We want to start our work with logarithms here, and logarithms are really just a fancy way to write exponents.
Let's look at the correspondence between logarithmic form and exponential form. Here, in exponential form, I have this true statement:
2³ = 8. If I write that in logarithmic form, what I get is log₂8 = 3.
So, you see this logarithm is an exponent—specifically it's the exponent I need on 2 in order to get 8. So, 2³ = 8.
This logarithmic form and this exponential form are equivalent. Here, I have logarithmic form: log₁₀0.01 = -2.
I'll write that in exponential form by saying 10⁻² = 0.01; 10⁻² = 0.01. Now, the question is,
is that a true statement? And the answer is, yes, that is a true statement. 10⁻² = 1/100, which is 0.01.
So, exponential form looks like this; I know this is a true statement; this is what it corresponds to in logarithmic form—
so this, and this, really say the same thing about the numbers 10, -2, and 0.01. Here's another one: 2⁻⁵ = 1/32.
That's in exponential form. I'll write it in logarithmic form as the log₂1/32 = -5.
And that tells me that 2⁻⁵ = 1/32. Next, I have this expression in logarithmic form:
log₉27 = 3/2. Let's check that out: 9—that's the base—raised to this exponent, 3/2, is equal to 27.
Now, is that a true statement? 9^3/2 = 27? The second root of 9 = 3³ = 27, so that's a true statement;
that's also a true statement. So, you just want to have memorized what logarithmic form looks like in this corresponding exponential form like this,
so that you can go back and forth between them. Of course, you're going to be more familiar and intuitive with exponential form
than you are with logarithmic form to start with; but once you get this memorized, everything is going to seem real easy.
Here's some more problems: let's solve for x. I have log₃x = 2. I'm not as familiar with logarithmic forms,
so I'm going to change this into exponential form. This is equivalent to 3² = x; 3² = x.
So, it's always this base to this exponent is equal to this number: 3² = x means that x = 9.
So, that's a pretty simple equation to solve, once I put it in exponential form. Next one: log₈2 = x.
That means that 8ˣ = 2. So, I'll change to exponential form, 8ˣ = 2;
so, what number do I raise 8 to in order to get 2? I'm going to write this as 2^3ˣ = 2; 2^3ˣ = 2;
let's move it over this way—that means that 2³ˣ = 2; 2¹—that means that x must be equal to (I'm just going to go down here);
3x =1; so, x = 1/3. Sorry that's so messy, but the idea is the same; I have this logarithmic form right here;
this base 8 raised to this exponent x must be equal to 2. So, I write that 8², sorry, 8ˣ = 2;
so, 8ˣ = 2. I want to solve this equation, so I write 8 as 2³; then, I have 2^3ˣ —
that's 2³ˣ = 2. And 2 is the same as 2¹. So, since these bases are the same, the exponents must be equal.
3x = 1, so x = 1/3. Here's our next problem. Let's graph y = log₂x. Well, I know that its inverse is y = 2ˣ;
so, y = 2ˣ is the inverse of this right here; so, first I'm going to graph this: y = 2ˣ.
When x = 0, y = 1; when x = 1, y = 2; and when x = 2, y = 4; when x = -1, 2⁻¹ = 1/2;
when x = -2, 2⁻² = -1/4; and so, I end up with this, for the graph of y = 2ˣ—
and I know that that is the graph of the inverse of y = log₂x; so, if I just draw in my line y = x right here,
and reflect this graph over this line, I'll have the graph of my function: y = log₂x.
So, when I reflect, this curve right here of about this line, I get this point, I'll get this point, I'll get this point right here, and it'll go down this way,
and this way; so, I reflect that; it looks like this; so that should be the graph of y = log₂.
So, graphing logarithmic functions is sometimes the easiest way to go, is to graph its inverse first, and then reflect the inverse
about the line y = x to get the graph of the logarithmic function.
Let's look at a few more problems. We want to simplify here: log₂16.
Now remember, logarithms are exponents, and this exponent in particular is the one I need to write on 2 in order to get 16.
So, 2 to what power is equal to 16? The answer to that is 4; 2⁴ = 16. Next problem: I have log₃3.
What exponent do I put on this base 3 to get 3? The answer is 1. Next problem: log₅1; I have log₅1.
What exponent do I put on 5 to get 1? The answer is 0: 5⁰ = 1. Now, my next problem down here: log₆6,
I'll start inside the parentheses. What exponent do I put on 6 to get 6? The answer to that is 1; 6¹ = 6.
Now, I can simplify this log₃1. What exponent do I put on 3 in order to get 1? And the answer to that is 0; 3⁰ = 1.
So, you see logarithms are exponents; and if you know that about them,
you can simplify expressions like this by thinking of logarithms in terms of exponents.