# Logarithmic Function

If you want to try some of the examples in this video on your own, grab a scientific calculator and be prepared to practice some skills and explore the power and properties of logarithmic functions. Let’s get started!

Before we start our work with logarithms, let’s take some time to review the basics of exponential equations. The components of an exponential equation are shown here in blue.

A base of 2 raised to the power of 3 is equal to 8. The answer, 8, is referred to as the argument. Rearranging these components allows us to write the inverse logarithm, as shown highlighted in green. This illustrates nicely that logs are the power of a named base.

Let’s practice identifying these components and re-writing a few exponential equations as logarithmic equations:

5^{2}=25

Normally we would say, “five squared equals 25.” If we were to use the terms we used in our first example, we could say, “a base of 5 raised to the power of 2 is equal to 25.”

Now, let’s rewrite this as a logarithm:

log_{5}25=2

Let’s try another one. If you’d like to try this out on your own, pause the video and give it a shot!

4^{3}=64

We would say this is a base of 4 raised to the power of 3 is equal to 64.

If we rewrite this as a logarithm, we get this:

log_{4}64=3

When we evaluate a logarithmic expression we ask ourselves, “What do I have to raise this base by to get the value of the argument?” Logarithms are the power that is needed.

Now, let’s focus on the notation of logarithms. In our first example, the base of the log was 5, and our second example had a base of 4. Note that the base is indicated as a subscript on the word, “log.” This will be true for logs with bases other than 10 and the irrational number, e.

Logarithms with a base of 10 do not indicate the base in the notation, and they are called “common logs.” There is a key on a scientific calculator for common logs. If you have a calculator on hand, try inputting the following:

log 1000

The answer is 3, which means that 3 is the exponent needed to raise a base of 10 to get to 1,000. See if you can guess the answer to this one before you see the answer:

log 10

Think you got it? The answer is 1. A base of 10 raised to the power of 1 will result in 10.

There is also a key for the “natural” log function, which reads, “ln.” Whenever you see this notation, you know that you are dealing with a base of e. Exponential functions that deal with continuously compounded interest or population growth models have a base of e. The natural log is used to solve such applications, but that’s a topic for another time.

As mentioned earlier, the Richter scale measures the amplitude of waves that result from seismic activity and uses a logarithmic formula, R equals log of A over B, \(R=log(\frac{A}{B})\), where A is the observed amplitude and B is a baseline measure.

If the amplitude of a wavelength is measured at 450 times the baseline, the formula would read as: R equals log of 450 times B, over B, R=log(\(\frac{450\times B}{B})\) , which simplifies to R equals log of 450. R=log(450). Entering this in your calculator results in a value of 2.7, rounded to the nearest tenth. This measurement reflects the exponent on a base 10, and would be considered a fairly minor earthquake on the Richter Scale.

Like all inverse functions, the graphs of exponential and logarithmic functions are symmetric about the identity line, y = x.

Graphing a logarithmic function is as simple as creating a table and plugging in values until the curve takes shape. Some students find it is easier to convert the logarithm to an exponential function and “flip” the domain and range values to get the mirror image graph. Let’s give that method a try:

Graph the exponential function,f(x) = 3^{x} and the logarithmic function, g(x)=log_{3}(x)on the same graph.

There are several properties of logarithms that allow you to rewrite expressions in order to simplify them. I will list the corresponding exponent rule that you are already familiar with that supports the logarithmic function. Given what we have already covered with respect to exponential expressions, the first three should seem rather intuitive:

Property #1 log_{b}1=0

Remember to ask yourself, “What base of b raised to the 0 power will give me an answer of 1?” Well, we know that any base raised to a power of 0 equals 1: b^{0}=1.

Property #2 log_{b}(b)=1

This also makes sense, because raising any base to a power of 1 is equal to the base: b^{1}=b.

Property #3 log_{b}(b^{n})=n log, base b, of b to the n is equal to n

This is similar to Property #2, but the exponent, n, can be any value: b to the n equals b to the n b^{n}=b^{n}. Thinking about the inverse relationship with an exponential also may be helpful here: taking the log of an exponential “undoes” the exponential. The result is the input, n.

The next three properties are more arithmetic and are used to simplify or expand logarithmic expressions. I will provide some examples for each of these properties.

Property #4 The Product Property log_{b}(m × n)=log_{b}(m)+log_{b}(n)

The key with this property is that you can “expand” the log of a product to the sum of logs with the same base. This relates to the exponent rule: a^{m}× a^{n}=a^{m+n}

Here is an example:

Write the logarithm as a sum and simplify, if possible.

log_{4}(24)→log_{4}(4 × 6)=log_{4}(4) + log_{4}(6)

For this, we’ll use Property #2 to simplify the first term of the expression on the right:

log_{4}(24)=1+log_{4}(6)

Here’s another example. Write the logarithm as a sum and simplify, if possible.

log_{3}(9x)

log_{3}(9) + log_{3}(x)

To simplify it, let’s rewrite 3^{2}.

log_{3}(3^{2})+log_{3}(x)

Then, we’ll use Property #3 to simplify further:

This gives us

2+log_{3}(x)

Let’s take a look at another property:

Property Property #5 The Quotient Property log_{b}(\(\frac{m}{n}\))=log_{b}(m)-log_{b}(n)

As you can see, the log of a quotient can be expanded into the subtraction of two logs of the same base. The associated exponent rule is \(\frac{a^m}{a^n}=a^{m-n}\).

Let’s practice expanding a logarithm using this property.

\(log_7(\frac{49}{3x})=log_7(49)-log_7(3x)\)

We need to be sure to keep the bases the same for the subtracted terms! The first term of the expression on the right can be simplified by writing 49 as 7 squared. 72.

\(log_7(\frac{49}{3x})=log_7(7^2)-log_7(3x)\)

Now, we can simplify with Property #3:

This gives us

\(log_7(\frac{49}{3x})=2-log_7(3x)\)

Now let’s work on the property backwards to condense the subtraction of logs with the same base to a quotient:

log_{5}(9)-log_{5}(3)

Now we’re gonna create the quotient.

\(log_5(\frac{9}{3})\)

Simplify the quotient.

log_{5}(3)

You have condensed the expression to one log!

The next property involves the exponent on the argument of a logarithm.

Property #6 The Power Property log_{a}(x)^{p}=p × log_{a}(x)

Note that when the argument is raised to a power, the expression is equal to the exponent being multiplied by the logarithm. The associated exponent rule is:a to the m, to the n power, is equal to a, to the (m times n) (a^{m})^{n}=a^{m × n}.

We will practice implementing this property with a more comprehensive example just to keep things interesting!

log(2) + 3log(3) – log(9)

Notice that there is no subscript indicated! What is the base? Remember, if there is no subscript indicated, the base is always 10.

Now, we need to identify the Power Property in the second term. The multiplier of the 3 before the log translates to the argument of 3 being raised to the third power:

So this is the same as saying,

log(2) + log(3^{3}) – log(9)

Use the Product Rule to combine the sum of log(2) and log(3^{3})

log(2 × 3^{3})-log(9)

Now, use the Quotient Rule to condense the subtraction of common logs to a single common log.

log(\(\frac{2\times 3^3}{9}\))

Simplify the quotient by canceling out the common factor of 9, for a final answer of

log(2 × 3) = log(6).

For our final practice problem, let’s go the other way and expand a single log into an expression of addition and subtraction:

log_{3}(\(\frac{3×2}{5}\))

First, we’ll expand the quotient into subtraction of two logs, base 3.

log_{3}(3x^{2})-log_{3}(5)

Then, we’ll expand the product in the first term to the sum of logs, base 3.

log_{3}(3) + log_{3}(x^{2})-log_{3}(5)

Now, we need to identify the Power Property in the second term. Move the exponent of 2 to the front of the log as a multiplier.

log3(3) + 2log_{3}(x) – log_{3}(5)

Simplify the first term using Property #2.

1 + 2log_{3}(x) – log_{3}(5).

Okay, that about wraps things up! I hope that the examples that we worked through have helped you better understand logarithms, their relationship with exponential functions, and the properties that allow us to simplify complex expressions.

Thanks for watching, and happy studying!