College Algebra
Answer/Discussion to Practice Problems
Tutorial 43: Logarithmic Functions
Answer/Discussion
to 1a
|
We want to use the definition:
if and only if .
First, let's figure out what the base needs to be. What do you
think? It looks like the b in the definition
correlates with 2 in our problem - so our base is going to be 2.
Next, let's figure out the exponent. This is very key, again remember
that logs are another way to write exponents. This means the log
is set equal to the exponent, so in this problem that means that the exponent
has to be 5.
That leaves 32 to be what the exponential expression is set equal
to.
Putting all of this into the log definition we get: |
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*Rewriting in exponential form |
Hopefully, when you see it written in exponential form you can tell
that it is a true statement. In other words, when we take 2 to the
5th power we do get 32. |
Answer/Discussion
to 1b
|
We want to use the definition:
if and only if .
First, let's figure out what the base needs to be. What do you
think? It looks like the b in the definition
correlates with b in our problem - so our
base is going to be b.
Next, let's figure out the exponent. This is very key, again remember
that logs are another way to write exponents. This means the log
is set equal to the exponent, so in this problem that means that the exponent
has to be 4.
That leaves 81 to be what the exponential expression is set equal
to.
Putting all of this into the log definition we get: |
|
*Rewriting in exponential form |
Answer/Discussion
to 2a
|
We want to use the definition:
if and only if .
First, let's figure out what the base needs to be. What do you
think? It looks like the b in the definition
correlates with 4 in our problem - so our base is going to be 4.
Next, let's figure out the exponent. In this direction it is easy
to note what the exponent is because we are more used to it written in
this form, but when we write it in the log form we have to be careful to
place it correctly. Looks like the exponent is -3, don't you
agree?
The value that the exponential expression is set equal to is what
goes inside the log function. In this problem that is 1/64.
Let's see what we get when we put this in log form: |
Answer/Discussion
to 2b
|
We want to use the definition:
if and only if .
Rewriting the original problem using exponents we get:
First, let's figure out what the base needs to be. What do you
think? It looks like the b in the definition
correlates with 9 in our problem - so our base is going to be 9.
Next, let's figure out the exponent. In this direction it is easy
to note what the exponent is because we are more used to it written in
this form, but when we write it in the log form we have to be careful to
place it correctly. Looks like the exponent is 1/2, don't
you agree?
The value that the exponential expression is set equal to is what
goes inside the log function. In this problem that is x.
Let's see what we get when we put this in log form: |
Answer/Discussion
to 3a
|
When we are looking for the log itself, keep in mind that logs are
another way to write exponents.
The thought behind this is, we are wanting the power that we would
need to raise 3 to to get 81. |
Step 1: Set
the log equal to x
AND |
Step 2: Use the definition
of logs shown above to write the equation in exponential form
AND |
|
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 3 to get 81 |
So the exponent we were looking for is 4. |
Answer/Discussion
to 3b
|
When we are looking for the log itself, keep in mind that logs are
another way to write exponents.
The thought behind this is, we are wanting the power that we would
need to raise 3 to to get 1. |
Step 1: Set
the log equal to x
AND |
Step 2: Use the definition
of logs shown above to write the equation in exponential form
AND |
|
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 3 to get 1 |
So the exponent we were looking for is 0. |
Answer/Discussion
to 3c
|
When we are looking for the log itself, keep in mind that logs are
another way to write exponents.
The thought behind this is, we are wanting the power that we would
need to raise 11 to to get 11. |
Step 1: Set
the log equal to x
AND |
Step 2: Use the definition
of logs shown above to write the equation in exponential form
AND |
|
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 11 to get 11 |
So the exponent we were looking for is 1. |
Answer/Discussion
to 3d
|
When we are looking for the log itself, keep in mind that logs are
another way to write exponents.
The thought behind this is, we are wanting the power that we would
need to raise 7 to get square root of 7. |
Step 1: Set
the log equal to x
AND |
Step 2: Use the definition
of logs shown above to write the equation in exponential form
AND |
|
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 7 to get square root of 7 |
So the exponent we were looking for is 1/2. |
Answer/Discussion
to 4a
|
First, we need to write in exponential form, just like we practiced
in examples 1 and 2 on the lesson page.
Looks like the base is 4, the exponent is y,
and the log will be set = to x: |
|
*Rewriting in exponential form
|
Step 2: Plug in
values for y (NOT x) to find some ordered pairs. |
Step 3: Plot points.
AND
Step 4: Draw a curve. |
Answer/Discussion
to 4b
|
Setting this up to be able to use the definition we get:
Next, we need to write in exponential form, just like we practiced in
examples 1 and 2 on the lesson page.
Looks like the base is 4, the exponent is y - 1, and the log will be set = to x: |
|
*Rewriting in exponential form
|
Step 2: Plug in
values for y (NOT x) to find some ordered pairs. |
Step 3: Plot points.
AND
Step 4: Draw a curve. |
Answer/Discussion
to 5a
|
Since x is part of the inside of the log
on this problem we need to find a value of x,
such that the inside of the log, 2 - x, is
positive. |
|
*Inside of log must be positive
*Solve the inequality
*Domain of this function |
The domain is x < 2.
That means that if we put in any value of x that is less than 2, we will end up with a positive value inside our log. |
Answer/Discussion
to 6a
|
Answer/Discussion
to 6b
|
We can either use the definition of logs, as shown above, or the inverse
properties of logs to evaluate this.
I'm going to use the first
inverse property shown on the lesson page: |
|
*Rewrite 1/e^5
as e to the -5th power
|
Answer/Discussion
to 7a
|
Answer/Discussion
to 7b
|
Last revised on March 17, 2011 by Kim Seward.
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