 In this final video for lecture 42, I want to talk about solving exponential and logarithmic inequalities. So instead of equations this time, how do you do it inequalities? And like we've seen, for the most part, inequalities, the way you solve them is how you solve the equations. But you have to be kind of cautious of the signs. You could try graphing these things. Graphical approaches worked really great because you know, you're often looking for are my above the x-axis or below the x-axis. That's approaches very appropriate here. But there's just a few things I want to point out when you're working with exponentials or logarithms. Because of the nature of exponential graphs, note that a to the x is greater than zero for all x and for all positive a's. So the point is, if you have an exponential expression, it has to always be positive. You can't ever have it be negative. So if you had to solve something like the following a to the x is less than negative one, there would be no solution in that situation. Your exponential can never be negative. So those are some things to look out for. When you're solving this thing, you have to make sure that you're within the range of the exponential, which the exponential always has to be positive, okay? And so this simplifies many inequalities involving exponentials because again, if you're trying to make an exponential be negative, it's not gonna happen. Also, if needing to solve an exponential of the form, let's say like a to the x is greater than equal to b or something like that, where b is some positive number, okay? Because I should mention that b with a negative number would be all real numbers would be x. But if b is some specific positive number, then we can switch the inequality to logarithmic form, right? We can move the exponential base a to the right-hand side becomes a logarithm. And the way we do that is we basically just apply the logarithm to both sides. So we take log base a on the left-hand side and we do log base a on the right-hand side as well, for which on the left-hand side it'll cancel, you would x is greater than or equal to the log base a of b. Now the thing you have to be careful about here is when you apply a function to both sides of an equation, the inequality sometimes slips around, but that happens if your function is decreasing. Your standard logarithm, log base a, is going to be an increasing function assuming that a is greater than one. So if a is greater than one, then the log base a is actually gonna be an increasing function, in which case then when you take the inequality of both sides, the direction will be reserved. Only if you have a decreasing logarithm, you're gonna switch the direction. And that would be if your base was actually between zero and one. So those are things to look out for. So the thing is if you have an increasing exponential, the base is greater than one, then the logarithm will be increasing as well. Now, if you have a decreasing, if you have a decreasing base, right, then your log will be decreasing. So accommodate accordingly. So when we look at this one right here, we have three times two to the x minus one plus five is less than 17. I'm gonna start solving this by performing inverse operations to both sides. Subtracting five from both sides does not change the inequality, right? So we're gonna get three times two to the x minus one is less than 12. Next I'm gonna divide both sides by three, dividing by a positive number does not change the direction of the inequality. So we get two to the x minus one is less than four. Next I'm gonna take the log base two of both sides, log base two of both sides, because again, my base is greater than one. This is an increase in exponential. It's inverse is an increasing logarithm. So that won't change the direction of the inequality. We get x minus one. It's gonna be less than the log base two of four, right? What power of two gives you four? That's gonna be two. And so then finally we're gonna add one to both sides. And so we end up with x needs to be less than three. x is less than three. So when we record that answer in interval notation, we get negative infinity to three, non-inclusive. And that gives us the solution to this inequality. Let's look at another example. Let's take two to the e to the x minus one is greater than equal to one. I wanna solve this inequality. I'm gonna add one to both sides. We end up with two e to the x is greater than equal to two. They're gonna divide by both sides. Now we divide, you have to be careful, right? If it's a positive number, the inequality will be reserved. If it's a negative number, the inequality will be reversed. It's positive two, so we go in the same direction. We get e to the x is greater than one. Now we're gonna take the natural log of both sides, the inverse operation to exponential base e. So we're gonna get x is greater than, base e is a base greater than one. It's about 2.7. So x is gonna be greater than equal to the natural log of one, which is zero. So we see that the solution this time would be zero to infinity, this time inclusive, because we're greater than or equal to. So equality to zero is perfectly fine in that situation. Now, okay, look at this one here. We wanna solve the inequality, one third to the x minus five is greater than or equal to 76. So the first step is quite natural. We're going to just take plus five on both sides. And so we end up with one third to the x is greater than or equal to 81, okay? And so the next approach that you could take, I certainly, you know, you could do it this way or there's another way I'll show you how to do it in a second. You could do the log base one third on both sides, log base one third on both sides, so that the left hand side is gonna become x. The right hand side will become the log base one third of 81, which we'll come back to that one in just a second. But because your base here, your base is one third that's less than one, this is actually a decreasing model. This is a decay model in terms of exponentials. And that also means it's gonna be a decay model in terms of the logarithms. So this will actually switch the direction around. So we get a x is less than log base one third of 81, which that equals negative four. So that would tell us that the solution would in fact be negative infinity to negative four bracket there, negative four being included in the solution. So that would be an acceptable way of solving this one. Now there are two reasons why one might not wanna do this, right? One is because since your base is decreasing, it's a decreasing model, that means you have to switch the things around. So that can sometimes be a little bit confusing, especially as you know, we're not as accustomed to exponentials and logarithms as perhaps other functions. So that's something to worry about. Then the other issue is like log base one third of 81, how'd you get negative four? I mean, we can work through the details of that, but that can be a little bit confusing. So an alternative approach one could take to try to avoid some of these confusions is when you have one third to the x is greater than or equal to 81. What you could do is you could actually rewrite the base on the left hand side as a positive base. We can change it to be base three, which notice is greater than one. The way to do that is that the one third becomes three to the negative one to the x greater than equal to 81, for which then when you combine the powers together, you get three to the negative x is greater than 81. Now you take the log base three of both sides, log base three, log base three of both sides, that'll give you negative x is greater than or equal to log base three of 81, which you're probably more comfortable with that. What power of three gives you 81? That's the fourth power. Notice because now your base is positive, it's a growth model, it's increasing, you actually don't have to switch the inequality around, you get negative x is greater than four, for which then you're gonna times both sides by negative one, and you'll get x is less than negative four. So you get the exact same expression that we did before, that leads to negative infinity to negative four, but the inequality switch direction, that happened, but it happened when you times by negative one, not when you had the decreasing logarithm. And the reason why that can be a little bit more advantageous for your typical algebra student is that's something you're used to by now, you're used to multiplying by negatives, switches the direction of inequalities, as opposed to the decreasing logarithms, things like that. So that's an alternative, there's a green approach, I'm not saying you had to do it, but it is an approach to solve these exponential inequalities here. I should also mention a little bit about logarithmic inequalities. Similarly, inequalities involving logarithms can be solved by switching to the exponential form. One also has to be careful when working with logarithmic inequalities, because the domain of the logarithm is often restricted, it's important not to include values of x into the solution set outside of the domain of the logarithm. We also have to worry about the base, right? When you switch to exponential form, if you have a base greater than one, that's increasing, the inequality stays the same. If the base is less than one, like between one and zero, then that would switch directions. That same principle is still true, but also pay attention to the domain, which notice here the domain, right? The domain of that logarithm would mean that x plus five has to be greater than zero, so x has to be greater than negative five. So that's sort of a restriction that's gonna be on the solution set right here. So then the next thing to say is that I'm gonna switch this to the exponential form, right? So basically I'm taking 10 to the log base x plus five on the left, and then on the left hand side, you're gonna get 10 to the first as well. So that those cancel out, you get x plus five is less than or equal to 10 to the first, which is 10, right? Then just minus five from both sides, and you get that x should be less than or equal to five. And so to solve this inequality, we have to put these two statements together. So the inequality says that, oh, x needs to be less than or equal to five, but the domain says x has to be greater than negative five. So we put this together, we get negative five to five. We will include the positive five because the positive five makes the right hand side equal, the left hand side equal to one, inequality is okay. We will not include negative five because negative five would make us take the log of zero, which that does not exist, it's undefined. And so we won't include negative five because that's the vertical asymptote, but we will include five because that gives us, well, that gives us equality and that's perfectly fine. So again, solving inequalities with logarithms is basically the same way that you solve inequalities with exponentials, you have to pay attention to your base. Do I have a growth model or a decay model, right? Is it increasing or is it decreasing? If you're careful about that, you'll be fine solving these things, but with the logarithms, you have the extra stipulation, you have to look at the domain. If you didn't look at the domain, you might think, oh, everything less than or equal to five is okay. You might've said the solution was negative infinity to five, which is actually it's not because we can't get anything less than negative five here.