100% organic, all natural logarithms
For many students (after the famous \( \pi \) and mysterious \( e \)), logarithms appear as a truly daunting creature. For one thing, it is likely the first time a student encounters a function given without its accompanying "rule". For example, \( f(x) = 3x+2 \) holds no surprises for anyone, we are explicitly told what to do with the input: "Multiply it by 3 and then add 2". However, logarithm just sits deviously, not letting you in on its secrets.
Let's explore exactly what it means. 
The logarithm has a strange name, but a natural motivation. Think back to when you were solving something like \( x+5 = 10 \). I wanted to get \( x \) by itself, so I needed a way to "undo" addition by 5. How did we accomplish this? Subtraction! Then we moved on to the slightly more complicated \( 2x = 4 \), and realized we needed to "undo" multiplication by 2 with division. Soon you will be solving \( x^3 = 8 \), and other expressions with exponents (powers). The logarithm is the "undo" of this procedure. Let's take a look at what it does.
\[ \log_b (x) \]
How is this like \( f(x) \)? Exactly like it actually. The name of this function is \( \log_b \) and the input is called \( x \) just as before. The \( b \) is a positive number called the "base" of the logarithm, and behaves in the following way. Time to pay attention:
\( \log_b(x) \), in English, says "to what power must I raise \( b \) to get \( x \)?"
I know I sound like a fortune cookie with that phrasing, so lets look at some examples.
\( \log_b(x) \), in English, says "to what power must I raise \( b \) to get \( x \)?"
I know I sound like a fortune cookie with that phrasing, so lets look at some examples.
\( \log_3(3) \) asks...
3 to what power is 3? 1! 
\( \log_4(16) \) asks...
4 to what power is 16? 2! 
\( \log_2(8) \) asks...
2 to what power is 8? 3! 
\( \log_{4}(2) \) asks...
4 to what power is 2? 1/2! 
Let's look at an important consequence or two of the meaning of logarithm.
\( \log_3(0) \) asks...
3 to what power is 0? Well...that's not possible. In fact, we can't raise ANY positive number to any power and get zero. Zero does not make sense as an input, that is, zero is not in the domain of a logarithm. Replace zero with a negative number and we get the same problem, leading to a critical fact: the logarithm function* will only accept positive arguments (inputs). *(For curious minds: This is only true of the real valued logarithm function. If we allow ourselves the full power of the complex valued logarithm function, we are only undefined at zero.) 
\( \log_b(1) \) asks...
b to what power is 1? Well, anything to the power zero is 1. \( 3^0 = 1 \) \( 4^0 = 1 \) \( \pi^0 = 1 \) \( 103481919481985^0 = 1 \)... So no matter what b is, \( \log_b(1) = 0 \)! 
And that's all there is to it. Lesson complete.....or is it? There are a few extra helpful facts about manipulating (in a healthy way of course) the logarithm, and a few quirks in the notation we need to cover. First...SPECIAL BASES.
There are two common special bases. 10 and \( e \). We write \( \log_{10}(x) \) simply as \( \log(x) \), and \( \log_e(x) \) as \( \ln(x) \). They may seem different, but they behave exactly the same way as we've described above. \( \ln(x) \) asks "to what power must I raise \( e\) to get \( x \)"?
Finally, for reasons we will talk about in another lesson, the following tools are incredibly useful.
There are two common special bases. 10 and \( e \). We write \( \log_{10}(x) \) simply as \( \log(x) \), and \( \log_e(x) \) as \( \ln(x) \). They may seem different, but they behave exactly the same way as we've described above. \( \ln(x) \) asks "to what power must I raise \( e\) to get \( x \)"?
Finally, for reasons we will talk about in another lesson, the following tools are incredibly useful.
\( \log_b(xy) = \log_b(x) + \log_b(y) \)

\( \log_b(x/y) = \log_b(x)  \log_b(y) \)

\( \log_b(x^k) = k \log_b(x) \)

It may strike you as interesting that the first identity turns multiplication into addition (provided you know the log), and people used enormous catalogs (no pun intended) for this very purpose. Before quick and dirty calculators, multiplying two mighty numbers from astrophysics required substantially more work than adding, and logs turned the process into something much more manageable.
NOW the lesson is complete, and we will be looking at using logarithms to solve equations involving powers in another lesson. 