There are several rules that you will need to remember when it comes to logarithms. We will detail these more general rules as well as how to use logarithms in basic algebraic equations in this chapter.

A logarithm is a function that describes the power to which a base must be raised in order to result in a certain number. This will make more sense as we understand how they are used in algebra.

For now, let’s examine the different parts of a logarithm that we will be talking about in this chapter:

$g_{2}(8)=3$

In this example, the $2$ is what is called a **base**. The **base** of a logarithm is the smaller number that rests between the word “$g$” and the parentheses. The $8$ is the number being operated on by the logarithm. $3$ is the answer we get when we solve “log base-2 of 8.”

When there is multiplication inside the parentheses of a logarithm, it can make our calculations very complicated. So, we can rearrange it by taking each number being multiplied and creating a new logarithm for each one of them. Then, we sum together each of the new logarithms.

So, we can start out with a logarithm like this:

$g_{2}(8∗x)$

And rearrange it to look like this:

$g_{2}(8)+g_{2}(x)$

You’ll notice that both of the logarithms have the same base, $2$. When you perform this expansion of logarithms, each term should have the same base.

When there is division inside the parentheses of a logarithm, we follow a similar process as above but we subtract logarithms instead of adding them. We will keep all the bases the same, too.

Let’s expand the following logarithm using the same rules as multiplication, but subtracting instead of adding the terms:

$g_{2}(x8 )=g_{2}(8)−g_{2}(x)$

As you can see, the numerator is the first logarithm, then the denominator is subtracted from it.

When there is an exponent inside the parentheses of a logarithm, we can reorganize it by taking the exponent **out of the logarithm** and multiplying it onto the front of the expression. The number that has the exponent will remain inside the parentheses; only the exponent will be shifted within the expression. Follow this in the example below:

$g_{2}(x_{8})=8∗g_{2}(x)$

We remove the exponent from inside the parentheses and put it in front of the logarithm. That’s all there is to it!

We have talked about three rules concerning logarithms. We can use these rules to take things in the parentheses and move them out of the parentheses, or we can use the rules in the reverse way. You must remember that these rules are a two-way street. If you can rearrange a logarithm in a certain way, you have to be able to change it back, too. Look at the following few examples to get an idea of how to recognize these reverse rules:

Rewrite the following expression as a single logarithm

$g_{10}(8)+g_{10}(7)+g_{10}(x)$

You should recognize the addition of multiple logs with the same base. Our multiplication rule is what causes this expansion into a sum, so let’s perform that step backward to get our final, simplified answer:

$g_{10}(8∗7∗x)=g_{10}(56x)$

Simplify the following logarithmic expression:

$3∗g_{6}(x)$

(spoiler)

Use the exponent rule here. The exponent rule says that we remove the exponent from inside the parentheses and multiply it outside the logarithm. The reverse of this would be restoring the exponent.

Thus, our final answer is:

$g_{6}(x_{3})$

If you must evaluate a logarithmic function with a base other than the usual $10$, you may have trouble putting it into your calculator. Some calculators like the TI-84 are able to calculate logarithms of different bases, but others are not. We will review both of these cases.

For a TI-84, perform a logarithmic calculation by pressing the $g$ button, entering the number **within** the $g$, a comma, and then the **base** of the $g$.

In a calculator, this would look like $g($number, base$)$. Let’s find the value of $g_{2}(30)$. We do this by typing (without quotation marks) into our calculator “$g(2,30)$.” This comes out to equal $4.9$.

If you need to calculate a logarithm with an abnormal base (a base other than $10$), use the following equation:

$g_{b}(a)=g_{x}(b)g_{x}(a) $

The base of the logarithms in the new fraction is unimportant, so it can be any number greater than $1$. It is best to make the base $10$ so that you can use the normal $log$ function in your calculator.

This being the case, we can edit the formula:

$g_{b}(a)=g_{10}(b)g_{10}(a) $

Let’s evaluate $g_{2}(30)$ using this formula:

$g_{2}(30) =g_{10}(2)g_{10}(30) =4.9 $

There will be times in which you will be asked to find the value of $x$ within a given logarithmic equation. This could be similar to the following form:

$g_{x}(9)=2$

To solve for $x$, we will go over one key concept of how to use logarithms in algebra without a calculator.

In order to solve this equation, we will get rid of the “$g$.” We do this by taking the base number, in the above example it is $x$, and moving it to the right side of the equation. On the right side of the equation, we take all the values that started there and raise them as the power of the base. At this point, we may remove the “$g$” from the equation. Follow along:

$g_{x}(9)99 3 =2=x_{2}=x=x $

A handy way to remember this rule is that you are swapping the **smaller number** in the equation. At first, $x$ is the small number because it is the subscript or the base. We want to make the $x$ normal sized, and then make the right side of the equation the new **small number** by raising it as a superscript, or the exponent. Try a few more short examples of how to apply this to algebraic problems.

What is the value of $x$?

$g_{2}(x)=4$

(spoiler)

Move the base ($2$) to the right side and raise the $4$ as its exponent:

$g_{2}(x)xx =4=2_{4}=16 $

Let’s try another!

Find the value of $x$ for the following equation:

$g_{4}(64)=x$

(spoiler)

Move the base ($4$) to the right side and raise the $x$ as its exponent:

$64=4_{x}$

Since $4_{3}=64$:

$x=3$

Solve for $x$ as an exponent by taking the log of both sides of an algebraic equation. This works out because if a log has the same value in its base as it does within the function, it will cancel out. Then, we are left with a logarithm on one side of the equation and an exponent lowered to be a normal number on the other side. Below is an example:

Find the value of $x$ for the following equation in terms of a logarithm:

$33=2_{x}$

Take the log of both sides, where $2$ is the base:

$g_{2}(33)=g_{2}(2_{x})$

This simplifies using our logarithmic rules to be:

$g_{2}(33)=x$

The important thing to learn from this example is that a logarithm may be used to lower $x$ from an exponent and solve for it algebraically.

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