First off, let me review what a logarithm is. I have explained it before, but once you understand the notation, you shouldn't need to have done Precalculus to understand this post. They are very easy to understand.

For instance, we know that 10

^{2}is 100.

10

^{2}= 100

If we take the logarithm of both sides, we are essentially bringing the two out of the exponent. Rather than doing an operation on the 10, we do an operation on the 100 to determine what the exponent is.

log(10

^{2}) = log(100)

2 = log(100)

In most situations, it is clear what type of logarithm you are using, especially in this one because ten is a common logarithm to use. However, many people will write a subscript to clarify. For instance:

log

_{10}(100) = 2

Logarithms become extremely useful when you need to solve an algebraic equation where the variable is in the exponent. For example:

2

^{x}= 64

As you know, algebra is about doing the inverse operation. If there is addition going on, you subtract. If there is multiplication going on, you divide. Similarly, if there is exponentiation going on, you use a logarithm. In this instance, it would be taking the log

_{2}of both sides.

c2

^{x}) = log

_{2}(64)

*x*= log

_{2}(64)

*x*= 6

In practice, there are three bases that are extremely popular to use in a logarithm. We just used two of them: log

_{10 }and log

_{2}, which are also known as the common logarithm and the binary logarithm. The third one is log

_{e}(using the number

*e*that is described here), which is called the natural logarithm, or the natural log. This one is found on most calculators, usually next to the common logarithm.

Though logarithms are not a part of most people's day-to-day life, they do have lots of practical applications. The common logarithm is the basis of the pH system which describes the acidity of water. The natural log is a huge aspect of finance and compound interest (as we saw with the Law of 72). And of course, they are all over nature.

Let's look at an identity of logarithms. Take the following problem:

log

_{6}(24) + log

_{6}(9) =

If you just used a calculator to do this, you would get:

log

_{6}(24) + log

_{6}(9)

1.773705614 + 1.226294386

3

That's odd. Two random numbers happened to have logarithms that summed to three. Let's look closer at this and see if we can figure out why. What number could you find the log

_{6}of and get 3?

log

_{6}(

*n*) = 3

First of all, let me point out that we just asked an algebraic question. We suddenly got curious about why something happened, so we asked a "what" question, which calls for an unknown quantity, which later becomes an algebraic variable. So when algebra seems like a drag, remember that it is all techniques for answering that "what" question. And "what" is a question asked in all branches of mathematics, science, and engineering.

Anyways, for this equation, we would want to do the inverse operation. We turn both sides into the exponent, and create a base of 6. This gives:

6

^{log6(n)}= 6

^{3}

The left hand side cancels, leaving just

*n*. The right hand side is six cubed, which is 216. So, we end up with:

*n*= 216

So, this means that the log

_{6}of 216 makes you end up with 3, or the sum of log

_{6}(24) and log

_{6}(9).

log

_{6}(24) + log

_{6}(9) = log

_{6}(216)

What is the relationship between these three numbers? Well, it shouldn't take to long to determine that 24 x 9 = 216, or:

log

_{6}(24) + log

_{6}(9) = log

_{6}(24 • 9)

In other words, the sum of the logarithms is the logarithm of the product. Wow! That's pretty cool! Is that always the case? Well, let's try to prove that it is for all logarithms.

Let's set a few terms equal to each other and see what happens. Since there are logarithms, we will need a lot of variables.

*x*= log

_{a}(

*p*)

*y*= log

_{a}(

*q*)

In other words:

*p*=

*a*

^{x}

*q*=

*a*

^{y}

Let's multiply those two equations together. Since they are both equal, multiplying the terms on each side by each other won't make a difference.

*p*•

*q*=

*a*

^{x }•

*a*

^{y}

*pq*=

*a*

^{x+y}

The right hand side was simplified using the Law of Exponents, which is explained very well here.

Now, we must take the logarithm of both sides, or specifically, the log

_{a }of both sides.

log

_{a}(

*pq*) = log

_{a}(

*a*

^{x+y})

log

_{a}(

*pq*) =

*x*+

*y*

*But what were*

*x*and

*y*? We defined them in terms of

*a*,

*p*, and

*q*earlier. So, let's substitute those values in and see what we get.

log

_{a}(

*pq*) = log

_{a}(

*p*)

*+*

*log*

_{a}(

*q*)

And this creates the identity that we were trying to prove: the sum of the logarithms is the logarithm of the product, and thus, completes our proof. There are other logarithmic identities like this one, but I will save that for another post.

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