welcome to another video we have a simple equation to solve all we need to do is find X but no matter how good your high school algebra is you can't solve this oh you can solve it if you can sketch the graph of 2 to the x plus X yeah just pick certain values of X and plot a graph and then find where the line Y = 5 crosses it like you see in this graph you can easily pick up the answer approximately or you can just use Desmos or your calculator and you'll be able to get what the point is absolutely however if you don't use a graph because actually graphs solve all equations if you can graph it you can find the answer but say you're not allowed to graph it or graphing it is not what you want to do you want to solve it like an equation like you would solve a normal equation then you will need something special and that's what I want to introduce in this video and in another video I'm going to go in depth um about the function which we call the Lambert W function it's like the natural log function and I'm going to explain how it is very similar to the natural log function that you're already used to before we continue please make sure you like this video you share it you leave a comment in the comment section and be subscribed if you're not let's get into it [Music] let me give you a good idea of what I'm talking about if I told you that e to the x is equal to five if this is all I have remember that you can solve this Yourself by taking the natural log of both sides so you can say that Ln of e to X is Ln n of five and then you can say that this is X because this undoes e because these are inverse functions natural log of e of anything is just that thing so equals Ln of 5 now will I be correct to box this and say this is my final answer the answer is yes because Ln of 5 is a number do I know exactly what Ln of 5 is okay I know because I've memorized it I've I've worked with with Ln of 5 so frequently that I know what it is but you don't really need to write the decimal part you just need to show that X is Ln of 5 and L5 is a number between 1 and two okay so I know that this is 1.609 I know that that's what Ln of 5 is but you're okay with writing the answer as the argument of a function because this is going to be a number the function I'm about to intr uced to you is going to behave like this also it's not going to be Ln it's just going to be a w with a bunch of stuff inside such that if you plug in those values into the function you'll always get your answer but you don't really need to plug in the values when you're solving it algebraically I just need you to show me that it's going to be Ln of something and that's it not Ln the W of something so let's go here so this is the Lambert w function and all it does is it undoes the product of a number with its exponential expression so what does this mean this just means that if I told you that there's a number two and I multiply it by e to the 2 this is not a function actually okay if I plug this into W my answer is going to be two that's what it means okay whatever you plug into it that looks like this something times e to that same thing if you plug if that's the argument of the lambra W function this is what you have this function W is called the product log function so if you have a computer system or you have a calculator that has the product log so this is called the product log product log function so this is the product log function and this is what it looks like so if you want to use the function it is most effective just as the natural log of e to the x is equal to X if you want to plug in anything here so that you can get what is here it has to look like this so you cannot plug in x * e into natural log and expect to get X this is not the correct format so the correct format that works in the lambra W function is something e to something and then you're going to get something back so what is the mission the mission is to be able to write whatever equation you have in such a way that what you plug in here will look like this so this is 2 to the X well this is very different from what we have here remember it must be e so this topic calls for a lot of algebraic manipulations until you can write it in this form so let's get rid of whatever we have and see how far we can go so this is just what we're about to do we will attempt to write this equation in this form there will be something here and this will be on the side and then we can just plug this into the function plug this into the function on the left we're going to end up with whatever is here which is going to be in terms of x and whatever is here is just something we plug into the function so let's begin I have to start thinking right from the beginning remember there is no e here so what I need is something that looks like something times 2 to a power of an exponent so what do I do here keep this here and move this over let's do that leave the exponential expression alone so you're going to have 2 to the X will be equal to 5 - x X so if I divide both sides by 2 to the X see what happens this is over 2 to the X this is over 2 to the X this gives me 1 and what I have here is 5 - x / 2 to the X which is the same thing as 2 to the X It's Beginning to Look a lot like the argument you're looking for why because remember what you're looking for needs to look like this something times an exponential function although we don't have e yet but we can write two in terms of e that's easy for us to do but we want what is up here to look exactly like this so um let me move this to this side so we're going to have 5 - x * 2 to thex is = to 1 what can I do what can I do what can I do look this is I can rewrite this ASX + 5 actually it's better for me to write it as - x + 5 that's what I have here so what can I do here so that what is here is the same thing as this I can add five to this and remember that adding five to an exponent is the same thing as multiplying the entire thing by 2 to 5 right so this is - x + 5 2 to thex multiplied by 2 to 5th because now when I put these two together this is going to become + 5 which looks like this but whatever you do to the left you also do to the right so I'm going to multiply 1 by 2 5 1 * 2 to 5 so watch this I have - x + 5 2 to thex + 5 what is here is now what is on top I can put this in parenthesis and on the right hand side I have 32 32 come on my mind is okay at this point we have a perfect recipe for the lambra W function unfortunately the W function only takes e it does not take two so we have to rewrite this two in terms of e and it's easy because we know that e of the natural log of 2 is equal to two because this always takes this out so we're going to replace two with all of this so watch what's going to happen this is going to be x + 5 instead of writing 2 I'm going to be writing e to the Ln of 2 but that then is raised to power-x + 5 which is equal to 32 I haven't changed anything I just changed the representation of the natural log I mean of two to e to the Ln of 2 but we know from our laws of exponents that this exponent multiplies this so that just by multiplying this by this that's the only thing we have modified and now we have this but what we have done has changed this what is here is no longer what is here in order to make this the same as this we have to multiply this by Ln of 2 and that means we have to go here and multiply it also by Ln of two ah so what we do here it is going to Bex + 5 Ln of 2 e to thex + 5 Ln of 2 is = to 32 Ln of two now we have come to our conclusion if you plug this into the W function you you're going to plug this also into the W function now I don't have enough space to write that I think I should write it but we know now that the W of --x + 5 Ln of 2 e to thex + 5 Ln of 2 will be equal to the W of the right hand side 32 2 Ln of 2 but when you plug this into W it just gives you your hibiscus I don't know what that flower is okay I thought I was sketching flowers but that's a Greek letter therefore we have --x + 5 Ln of 2 will be equal to the W or the product log of 32 ln2 so I just keep calling this the product log or the W okay so how do you get X we know how to solve for x divide both sides by L and two and then we're going to put a minus sign and then we're going to subtract five that's it right so I'm just going to write The Final Answer here because if we divide both sides by ln2 it's going to get rid of this and then this is min - x you move the five there and then when you switch it your answer is going to be X will be equal to 5 minus 32 ln2 ided by the natural log of two okay now I told you before you can leave your answer like this or you can plug in this value 32 ln2 into a product log calculator and you get an answer like this and then you can divide that by the natural log of two and guess what your ultimate answer is going to be it's going to be the same answer we got when we plotted the graph in the beginning so we'll take 1. 716 as the approximate answer to this nice problem and that's it X = 1716 that's if you want to deal with the decimals otherwise just stick to this x = 5 minus the product log of 32 L natural log of 2 wow over natural log of two this answer has a product log and the natural log of all of them logging together never stop learning those who stop learning have stopped living bye-bye