 This video is going to talk about exponential equations. So if we look at these problems, we're going to solve them either by inspection or trial and error. So we're thinking 5 to the what is 125. Actually if we wanted to, we could go to our calculator and say y equal and clear it out and say 5k at x and then look at our table. That would be inspection. And these are way too big. But when we get here, we see that, oh, here at 3, x is 125. Now if we look at this one, 1 over 100, that'd be a little bit harder to see in our calculator, although we know what that decimal is. But if you think about it, let's think about this intuitively. This is a base of 10 on my, with my exponent. And on the other side, I have 1 over 100. And you could think of this as the same thing as 1 over 10 squared is equal to 10 to the t. Well, if I have 1 over 10 squared, that's really 10 to the negative 2. And that's the same thing as 10 to the t. So that must tell us that t is equal to negative 2. So let's look at this example. It's the same kind of idea. 3 to the a is 1 over 27. So we have 3 to the a, but I want it to be a 3. If I can think of it that way, and 27 would be the same thing as 3 to the third. So 1 over 3 to the third would actually be 3 to the negative 3. That would be the same thing as 1 over 27. And if 3 to the a is equal to 3 to the negative third, if my bases are the same, then my exponents must be equal. Now, if I have a problem like this one, I want to do the same thing that I've been doing so far, but I've got this 3 in front of it. So I've got to do something with that 3. I've got to divide it off. So 4 to the x is going to be equal to 192 divided by 3 is going to be 64. And so now you can either say 4 to the first is 4, squared would be 16 to the third. Oh, there it is. 4 to the third is 64. And if you're not really not sure about that, you want to double check it again, come back into your calculator. 4 to the x and look at your table. And there it is. 3 is 64. One last one. Now, we've got a lot of things going. But remember, we are trying to get to 3 to the t all by itself. So we have to pull all the layers off. So the first thing we want to do is subtract 12, just like we are solving any equation. So we have 81 times 3 to the t on this side. And then if we subtract the 12, we're left with 3. And then we're going to divide by 81. So 3 to the t is going to be 3 over 81. But that reduces. And the reduced form might be more helpful. So let's see what happens if we reduce it. 3 goes into 81 perfectly. So it would be 1 and it goes in 3 times 27 is 81. So now we're back to a problem that we've actually already solved earlier. 3 to the t is 1 over 27. 3 to the t is then the same thing as 3 to the negative 3. That would be 1 over 27. And my exponents are the same since my bases are the same. So t for negative 3. Now let's look at some problems where the exponent isn't a variable. This time the exponent is a number. Well, how do you solve those differently? If we solve these, remember we want to take and get this to be an exponent of 1. One way to do it would be just to say, well, if I use my exponent properties, I could multiply my exponents five times one-fifth would give me x to the first. And if I take that side to the one-fifth, I can take this side to the one-fifth. And in my calculator, 7, oops, I have to get on my home screen first, 7,776 carat, and then it's a fraction. So I have to put it in parentheses 1 divided by 5. And that is 6. Or some of you might have thought, what is she doing? Why didn't she just take the fifth root? Because the root and the exponent will cancel each other out. And that's exactly the same thing as what we did with the exponents. If you remember when we were talking about exponent properties, the denominator of a fraction is the root. So if I really was taking a fifth root, I was just using exponents. Exponent of five cancels out the root of five because they're opposite operations. So that's just x. And then in our calculator, if you remember how to do these, you first have to put the root number in, and then you go to math, and you choose option five because that's the x-root. We've already put the x-part in. And then we can put our 7,776 in there. And we still get 6. Alright, x to the fourth, but it's got a 36 on the front of it. You have to divide off the 36. You've got to get to that base by itself, just like we did when we were solving with the exponent as the unknown. And I don't know what that is right up top of my head. So 182.25 divided by 36 is going to give me x to the fourth being equal to, isn't that beautiful, 5.0625. But that's okay because I can just take either the fourth root or I can take to the 1 fourth, depending on which one's more comfortable for you. The fourth root of x to the fourth is going to be x. The root is 4. Math, option five. And since I already have 5.0625 here in my calculator, I actually, since I'm using my calculator, could just say second. And then the negative down here in the blue says answer. So it's automatically going to put my 5.0625 in there for me. And I find out that it's 1.5. And if we just want to double check that, we could just plug and check. 36 times 1.5 to the fourth. Let's see if we get 182.25. 36 times 1.5, care at four. And sure enough, 182.25. So we know that x is equal to 1.5. There's two more problems. So I need to subtract the 128. And when I subtract 128 from 3.45, I'm going to get 217. So x to the eighth is equal to 217. And I'm going to take the eighth root again. And the eighth and the eighth root cancel each other out since they're opposites. So I have x and then 8. Math, option five, 217. And I have 1.959. So we'll call it 1.96. That's an approximation. So if I plugged that one back in, 1.96 to the eight and then plus the 128, I probably won't get exactly 345. But I should be close if I found the right answer. So let's try it out. 1.96, care at eight and then plus my 128. And that gives me 345.79. So I'm a little bit over. But remember, I rounded my 1.96 and that's y. Last problem, it says we have 2 to the x plus 7 equal 35. And we want to know how that's different from the ones that we just solved. Well, it's different because the x is in the exponent. It's not the base like it was in the previous problems. It's the exponent. So how might we solve this one? Well, I could try and subtract my 7. So I'd have my 2 to the x and 35 minus 7 would be 28. But I can't say 2 to the something is 28. If you go and look in your calculator and make this a 2 to the x and go look at our table, I can estimate it. I can say, well, it's somewhere between 16 and 32. So I know it's going to be four point something. So x is approximately four point something. But I don't really know specifically. The only way I know how to solve this one right now is to actually just graph it. So I'm going to graph it. And since I want to stay in a standard window, I'm just going to say that this is 2 to the x and I'm going to bring everything to the other side. So minus 28 equal to 0. Then I'm going to look for my x intercept. So if I graph it, I have 2 to the x again, and then minus 28 and zoom 6 for a standard window. And I get a graph that looks like something like this and my intersection point second trace 5, enter, enter, enter, enter for the second. I don't have a second equation in there. That happens a lot. So I'm glad I did that. It kept stuck on second curve because I didn't have a second equation in there. I needed to tell it that I wanted it to be 0. So now second trace and 5. Now I'm going to press enter for the first curve and enter for the second curve. It passes the second curve and goes to guess and it tells me that that is 4.8 and 0. So x must be approximately 4.8.