##### Example4.1.1

\begin{align*} 9^{-1}\amp= \frac{1}{9}\amp 9^{1/2}= \sqrt{9} = 3 \end{align*}

For additional practice and information follow this link to Khan Academy.

Logarithms can sometimes be the source of much turmoil and angst to students. To some members of the English department, the word “logarithms” is used as a joke to respond to something that is confusing, “Is that like logarithms?” (laughter ensues).

But you will not have these issues, because logarithms are not bad, they're just spelled that way. Remember this, in short, a logarithm is essentially an exponent.

This activity is an introduction to logarithms and the logarithmic function. And since logarithms are exponents, it is a good idea to review some key properties of exponents, in particular negative and fractional exponents.

Remember that negative exponents mean “reciprocal” and fractional exponents mean “roots”.

\begin{align*} 9^{-1}\amp= \frac{1}{9}\amp 9^{1/2}= \sqrt{9} = 3 \end{align*}

For additional practice and information follow this link to Khan Academy.

In the next set of exercises, practice the idea of using the power (exponent) of a number (the base) to get a particular result.

Earlier in the study of exponential functions, we solved exponential equations graphically. Another way to solve an exponential equation is to “guess-and-check”. It is not a very efficient method, but in this case it may help clarify what a logarithm does.

Notice that for each ordered pair, we could say “\(10\) raised to the (input) power is (output)”. For instance, \(10\) raised to the \(-1\) power is \(\frac{1}{10}=0.1\)

In the following exercise, our goal is to find the power of \(10\) that equals \(36\text{.}\) In other words, to solve the equation \(10^x=36\) in which we want to know “What power of \(10\) gives us \(36\text{.}\)”

From the results in the previous table we realize the exponent is not going to be an integer like \(0\) or \(1\) or \(2\text{.}\) It will have to be some decimal value.

Next, we can try a graphical approach.

So far the Geogebra graphing tool has been sufficient in helping us create and manipulate graphs. However, to get more accuracy in our solution we can use another free, online graphing tool, Desmos. Although you may also use your graphing calculator.

As you experienced in the last two exercises, different powers of \(10\) give us different results. In fact, we can get any positive result we want using only powers of \(10\) (or any positive base for that matter).

One use for logarithms is to answer the question,

I wonder what power of (choose a base) I need to get (whatever result you want)

In English we can ask an exponential question and get a logarithm answer:

Exponential question,

What power of \(3\) gives you \(81\text{?}\)Logarithm answer,

The power of \(3\) that gives you \(81\) is the \(4\)th power.(Because \(3^{4}=81\))

In Math that same question and answer look like this:

Exponential question, \(3^{x}=81\)

Logarithm answer, \(x= \log_{3}(81)=4\)

The structure of a logarithm is that it must tell the reader the base number and it has to state what value you are trying to get.

Base \(10\) means you are using powers of \(10\) as in \(10^{x}\text{.}\) Base \(3\) means you are using powers of \(3\) as in \(3^{x}\) etc.

Below is a table that shows how our previous example using base \(10\) looks when written in English versus what the same statement looks like in Math.

English | Math | How you “say the math” |

The power of \(10\) that gives you \(36\) | \(\log(36)\) | log base 10 of 36 |

The power of \(10\) that gives you \(0.6\) | \(\log(0.6)\) | log base 10 of 0.6 |

For example, \(\log(100)\) means “The power of \(10\) that gives you \(100\text{.}\)”

The math word *log* is short for *logarithm*. It means “log base \(10\)”. But we can use any positive number as a base, so \(\log_b\) means “log base \(b\)”.

Since \(\log\) is base \(10\) (powers of \(10\)) we technically should write \(\log_{10}\text{,}\) but base \(10\) is so often used, we leave out the \(10\) for convenience.

Another often used logarithm is base \(e\text{.}\) It's called the *natural logarithm* and is written \(\ln\text{.}\) Again, technically we should write \(\log_e\text{,}\) but we write \(\ln\) for convenience.

All other logarithms are written in standard format. The base, \(b\text{,}\) indicates what number is being multiplied over and over. So if \(2\) is the base, we write \(\log_2\) which indicates powers of \(2\text{.}\)

At this point of the activity we should be able to estimate the value of a given logarithmic expression with the understanding that a logarithm is just an exponent.

Try to evaluate \(\log(-100)\) on your calculator. Your calculator should read `error` or `quit` or something indicating it cannot complete the calculation. To understand why this is happening, write out the meaning of \(\log(-100)\) in English.

If you have a bunch of positive \(10\)s then how many do you have to multiply to get a *negative* \(100\text{?}\) In fact, you *can't* get a negative result if you only have positive \(10\)s! You would need at least one negative \(10\) and you don't have any.

Try to evaluate \(\log(0)\) on your calculator. Your calculator should read `error` or `quit` or something indicating it cannot complete the calculation. To understand why this is true, write out the meaning of \(\log(0)\) in English.

If you have a bunch of positive \(10\)s then how many do you have to multiply to get \(0\) as a result? In fact, you *can't* get \(0\) as the result if you only have positive \(10\)s! At least one multiple would have to be \(0\text{,}\) but you have is \(10\)'s.

It turns out that the \(\log\) function (with any base \(b \gt 0\)) is *not defined for values that are less than or equal to \(0\)*. Therefore the domain of the \(\log\) function (with any base \(b \gt 0\)) is *only positive numbers* (you can't use negatives or zero).

Earlier we used graphs to answer the questions: “What power of \(10\) gives you \(36\text{?}\)” and “What power of \(10\) gives you \(0.6\text{?}\)”

We answered the questions using logarithm statements: “\(\log(36) \approx 1.556\)” and “\(\log(0.6) \approx -0.222\)”

Notice with just a base \(10\) we can use positive or negative exponents to get any positive result we want.

You want to get a result of \(36\text{?}\) Use a power of about \(1.556\) as in \(10^{1.556} \approx 36\)

You want to get a result of \(0.6\text{?}\) Use a power of about \(-0.222\) as in \(10^{-0.222} \approx 0.6\text{.}\)

Remember logarithms are exponents and clearly we can use both positive and negative exponents.

Logarithms tell you the power you need to get a particular result. You input the result you want, the function tells you the exponent you need. The exponent is the output.

Given an input from the domain of a logarithmic function, the range can be positive, negative or zero. In fact, the range is all real numbers.

If \(\log_{b}(x)=y\) then \(y\) can be any real number.

Our discoveries about the domain and range of a logarithmic function lead us to wonder what the graph might look like. So far we know:

The domain is limited to only positive input values. So there should be some kind of boundary that separates the inputs the function can use from the ones the function cannot use.

For a base greater than \(1\text{,}\) the bigger the result we want to get, the bigger the exponent needs to be. So the logarithm should be an increasing function.

The output can be any real number, so we expect the graph to exist both above and below the horizontal axis as well as having a horizontal intercept.

In order for this part of the activity to make sense, we must remember what a *base* and an *exponent* are. Remember in the expression \(b^x\text{,}\) \(b\) is the *base* and \(x\) is the exponent. The exponent literally counts how many times you multiplied the base. For instance \(b^7\) means \(\overbrace{b\cdot b\cdot b\cdot b\cdot b\cdot b\cdot b}^{7\text{ times}}\text{.}\)

An exponential equation \begin{equation*} b^{y}=x \end{equation*} can be rewritten as a logarithmic equation.

\begin{equation*} \log_b(x)=y \end{equation*}

We start with a very simple property of exponents: \(x^a\cdot x^b=x^{a+b}\text{.}\) When you multiply numbers of the same base you add the exponents.

Now let's take some obvious information that you already concluded above to discover a not-so-obvious property of logarithms.

Start with the simple equation \begin{equation*} 8\cdot16=128\text{.} \end{equation*} Apply the \(\log_2(\phantom{x})\) function to both sides of the equation to get \begin{equation*} \log_2(8\cdot16)=\log_2(128)\text{.} \end{equation*} Notice on the right side we already know the answer (i.e. the power of \(2\) that gives you \(128\))

Now you can write the equation \(\log_{2}(8\cdot16)=\log_{2}(128)\) as \begin{equation*} \log_{2}(8\cdot16)=7\text{.} \end{equation*}

Notice on the left side of the equation:

- There are \(3\) factors of \(2\) in \(8\text{.}\) That's why \(\log_{2}(8)=3\text{.}\)
- There are \(4\) factors of \(2\) in \(16\text{.}\) That's why \(\log_{2}(8)=4\text{.}\)

So the simple equation \(7=3+4\text{,}\) can be written using logarithms \begin{gather*} 7=3+4\\ 7=\log_{2}(8)+\log_{2}(16)\\ \log_{2}(128)=\log_{2}(8)+\log_{2}(16)\\ \log_{2}(8 \cdot 16)=\log_{2}(8)+\log_{2}(16) \end{gather*}

*What???*

Yes, you just discovered that \begin{equation*} \log_2(8\cdot16)=\log_2(8)+\log_2(16)\text{.} \end{equation*} In fact this property applies to all \(\log_{b}(x)\text{.}\)

In general, *for any base*

\begin{equation*} \log_b(A\cdot B)=\log_b(A)+\log_b(B)\text{.} \end{equation*}

We start with something obvious. You already know that instead of writing \(8+8+8+8\) you could just write \(4\cdot8\text{.}\)

Maybe you never thought about it, but multiplication is just repeated addition. It's a shortcut way to write the same thing *added* over and over.

On the other hand, exponents are just repeated multiplication. Exponents are a shortcut way to write the same thing *multiplied* over and over.

Consider the expression \(\log(8\cdot8\cdot8\cdot8)\text{.}\) We can use some shortcuts to write the expression in at least two different ways. \begin{equation*} \log(8\cdot8\cdot8\cdot8)=\log\mathopen{}\left(8^4\right)\mathclose{} \end{equation*} and using the last rule we discovered about the log of a product \begin{equation*} \log(8\cdot8\cdot8\cdot8)=\log(8)+\log(8)+\log(8)+\log(8)\text{.} \end{equation*}

Since both shortcuts must give the same answer, we get \begin{equation*} \log\mathopen{}\left(8^4\right)\mathclose{}=\log(8)+\log(8)+\log(8)+\log(8) \end{equation*}

A very important rule: \begin{equation*} \log_{b}(N^{x})=x \log_{b}(N) \end{equation*}

This rule you will discover using the two rules you already know.

Remember that a negative exponent means “reciprocal” so a quotient can be rewritten as a product \begin{equation*} \frac{Q}{P}=Q \cdot P^{-1} \end{equation*}

\begin{equation*} \log_{b}\mathopen{}\left(\frac{A}{B}\right)\mathclose{}=\log_{b}(A)-\log_{b}(B) \end{equation*}