 Welcome back to our lecture series math 1050 college algebra for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In lecture 39, we want to introduce the idea of a logarithmic function for which we will do that in just a second by the end of this short little video right here. So what I want to do is remind us what we've learned about exponential functions so far. So with the stipulation that we have some exponential function f of x equals a to the x where a is a positive number and other than one, then the graph of the exponential function will look something like this. This is a growth model. If we had a value less than one, we would get a decay model like so that's acceptable, but we'll just stick with the growth model going on right now. You'll notice that when you look at this exponential graph that this function passes the horizontal line test and therefore exponential functions are in fact one to one. They're one to one functions, which the consequence of that is actually very important for solving equations. Because the functions one to one, we know that if a to the u equals a to the v, then u equals v. So if two exponential expressions are equal to each other, it turns out that the powers must have been equal in the first place. And so we can basically cancel out the bases and just equate to the exponents equal to one another. This can be a useful strategy when trying to solve equations involving exponential expressions. Take us case in point the equation three to the x plus one is equal to 81. So the left hand side is an exponential expression. If I could rewrite the right hand side as an exponential base three, then you can cancel out the bases and then equate the exponents equal to each other, which we can do that here of course. Notice 81 is actually three to the fourth. You get three times three, which is nine times three, which is 27 times three, which is 81. And so therefore because the exponential function three to the x is one to one, we essentially can cancel out the bases. And then we have the linear equation x plus one equals four, for which when you subtract one from both sides, you end up with x equals three, the solution to this equation. And then in retrospect, it's easy to check, right? If you plug in three, three plus one is four, three to the fourth is 81. Great. We are able to solve that equation. Let's look at one that's a little bit more involved, but still. Take the number e, right? This is an irrational number approximately 2.7. Could I write both sides of the equation as a power of e? Well the left hand side is already good to go. We have e to the negative x squared. So I'm just going to record that down right there. On the right hand side, it's like, okay, notice you have e to the x squared. When you have a chain of exponents like this, we actually, by exponent laws, know to multiply those together. So that gives us e to the 2x, like so, multiply. Then we have this one over e cubed, which when you divide by an exponential, that's the same thing as having a negative exponent. And so we could rewrite that as e to the negative three. And then as we have a product of two exponential expressions, yet another exponential rule comes into play for which we know to add together the exponents. So we get that e to the negative x squared is equal to e to the 2x minus three. And since we have both the left hand side and the right hand side as base e, we can cancel it out. So the fact it was base e was really not of much consequence here. We end up with negative x squared is equal to 2x minus three. This is the now a quadratic equation. I can just have one side equal to zero. And I also like my leading coefficient to be positive. So I'm going to add x squared to both sides. So that cancels giving us x squared plus 2x minus three is equal to zero. And a quick factorization shows us that we could factor this thing as x plus three times x minus one equals zero. Notice that three times negative one is negative three and three minus one is equal to positive two. This would give us two solutions of x equals negative three and x equals one. For which you can check those in the original equation and see that they in fact work out. All right. Well, that was great. We were able to solve some exponential equations using this one to one property. But what happens when you run across an equation like the following? What do you have like two to the x is equal to five? As you start scanning for powers of two, it's like two to the first is two. Two squared is four. Two cubed is eight. Well, five was in between the two. So the solution x, I can tell you, since the function is increasing, it's got to sit somewhere between two and three. Great. But can I do better than that? Well, you could try to use some type of like maybe rational exponent. Like what if I take two to the two and a half power or something like that? Be aware, if I'm talking about two to the two and a half, we could rewrite that as, of course, as a fraction two to the, let's see, that would be five halves, right, which is the same thing as the square root of 32, right, which isn't quite five either. So you could try to, you could try to estimate it perhaps. But the point is, there's numbers that are kind of missing, right? How do you get something like five when the answer appears to actually be irrational? Because after all, we could find that on our picture. Looking at this graph right here, I mean, we could see that there is a horizontal line that's going to hit the graph at five. Now, middle of this graph is not two to the x. It looks more like it's supposed to be three to the x. But nonetheless, I mean, you look at the graph, there's going to be a location, whoops, there's going to be some location for which there should be some x-coordinate that produces the y-coordinate. That is, that produces five. And so since the function's one to one, there's got to be, and because five's in the range, there's got to be a unique number that works there. And in fact, we can, we can find that number by using the inverse function. Remember, if f of x equals y and the function's one to one, then there's an inverse function such that x equals inverse of y. That is, there should be some function, which if we insert five will produce the number we're seeking right now. And this number is referred to as the logarithm. If we have a number, a is greater than zero and not equal to one, then the logarithm, what we call the logarithm base a, is the inverse function to the exponential. So log base a of x is equal to y whenever x is equal to a y. Coming back to our picture right here, we would anticipate, let me clean this up. Since the function is in fact one to one, we could reflect the graph across the diagonal line y equals x. And so the graph of a to the x is provided here. Its inverse function is provided as well. And the inverse function here is what we call the log base a of x. We'll talk about more of this in the forthcoming videos in this lecture in the next several lectures.