
Titre:
Rules of Logs  MIT 18.01SC Single Variable Calculus, Fall 2010

Description:

Welcome back to recitation.

Today we're going to talk about some rules of logaritms that you're going to need to remember.

We're going to prove why one of them is true.

And then I'm going to ask you to use these rules to take the derivative of a function.

So let's just look at these rules first.

So I want to point out as I'm talking about these rules

the first three are written with the natural log (ln)

But one can also write them in any base, as long as the base is the same all the way across.

In any legitimate base that one is allowed to use, so with a positive base

one can use it all the way across instead of the natural log.

The first one says that the natural log of a product is equal to the sum of the natural logs.

So natural log of M times N is equal to the natural log of M plus the natural log of N

The second one says that the natural log of a quotient is equal to the difference of the natural logs

So natural log of M divided by N is equal to natural log of M minus the natural log of N

The third one says that the natural log of "something" raised to a power

is that power as a coefficient times the natural log of the "something".

So natural log of M to the K is equal to K times the natural log of M

And what I want to point out is that there is a distinct difference where the power is...

If the power is inside the argument, then this rule holds.

But if the power is outside the argument, if it's natural log of M the whole thing raised to a power...

this does not work. This is NOT equal to what's written above.

And then the fourth one is a changeofbase formula.

So if I have log basesomething b that maybe I want to change the base of M

I can rewrite that in the base e.

I can as natural log M divided by natural log of b

And I want to point out a common mistake people make

is sometimes they confuse the second and the fourth

because they both have quotients.

But notice that the second one is the natural log of a quotient

And the fourth one is about the quotient of natural logs.

So that's a disctinct difference.

And hopefully then you see that these two statements are not in fact the same statement.

So now what I'd like to do is using what we know about exponential and log functions

I want to prove number one.

So let's set out to do that.

In order to make this top line make sense we know that M and N have to be positive

and so I can find (actually, let me write first what we're doing. We're going to prove 1)

So with M and N both positive I can find values a and b such that

e^a equals M and e^b equals N.

And let me just write out also what that means...

because exponential and log functions are inverses of one another.

This means that "a" is equal to natural log of M, and b is equal to natural log of N.

So these are equivalent statements.

This statement and this statement are equivalent...

This statement and this statement are equivalent.

So now let's use that information to try and solve the problem.

Or to try and prove number 1.

So the natural log of M times N, well what is that?

M is e^a. N is e^b. So I can write this as...

natural log of e^a times e^b.

What's e^a times e^b? This is where we use our rules of exponents.

e^a times e^b is e to the (a+b).

So this is the natural log of e to the (a+b)

And now, what's the point?

The point is that natural log and exponential functions are inverses of one another.

Or natural log of e to the x is x.

So the natural log of e^(a+b) is just (a+b)

And I've already recorded for you what those are.

That's the natural log of M plus natural log of N.

So notice we've done what we've set out to do.

Natural log of the quantity M times N is equal to natural log of M plus the natural log of N.

And in a similar flavor one could immediately do number two.

And number three follows quite similarly as well.

Obviously these are going to use different rules for exponents besides the

product of two exponential functions is equal to the sum of the powers.

It's going ot use some of those other rules.

And I believe that some of these other things might actually also to be proven

in a later lecture in the actual course.

So you'll see these, but I would say you might want to try and prove two and three

at least on your own. It might be helpful to look at how those work using the same kind of rules here.

So now what I'd like us to do, is, using these rules, I'd like us to take a derivative.

So what I want us to look at is y equals the square root of x times (x+4)

and we'll just assume that x is bigger than zero.

And I want you to find y'.

Now you could do this just brute force, cranking it out.

But I'd like you to try and use the log differentiation technique in order to find this derivative.

I'll give you a moment to do it and then I'll come back and show you how I do it.

OK, welcome back.

I'm going to use the log differentiation and the rules I have on the side of the board there

to take a derivative to find y'.

So first what we do is we take the log of both sides

and then we use some of the rules of logaritms to simplify the expression on the righthand side.

I will take natural log of y is equal to natural log of the square root of x times (x+4)

Now square root (wow, sorry) square root is the power of something raised to the 1/2, right?

That's what it means to take the square root.

You can take this whole product and raise it to the 1/2.

So I'm going to use rule number three...

and I'm going to bring that 1/2 that's a power, out in front of the log.

So I can rewrite this expression as 1/2 log of this product.

So I have 1/2 ln(x(x+4))

So now I'm going to use rule number one which changes the natural log of a product into

the sum of the natural logs.

And I can rewrite this as 1/2 natural log x plus 1/2 natural log of the quantity (x+4)

Essentially what I'm doing here is I have to distribute this 1/2 because

I had one term and then I'm going to have two terms that are added together

But the 1/2 applies to both of them.

So now I have this nice setup.

I have natural log of y is equal to something in terms of x

and now I can take the derivative of both sides.

Now remember, I want to find y'

So there's some implicit differentiation going on.

So let's just be careful when we do that.

If I take the derivative of this side I don't just get y'.

I get y'/y

Where does that come from?

Well d/dx of this expression is the derivative of the natural log evaluated at y

then times the derivative of y.

You've seen this I think a lot by now, but just to make sure

you understand where both of those comes from.

So when I take the derivative here I get y'/y.

When I take the derivative here with respect to x, the derivative of the natural log of x is just

1/x, so i get 1/2x

And then the derivative of the natural log of (x+4)..

if I use the chain rule I get 1/(x+4) times the derivative of (x+4)

which is still just 1.

So I get 1/2(x+4)

So now I wanted us to find y'.

So to find y' I'm going to move over a little bit more

and just notice that y' is going to equal y times all of that.

Well, I know y.

I'm going to write what y is.

y is the square root of x times (x+4) times this quantity, 1/2x plus 1/2(x+4)

That's actually one way to write the derivative of y.

I could combine these two fractions into a single fraction

and try to make it look a little bit nicer,

or I could just leave it this way.

This is technically, a derivative.

If I started trying to combine things I might find out that I could have just taken

the derivative the long way.

So this is a nice short way to just get to a place

where I can start to find out something about the derivative of y.

So I guess I'll stop there.