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:**

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.

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:**

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:**

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:**

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. **

***Setting the log = to x**

***Rewriting in exponential form**

*** x is the exponent
we need on 3 to get 81**

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. **

***Setting the log = to x**

***Rewriting in exponential form**

*** x is the exponent
we need on 3 to get 1**

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. **

***Setting the log = to x**

***Rewriting in exponential form**

*** x is the exponent
we need on 11 to get 11**

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. **

***Setting the log = to x**

***Rewriting in exponential form**

*** x is the exponent
we need on 7 to get square root of 7**

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:**

I have found that the best way to do this is to do it the same each
time. In other words, put in the same values for *y* each time and then find it's corresponding *x* value for the given function.

The first two columns just show what values we are going to **plug
in for y.**

The last three columns show the corresponding values for *x* and *y* for the given function.

*x**y**y***( x, y)**
-2
-2
(1/16, -2)
-1
-1
(1/4, -1)
0
0
(1, 0)
1
1
(4, 1)
2
2
(16, 2)

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:**

I have found that the best way to do this is to do it the same each
time. In other words, put in the same values for *y* each time and then find it's corresponding *x* value for the given function.

The first two columns just show what values we are going to **plug
in for y.**

The last three columns show the corresponding values for *x* and *y* for the given function.

*x**y**y***( x, y)**
-2
-2
(1/64, -2)
-1
-1
(1/16, -1)
0
0
(1/4, 0)
1
1
(1, 1)
2
2
(4, 2)

Based on the **definition
of logs**, the inside of the log has to be positive.

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.

***Domain of this function**

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.

I'm going to use the **second
inverse property** shown on the lesson page:

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:

I'm going to use the **second
inverse property** shown on the lesson page:

I'm going to use the **first
inverse property** shown on the lesson page:

Last revised on March 17, 2011 by Kim Seward.

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