In this section we now need to move into logarithm
functions. This can be a tricky function
to graph right away. There is going to
be some different notation that you aren’t used to and some of the properties
may not be all that intuitive. Do not
get discouraged however. Once you figure
these out you will find that they really aren’t that bad and it usually just
takes a little working with them to get them figured out.
Here is the definition of the logarithm function.
In this definition 
is called the logarithm form and is called the exponential form.
Note that the requirement that is really a result of the fact that we are
also requiring . If you think about it, it will make sense. We are raising a positive number to an
exponent and so there is no way that the result can possibly be anything other
than another positive number. It is very
important to remember that we can’t take the logarithm of zero or a negative
number.
Now, let’s address the notation used here as that is usually
the biggest hurdle that students need to overcome before starting to understand
logarithms. First, the “log” part of the
function is simply three letters that are used to denote the fact that we are
dealing with a logarithm. They are not
variables and they aren’t signifying multiplication. They are just there to tell us we are dealing
with a logarithm.
Next, the b that
is subscripted on the “log” part is there to tell us what the base is as this
is an important piece of information.
Also, despite what it might look like there is no exponentiation in the
logarithm form above. It might look like
we’ve got in that form, but it isn’t. It just looks like that might be what’s
happening.
It is important to keep the notation with logarithms
straight, if you don’t you will find it very difficult to understand them and
to work with them.
Now, let’s take a quick look at how we evaluate logarithms.
|
Example 1 Evaluate
each of the following logarithms.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
(f) [Solution]
Solution
Now, the reality is that evaluating logarithms directly
can be a very difficult process, even for those who really understand
them. It is usually much easier to
first convert the logarithm form into exponential form. In that form we can usually get the answer
pretty quickly.
(a)
Okay what we are really asking here is the following.
As suggested above, let’s convert this to exponential
form.
Most people cannot evaluate the logarithm right off the top of their head. However, most people can determine the
exponent that we need on 4 to get 16 once we do the exponentiation. So, since,
we must have the following value of the logarithm.
[Return to Problems]
(b)
This one is similar to the previous part. Let’s first convert to exponential form.
If you don’t know this answer right off the top of your
head, start trying numbers. In other
words, compute ,
,
,
etc until you get 16. In this case we need an exponent of 4. Therefore, the value of this logarithm is,
Before moving on to the next part notice that the base on
these is a very important piece of notation.
Changing the base will change the answer and so we always need to keep
track of the base.
[Return to Problems]
(c)
We’ll do this one without any real explanation to see how
well you’ve got the evaluation of logarithms down.
[Return to Problems]
(d)
Now, this one looks different from the previous parts, but
it really isn’t any different. As
always let’s first convert to exponential form.
First, notice that the only way that we can raise an
integer to an integer power and get a fraction as an answer is for the
exponent to be negative. So, we know
that the exponent has to be negative.
Now, let’s ignore the fraction for a second and ask . In this case if we cube 5 we will get
125.
So, it looks like we have the following,
[Return to Problems]
(e)
Converting this logarithm to exponential form gives,
Now, just like the previous part, the only way that this
is going to work out is if the exponent is negative. Then all we need to do is recognize that and we can see that,
[Return to Problems]
(f)
Here is the answer to this one.
[Return to Problems]
|
Hopefully, you now have an idea on how to evaluate
logarithms and are starting to get a grasp on the notation. There are a few more evaluations that we want
to do however, we need to introduce some special logarithms that occur on a
very regular basis. They are the common logarithm and the natural logarithm. Here are the definitions and notations that
we will be using for these two logarithms.
So, the common logarithm is simply the log base 10, except
we drop the “base 10” part of the notation.
Similarly, the natural logarithm is simply the log base e with a different notation and where e is the same number that we saw in the
previous section and is defined
to be .
Let’s take a look at a couple more evaluations.
So, when evaluating logarithms all that we’re really asking
is what exponent did we put onto the base to get the number in the logarithm.
Now, before we get into some of the properties of logarithms
let’s first do a couple of quick graphs.
|
Example 3 Sketch
the graph of the common logarithm and the natural logarithm on the same axis
system.
Solution
This example has two points. First, it will familiarize us with the
graphs of the two logarithms that we are most likely to see in other
classes. Also, it will give us some
practice using our calculator to evaluate these logarithms because the
reality is that is how we will need to do most of these evaluations.
Here is a table of values for the two logarithms.
|
x
|
|
|
|
|
-0.3010
|
-0.6931
|
|
1
|
0
|
0
|
|
2
|
0.3010
|
0.6931
|
|
3
|
0.4771
|
1.0986
|
|
4
|
0.6021
|
1.3863
|
Here is a sketch of the graphs of these two functions.
|
Now let’s start looking at some properties of
logarithms. We’ll start off with some
basic evaluation properties.
Properties of
Logarithms
- . This follows from the fact that .
- . This follows from the fact that .
- . This can be generalized out to .
- . This can be generalized out to .
|
Properties 3 and 4 leads to a nice relationship between the
logarithm and exponential function.
Let’s first compute the following function
compositions for and .
Recall from the
section on inverse functions that this means
that the exponential and logarithm functions are inverses of each other. This is a nice fact to remember on occasion.
We should also give the generalized version of Properties 3 and 4 in terms of both the
natural and common logarithm as we’ll be seeing those in the next couple of
sections on occasion.
Now, let’s take a look at some manipulation properties of
the logarithm.
More Properties of
Logarithms
We won’t be doing anything with the final property in this
section; it is here only for the sake of completeness. We will be looking at this property in detail
in a couple of sections.
The first two properties listed here can be a little
confusing at first since on one side we’ve got a product or a quotient inside
the logarithm and on the other side we’ve got a sum or difference of two
logarithms. We will just need to be careful
with these properties and make sure to use them correctly.
Also, note that there are no rules on how to break up the
logarithm of the sum or difference of two terms. To be clear about this let’s note the
following,
Be careful with these and do not try to use these as they
simply aren’t true.
Note that all of the properties given to this point are
valid for both the common and natural logarithms. We just didn’t write them out explicitly
using the notation for these two logarithms, the properties do hold for them
nonetheless
Now, let’s see some examples of how to use these properties.
|
Example 4 Simplify
each of the following logarithms.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
The instructions here may be a little misleading. When we say simplify we really mean to say
that we want to use as many of the logarithm properties as we can.
(a)
Note that we can’t use Property 7 to bring the 3 and the 5
down into the front of the logarithm at this point. In order to use Property 7 the whole term
in the logarithm needs to be raised to the power. In this case the two exponents are only on
individual terms in the logarithm and so Property 7 can’t be used here.
We do, however, have a product inside the logarithm so we
can use Property 5 on this logarithm.
Now that we’ve done this we can use Property 7 on each of
these individual logarithms to get the final simplified answer.
[Return to Problems]
(b)
In this case we’ve got a product and a quotient in the
logarithm. In these cases it is almost
always best to deal with the quotient before dealing with the product. Here is the first step in this part.
Now, we’ll break up the product in the first term and once
we’ve done that we’ll take care of the exponents on the terms.
[Return to Problems]
(c)
For this part let’s first rewrite the logarithm a little
so that we can see the first step.
Written in this form we can see that there is a single
exponent on the whole term and so we’ll take care of that first.
Now, we will take care of the product.
Notice the parenthesis in this the answer. The multiplies the original logarithm and so it
will also need to multiply the whole “simplified” logarithm. Therefore, we need to have a set of
parenthesis there to make sure that this is taken care of correctly.
[Return to Problems]
(d)
We’ll first take care of the quotient in this logarithm.
We now reach the real point to this problem. The second logarithm is as simplified as we
can make it. Remember that we can’t
break up a log of a sum or difference and so this can’t be broken up any farther. Also, we can only deal with exponents if
the term as a whole is raised to the exponent. The fact that both pieces of this term are
squared doesn’t matter. It needs to be
the whole term squared, as in the first logarithm.
So, we can further simplify the first logarithm, but the
second logarithm can’t be simplified any more. Here is the final answer for this problem.
[Return to Problems]
|
Now, we need to work some examples that go the other
way. This next set of examples is
probably more important than the previous set.
We will be doing this kind of logarithm work in a couple of sections.
|
Example 5 Write
each of the following as a single logarithm with a coefficient of 1.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
The instruction requiring a coefficient of 1 means that
the when we get down to a final logarithm there shouldn’t be any number in
front of the logarithm.
Note as well that these examples are going to be using
Properties 5 7 only we’ll be using them in reverse. We will have expressions that look like the
right side of the property and use the property to write it so it looks like
the left side of the property.
(a) The first step here is to get rid of the coefficients on the
logarithms. This will use Property 7
in reverse. In this direction,
Property 7 says that we can move the coefficient of a logarithm up to become
a power on the term inside the logarithm.
Here is that step for this part.
We’ve now got a sum of two logarithms both with
coefficients of 1 and both with the same base. This means that we can use Property 5 in
reverse. Here is the answer for this
part.
[Return to Problems]
(b) Again, we will first take care of the coefficients on the
logarithms.
We now have a difference of two logarithms and so we can
use Property 6 in reverse. When using
Property 6 in reverse remember that the term from the logarithm that is
subtracted off goes in the denominator of the quotient. Here is the answer to this part.
[Return to Problems]
(c) In this case we’ve got three terms to deal with and none of
the properties have three terms in them.
That isn’t a problem. Let’s
first take care of the coefficients and at the same time we’ll factor a minus
sign out of the last two terms. The
reason for this will be apparent in the next step.
Now, notice that the quantity in the parenthesis is a sum
of two logarithms and so can be combined into a single logarithm with a
product as follows,
Now we are down to two logarithms and they are a
difference of logarithms and so we can write it as a single logarithm with a
quotient.
[Return to Problems]
|
The final topic that we need to discuss in this section is
the change of base formula.
Most calculators these days are capable of evaluating common
logarithms and natural logarithms.
However, that is about it, so what do we do if we need to evaluate another
logarithm that can’t be done easily as we did in the first set of examples that
we looked at?
To do this we have the change of base formula. Here is the change of base formula.
where we can choose b
to be anything we want it to be. In
order to use this to help us evaluate logarithms this is usually the common or
natural logarithm. Here is the change of
base formula using both the common logarithm and the natural logarithm.
Let’s see how this works with an example.
|
Example 6 Evaluate
.
Solution
First, notice that we can’t use the same method to do this
evaluation that we did in the first set of examples. This would require us to look at the
following exponential form,
and that’s just not something that anyone can answer off
the top of their head. If the 7 had
been a 5, or a 25, or a 125, etc.
we could do this, but it’s not.
Therefore, we have to use the change of base formula.
Now, we can use either one and we’ll get the same
answer. So, let’s use both and verify
that. We’ll start with the common
logarithm form of the change of base.
Now, let’s try the natural logarithm form of the change of
base formula.
So, we got the same answer despite the fact that the
fractions involved different answers.
|