, checking out 'y' equals 'e to the 3x', we
get 'y prime' is '3e to the 3x', 'y double prime' is '9e
to the 3x'.
Therefore, '9e to the 3x' minus '15 e to the 3x' plus
'6e to the 3x' is again, identically 0.
And again, you see what this powerful technique is.
In general, if 'a' and 'b' are constants, the substitution 'y
sub t' equals 'e to the rx'.
And there'll be a problem in the exercises on this to give
you additional drill.
But that substitution transforms the second or the
differential equation, 'y double prime' plus 'ay prime'
plus 'by' equals 0, into an equivalent quadratic equation,
'r squared' plus 'ar' plus 'b' equals 0.
And you see from this equation, using the quadratic
formula, we can find the values of 'r' that satisfy
this, and that gives us a couple of
solutions to the equation.
Well, at any rate, I deliberately want this lecture
to stay short to make the most important impact.
And that is again what?
That once we knew what the natural log function was, and
we defined the inverse natural log function, everything that
we wanted to know about the inverse natural log followed
from the properties of the log itself.
And this is basically the lesson for today.
We will continue the discussion of exponential
functions from a different point of
view in our next lecture.
But until that next lecture, goodbye.
ANNOUNCER: Funding for the publication of this video was
provided by the Gabriella and Paul Rosenbaum Foundation.
Help OCW continue to provide free and open access to MIT
courses by making a donation at ocw.mit.edu/donate.