 Hi, I'm Zor. Welcome to Unizor Education. I would like to continue solving problems related to logarithms and exponential functions. Maybe these problems just may be a bit more difficult than the previous lecture. Well, in any case, I think it's very important to basically be prepared for any kind of ingenuity the problem creators came up with. So let's not be afraid of this. It looks a little scary maybe in the beginning, but believe me, it's not really very difficult at all. All right, so I'm sure you have spent some time looking at these problems before listening to the lecture. If you didn't, just pause this video and I strongly recommend you to try to solve these problems yourself. They are all in the notes for this lecture on Unizor.com, so go ahead and try to do it yourself first. Then listen to this. All right, so problem number one, 0.5, 5 over 3 minus x times square root of 2 minus 0.25 to the power of 2 minus x equals to 0. Well, what you really have to notice here, this is one-half, well, somehow related to number two, right? This is square root of 2, also related. This is one-quarter, also related to number two. So let's bring everything to the powers of 2. I mean, that's exactly what kind of unifies the whole equation and will make it much better. So what is 0.5? It's 2 to the power of minus 1, right? That's one-half. And then I multiply it, I mean, I raise it to the power of 3 minus x, okay? Square root of 2 is 2 to the power of one-half, minus. One-quarter is 2 to the power of minus 2. Minus will inverse it to one-half and 2 will square it, it will be one-quarter. And this is to the power of 2 minus x equals to 0. Now, when you raise to the power and then raise to another power, you multiply the powers. Remember this, right? So it's basically 2 to the power 5 over x minus 3. I reverse the sign here. Minus 1 times 5 over 3 minus x is 5 over x minus 3 times 2 to the power of one-half. Now, instead of having minus and then equals to 0, I will add this thing to both sides. So I will have nothing here and it will be on that side of the equation 2 times 2 minus x. Now, if you multiply two exponential expressions with the same base, you know you have to add the exponential part. So it's 2 to the power 5 over x minus 3 plus one-half equals 2 to the power minus 2 times this. It's 2x minus 4. And now, we basically go to a very easy thing. Since these two exponential expressions with the same base are equal to each other, it means their powers, exponential parts, should be equal to each other, which means we can go to this particular equation. Obviously, you can multiply both parts by x minus 3 and you will get a quadratic equation. Now, what's wrong with this solution? Well, let's finish it up, actually. So it will be 5 equals 2xx minus 3 minus, one-half would be here, so it's 9 seconds times x minus 3, right? Well, you probably should multiply by 2. So it will be 10 equals 2x square minus 6x. Well, we have to multiply by 2, so it's 4 minus 12x minus 9x plus 27. 9x 27 to multiply here 4 seems to go like the right equation. Alright, so simplifying it further, we will see that this is 4x square minus 21x plus 17 equals 0 and solutions are 21 square is 400, 441 I guess, 441 minus, 4 times 17 times 4 is 40, 28, 68 times 4, 70 times 4 is 80 minus 8 is 72, something like this, if I'm not mistaken. And 441 minus 272, 441, 272, it's 9, 3, 6, 1, 169. 169 is 13 square, which is very convenient, so we can actually have plus minus 13. So what are our solutions and what's wrong with them? One solution is 34, 8, or 17 fourths, okay, 17 fourths, and another solution is, that's 8, so it's 1. So we have two solutions, okay. I'm not saying something wrong with them, but what's important is to analyze these solutions against the domain of the original equation. So let's go back to original equation, which is this one, and think about if there are any restrictions on these things. Well, the exponential function can have any argument. So the question is, what's the values of arguments? So x not equal to 3, that's the condition here. Now here, we don't have any restrictions. So basically both are fine, and both are satisfied this particular condition, which means that's the right solutions for this particular equation. That's it. It's very important to check your solutions against the domain of all the functions participating in the equation. 4 to the power of 3x equals 3 to the power x square times 2 to the power 3 minus x. Well, incidentally, we don't have any restrictions on the value of argument x here, because these are all exponential functions, so everything is cool. Right, okay, what should we do about this? Well, the best thing is to apply logarithm to both sides, left and right. But before doing this, I would probably convert 4 into 2 to the power of 2 to the power of 3x, which is I have to multiply the exponential part. So it's 2 to the power of 6x equals 3x square times 2 to the power of 3 minus x. Now I will apply logarithms, base 2 probably would be the most convenient thing. So if I apply logarithm to both sides, left and right, I will have legitimate equation with logarithms. Now log 2 of 6x, sorry, 2 to the power of 6x equals log of 3 to the power x square, now this is base 2, times 2 to the power of 3 minus x. Now, what this is? Which power should I use to raise 2 to get to the 6x, obviously 6x. Now here, first what I should do is obviously represent logarithm of the product as a sum of 2 logarithms. So it will be logarithm 2, 3 to the x square plus log base 2 of 2 to the power of 3 minus x. Alright, so that's obvious. Now, this is also easy. This is 3 minus x. Now, how about this? Well, this is not actually the easiest thing to do. We would rather prefer to have logarithm base 3, not base 2. But we can very easily convert it, one into another. You remember that log A B times log B C equals to log C base 8, right? So in this particular case, I would like to convert log base 2 into log base 3. So I have to basically divide these two things by log A B, log A B. Now A is our 2, B is the 3, and C is this thing. So it will be 6x is equal to, so instead of log 2, I will use log 3 of 3 to the x square. Which is this one. But then I have to divide it by log base 3 to the plus, as I was saying, this is 3 minus x. Now, what should I do with this? Well, obviously if this is, the power can always be taken out from the logarithm as a multiplier. So my new equation would be 6x equals log 2 base 3 x square log 3 base 3, which is 1, right? So I leave it as is. Now what is this? Well, this is a quadratic equation, which we can always solve. This is just a constant, not a big deal. So I'm not going to do it. Obviously it doesn't have integer solutions or anything like that. But whatever it is, it is. Now, since original equation does not present any restrictions on the argument, both solutions to this quadratic equation actually are good. If there are any solutions in this particular case. Whatever this quadratic equation presents us as solutions, are solutions to original equation. I'm not going to go through this quadratic equation for obvious reasons, but you should do it. Let me just say that whatever I'm doing here is just a help. You really have to drive the whole thing to the conclusion and get the number, and it would be great if you can even check the results. Next. Okay. Absolute value of x plus 3 to the power of x. And then the whole thing to the power of x plus 2 minus absolute value of x plus 3. The same thing, which is important to the power of x plus 4 equals to 0. Well, again, you are raising to the power and then to another power, which means you multiply the powers. So it would be x times x plus 2. Now, this thing can be moved to the right, so it's equal. This is minus, so it will be plus on this side. x plus 3 to the power of 3 times x plus 4. And obviously, since you have the same expressions for exponential functions, the bases are the same. It means that your exponents must be the same. So x times x plus 2 equals 3 times x plus 4 quadratic equation. And now, back to the original. Are there any restrictions on the argument x? Well, no. Because this is absolute value, as you see. Which means x can be anything because this thing will always be positive. So we can always raise any number to the positive. Well, there is one restriction there. The exponential function usually is considered not only for positive base, but also the base not equal to 1, because then it makes it trivial. I'm not sure if minus 2, that what makes the base equal to 1, if minus 2 is the solution. Well, actually not, because minus 2 will nullify this x plus 2 and will not nullify this. So minus 2 is not a solution. So even if you want to restrict our solutions to non-trivial cases, we still have this quadratic equation as the good one to represent solutions to original. But you see, all these problems, they really look a little scary from the beginning. But when you think about them and you find that there are certain ways to approach them, it doesn't look as scary at all. Next, log x. Now, this is a common or decimal logarithm, since I don't use the base. Minus 2 log 4 third equals 2 log square root of x minus 1. Well, here we will have to do the reverse instead of using the log to convert exponential to the plane one. We will use exponential to convert log into a plane equation. Now, here is how to do it. The difference between two logs is a logarithm of their result of their division. So this is log of x divided by, here I will immediately insert that to inside, so it will be log of 4 third square, right? And which is 16-19ths, so it will be 16-9ths, I mean, 16-9ths. That's how it will be, right? Because 2 log 4 third is equal to log 4 third square, which is log of 16-19. So 16-19ths should be the denominator, so 16 goes down, 9 goes up. That's simple, so that's on the left. What do we have on the right? Well, similarly, we will put these two inside the log, so it will be log square root of x minus 1 square. Now, two logs are equal, which means the arguments are equal. 9x over 16 is equal to, and this I can always represent as x minus 2 square root of x plus 1. How to solve this equation? Well, you assign this, and what you will have? You will have 9y square 16s equals y square minus 2y plus 1. Now, this is a quadratic equation. Now, what's wrong with this picture? Well, in this case, we are actually restricted by the original equation. What's the restrictions on? Forget about this quadratic equation. We're not going to solve it. You do it yourself, but whatever the roots of this equation are, you really have to look back at the original equation. First, log exists only for positive arguments. So that's what this gives us. Here, not only x should be positive, so square root of x actually exists. Also, square root of x minus 1 should be positive for log to exist, which means square root of x should be greater than 1, and considering x is a positive thing, so it's x greater than 1. This is the final restriction on the argument. Whatever roots that particular quadratic equation produces, whatever solutions you can find, they always should be checked against this. And only the solutions which satisfy this inequality are real solutions to this equation. So you should never forget that certain equations actually restrict your solutions. And that's why you have to do the checking and the verification of the domain, etc. x to the power a log x plus b, now it's also a decimal or common logarithm, equals to cx. Well, so far as you see, the only restriction is for this log, x should be greater than 0. And no other restriction succeeds because you can... Well, I'm sorry, there's another restriction. This is the exponential function, but it actually gives exactly the same thing, x is greater than 0. So this is the only restriction. Now, how can we solve this equation? Well, let's just logarithm both sides. Can we do it? Well, a little nuance, c must be... It's just a constant, right? c is a constant, but the whole solution exists only in case c is greater than 0, obviously, right? Because otherwise, the left part is always positive and the right part should be positive as well, therefore. So whatever constants are provided in the beginning as conditions for this particular equation, this is one of the conditions. We don't have to check our solutions against this because this is a constant and supposed to be given, but it's supposed to be given even in this particular range of ways. Now, we can apply the logarithmic function to both sides, left and right. If arguments are equal, logarithms are equal. So what's the logarithm of this thing? Well, this is something to the power of something. The power goes outside of the logarithm, so we will have a log x plus b and then log x of this one, right? So that's on the left. On the right, we have log c plus log x because logarithm of product is sum of the logarithms. Now, what do we have now? Obviously, a quadratic equation. If you will assign the value of y to log x, you will basically have a y square plus b y equals log c plus y. So this is a quadratic equation, so you can find the values of y. Now, obviously, values of y must be converted into values of x. x obviously is equal to 10 to the power of y, right? From here, this is the base 10. So 10 to the power of y is x. So once you find y in this particular equation, this is the solution. And again, you really have to worry about c being greater than 0, and that's supposed to be given. And as a result, by the way, as a result, since x would be 10 to the power of y and y is whatever it is, this is always positive. So the condition x greater than 0 will always be satisfied because of this. Alright, move further. We have one more problem for this lecture. Log base 2 of x plus 12 times log x of 2. What's obvious here is this thing, as I'm sure you understand, can be converted into y. Well, obviously, log 2x times log x. 2 is equal to log 2, 2, which is 1. This standard conversion, which I have used many times and proved that particular conversion in one of the lectures, right? In this case, a is 2, b is x, and c is 2 as well. Now, using this, we can rewrite this as or multiplying by log x. You have log 2x plus 12 equals log 2x times 2. Now, what is 2 times log x? If you want to put this to under the logarithms, it's x squared because the power is factored out in this particular case. And since logarithms are equal to the same base, then you have an equation x plus 12 equals x squared, which is a quadratic equation, and you can always find it. Now, again, what kind of restrictions we have? Well, obviously, this one, x should be greater than 0 and x should not be equal to 1 because only these bases are allowed for logarithms. Now, this x plus 12 should be greater than 0, but this actually is enough to satisfy this equation so there are no other restrictions. Now, here, you have to find out if these two are satisfied. Now, what's the solution? x squared minus x minus 12 equals to 0. So, that's what? 4 and minus 3, right? So, x equals 4 and x2 is equal to minus 3. Am I right? 4 is 16 and 9 plus 3. So, this one is not good because of this condition. Well, that was the last problem I wanted to present today. I do suggest you to go again through these problems. Solve them yourself completely from the beginning to the very end. Make sure you check against the domain of the original equation. Check the results of your calculations, whatever the solutions you have. Put it into the original equation and see if it really satisfies because you see we are doing certain transformations of these equations which are not always invariant. So, there is an interesting lecture about invariant transformations which I put into the equations category. If transformations are invariant, they do not change the solutions. But if transformations are not invariant, that might actually present extra solutions. Just as an example, let me deviate slightly. If you have an equation x minus 1 is equal to 0, the only solution is x equals to 1, correct? Now, what if you will do it differently? What if you transfer into this and then squared? And you will get this equation. Now, raising to the power of 2 is not invariant transformation. And this equation already has two solutions. And this one is not yours. If you will put it here, you will see that this is not really checked. So, non-invariant transformations might actually add you extra solutions. And in some cases, you might even lose certain solutions if you are performing an invariant transformation. So, you have to be very, very careful about this. I might actually spend a lecture about non-invariant transformations, about how they add or remove certain solutions if you are making non-invariant transformations. But this is not about this particular lecture. That's for the future. Thank you very much. Using this website unizord.com, I do suggest you to register and take exams. Then you will need a supervisor or a parent who will enroll you in the course where you will take exam. So, that would allow you to much better control the educational process. Thank you very much. That's it for today.