An Interesting Exponential Equation

Hello everyone. In this video we are going to be solving an interesting exponential equation. We have 8 to the power x plus 2 to the power x equals 130. We are going to be solving for x values, both real and non-real. So we are going to be looking at the solutions first and then I'll show you a graph. Alright, great. So let's go ahead and use one of the most powerful tools that we have been using pretty much all the time and that is substitution. So let's go ahead and set 2 to the power x equal to something like y maybe. And then 8 to the power x can be written as 2 to the power 3 to the power x which can also be written as 2 to the power x to the third power. Therefore 8 to the power x can be written as y cubed. So that's the critical part. So now we get y cubed plus y equals 130. Now at this point you could probably guess and check your solution but I'm going to show you a way to factor both sides so we can kind of see the solution. And this is helpful especially if you are dealing with larger numbers. Suppose instead of 130 you had a 5 digit number, you could try to factor it and use the same approach. Okay, so let's go ahead and write the left hand side as y times y squared plus 1. So if I can get something similar to that, hopefully I can make a one to one correspondence. So 130 is 5 times 26, right? And remember the shortcut for multiplying by 5, especially with even numbers, you cut the even number in half and add a zero. I think we talked about this in one of the shorts. Anyways, 26 and 5 are related because 26 is 1 more than 25 so I can write it as 25 plus 1 and then I can write this as 5 times 5 squared plus 1. Now notice the similarity between these two expressions and when you set them equal to each other, y times y squared plus 1 equals 5 times 5 squared plus 1, you can safely say that y equals 5 is a solution. We're not necessarily saying it's the only solution, but we're saying that y equals 5 works. So, what about the rest? Well, once you know that y equals 5 is a solution, the rest is easy. You can do polynomial division or, you know, just manipulate the expression or long division or short division, whatever. You can do something. But let's go ahead and do the following. Let's distribute y cubed plus y and same thing here. We're going to get 5 cubed, which can be written as 125 and 5 times 1 equals 5. Obviously, we could do this at the beginning, but it's not always easy to see that y cubed plus y can be written as 125 plus 5, okay? Because there's more than one way to do it. In this case, it's obvious, sort of. Anyways. So, here's what we're going to do. We're going to put everything on the same side. So, bring the y cubed and 125 together plus y and the 5 together. And that equals 0. Now, this is a cubic equation, but it's grouped, and we're going to do factoring by grouping. So, let's go ahead and take a look at this first. What does this look like? And if you said difference of two cubes, you got it right. So, this can be written as y minus 5 multiplied by y squared plus 5y plus 25 by using difference of two cubes. And y minus 5 can be written as 1 times y minus 5. We just put the extra 1 there so, you know, to make factoring easier. Great. So, let's go ahead and take a look at this expression now. y minus 5 is a common factor. So, I can kind of take it out, y minus 5. And then inside, we're going to have y squared plus 5y plus 25 plus 1, which is 26. And that concludes the factoring. Now, we already know y equals 5 is a solution. We don't need to redo it. But let's go ahead and focus on the quadratic. This quadratic can be solved by using the quadratic formula. You can also do, it's not factorable, by the way, you can also do completing the square, but I don't recommend it, especially because 5 is an odd number. Anyways, just use the formula, it's the same thing. Negative 5 plus minus the square root of b squared 25 minus 4 times 26, which is 25 minus 104, which is negative 79. Well, that's a negative number. So, we have to use an i here. So, we're going to write it like this, which means these solutions are non-real. Right? Wow. Interesting. So, I found the complex non-real solutions as well in addition to 5, but we're not looking for the y values, right? We're looking for the x values. What is x? Okay. Great. Let's go ahead and take a look at the x values. So, to find the x values from the y values, we have to go back and back substitute. Remember, 2 to the power x equals y. Right? We know that one of the values of y is y equals 5. So, if you set 2 to the x equals 5, from here you get x equals log 5 with base 2, or you can ln both sides if you want. So, you can kind of write it like, let's see, how about ln-ing both sides like natural log? I like that better. Move the x to the front, x ln 2 equals ln 5. And from here, x is going to be ln 5 over ln 2. By the way, the graph doesn't show the exponential. It just shows the polynomial. So, the root that you see is going to be the solution for y, not for x. I just realized that later on after I made the graph and all these things. Anyways, let's go ahead and take a look at the other solution, 2 to the power x equals negative 5 plus minus the square root of 79i over 2. Obviously, if you ln both sides, ln 2 to the x equals ln this. And this is a complex number, so ln-ing it is going to be a little complex. So, I'm just going to write it like this and leave it at that, but, you know, it can be done. How is it done? Okay, something to think about. Anyway, so this is the other two solutions that are not real, of course. Now let's go ahead and, oh, one thing that I wanted to bring up before we start looking at the graph, why is there only one real solution to this equation? That's something that I wanted to bring up because it's a really cool idea. So suppose we have the function f of t equals t cubed plus t. So I'm going to show you that, I'm going to explain why this function has, or why this equation has one real solution. And when you look at the graph, obviously, it's going to make more sense. If you differentiate f with respect to t, you get 3t squared plus 1. What do you know about this? This expression is always positive, which means f is always increasing. Remember, the first derivative tells you whether the function is increasing or decreasing or it has a horizontal tangent. Of course, you have to look at the change of sign, the second derivative, so on and so forth. But basically, we have a function that's constantly increasing, and it's only going to intersect the horizontal line y equals 130 at one point. And sorry, I had to mess with the scale so I could show you the graph because 130 is actually pretty large, and I didn't want to zoom out that much. And this brings us to the end of this video. Thank you for watching. I hope you enjoyed it. Please let me know. Don't forget to comment, like, and subscribe. I'll see you tomorrow with another video. Until then, be safe, take care, and bye. Bye-bye.