 Hi, I'm Zor. Welcome to Unizor education. We will continue solving problems related to algorithms today. Just a little bit more difficult maybe than the previous lectures. Obviously, it's part of the course. You can't really start from here. You have to go from the very beginning, listen to the lectures, first couple of problem solving lectures, and then we will proceed to this one. All right, so one more comment. You definitely have to try to solve these problems yourself first. Go to the Unizor.com. The comments for this lecture contain all the problems, so try to solve them yourself. Only after that, we'll listen to this particular lecture. And one more detail. In many cases, we are expressing the solution to some equation, let's say, not as a number in some kind of a representation related to either integer number or let's say decimal number. Now, there are other representations of the numbers. For instance, this is also a number. Now, obviously, we can approximate this number with some decimal notation, 1.4, blah, blah, blah, etc. But this is the number in its own right. So, if the solution to equation is written as this or as this, this is it. These are final forms. Same thing applies to logarithms. Sometimes we can say that the result of the equation, the solution to the equation is something like, well, this is a number. You don't have to express it in any better form. There is no better form for this one. I mean, if it's something like this, you can express it better because you know that 3 squared would give you 9, which means that this is, which is an exponent, 3 should be raised into to get 9, then you know this is 2. This is definitely better. So, this is not an appropriate, so to speak, form of solution. But this is because it cannot be any simpler than that. Unless you start approximating with decimal numbers, etc., but we're not talking about this, and these are not really representing a true solution. If solution is this, any kind of a decimal representation would probably be just an approximation, not the exact solution. Okay. Having said all that, we will try to solve a few equations. All right. Question number one. 5 to 2x minus 5 minus 2x minus 2 is equal to 0. All right. This unusually looking equation is really a relatively simple one. And here is why. You know that something like this is actually representable as this, or alternatively as this. So, power to the power is basically a multiplication of powers. So, you know this. Therefore, 5 to the 2x can be represented as 5 to the x to the 2. Now, we will use a special technique, which we can call a substitution. What if we will substitute a new variable called y, which is actually 5 to the power of x? If I will be able to find y, then obviously x is log y base 5, right? So, which power should I raise 5 to get y? Well, this is the power. So, why did they do this? Why did they do this substitution? Well, for a very simple reason. Since 5 to the power 2x is 5 to the x square, 5 to the x I have substituted as y, I can rewrite this equation as y square minus 5 to the x is y, minus 2 is equal to 0, which is at regular quadratic equation, which everybody knows how to solve this equation. Now, obviously solutions are minus 1 and 2. Are these both solutions good if I consider this? Obviously not, because 5 to the power of x, this is an exponential function. It has only positive values, which means this we have to completely reject. This is not a candidate for our solution. But this is good, so I can always say that x is equal to log 2 base 5. And that's the answer. Log base 2 of x minus 2 plus log base 2 of x plus 3 is equal to log 2 x plus 10. This is a logarithmic equation. And for obvious reason, we can use this property of logarithms. Log a of b times c is equal to log base a of b plus log base a of c. So you know this property of logarithms. We can use it here. b would be x minus 2, c would be x plus 3. So on the left, we will have log 2 of x minus 2 times x plus 3. That's what's on the left. On the right, we have log 2 x plus 10. Now, is it better than this? Oh, it's much better, obviously. Now, if the logs are equal with the same base, then the expressions under the log are supposed to be equal as well. Because as you remember, logarithm is a one-to-one correspondence between its domain and its range. So the only thing we have to really be concerned about that the domain for log is actually only positive numbers. So we have to really be very careful when we will get rid of the logarithms. Because when we get rid, we will get x minus 2 times x plus 3 equals to x plus 10. But we should not forget that this should be greater than 0. And this should be greater than 0. So in addition to this particular equation, which is obviously a quadratic equation, we also have to say that x minus 2 times x plus 3 should be greater than 0. And x plus 10 should be greater than 0. So these are restrictions on whatever the roots of this particular equation we can find. All right, so let's do this. And let's do it carefully, always thinking about the proper domain. Now, this one gives me x square minus 2x plus 3x is plus x minus 6 equals x plus 10. Obviously, this is reduced. This is x minus, so x square is equal to 16. x1 equals 4, x2 equals minus 4. So that's what we've got from this quadratic equation. Now, how about this? Well, let's try. 4, obviously this is greater than 0 and this is greater than 0, minus 4. Minus 4 minus 2 is minus 6 times minus 1. It's 6 and x plus 10 is also 6, so it's also fine. So in both cases, both roots of this equation, both solutions are satisfying these conditions, these inequalities, which must be applied to make sure that the whole logarithms make sense. However, I did actually skip one particular step, and which is, you see, I have already started solving the second equation when I have log of x minus 2 times x plus 3. But original equation has them separately. So not only this, but actually I have to apply separately. x plus 2 should be greater than 0 and minus. And x plus 3 should be greater than 0. So if I really apply my domain restrictions to the original equation, not the one which I've got like intermediary one, I have a little bit more strict conditions. And if you will check it now, this one is satisfying, but this one doesn't satisfy this, which means this is not a true solution to our original equation. Because then, this particular logarithms would not be this one. Yeah, actually, yes, both. Both would be negative, so both would not really exist and cannot be a solution. So I have only one solution, which is 4. OK, done this. So jumping to quick solutions in these cases is really kind of not really advisable. So you really have to think about original equation and its original domain of whatever restrictions it puts on the variable. So next one, 4 to the power of square root of x is equal to 256. OK, obviously x should be greater than 0, otherwise the square root would not exist. Now, obviously, the first thing which we can do is we can convert 256 into 4 into which degree? Square is 16. Cube is 64, so it's the fourth degree, 4 to the fourth degree. So 4 to the power of square root of x is equal to 4 to the fourth degree. And as we know, since exponential function establishes one-to-one correspondence, if function values are equal, then the arguments must be equal. So square root of x is equal to 4 and x is equal to 16. That's the solution, and it satisfies our original inequality x greater than 0. Actually, x can be greater or equal to 0 in this case. But since we have obtained a 16 as a solution, that's OK. Now, 4 to the power x plus 3 minus 2 power x plus 2 is equal to 14. All right? Well, we obviously notice this is 4, which is 2 square. So we can always write it as 2 to the power of 2 and then to the power of x plus 3, right? Which is actually multiplication, which is 2 to the power of 2x plus 6. I have to multiply this exponential expression by this. Exponents are multiplied. So that's how I can represent this, which is good. Now, this has 2x plus 6. This has x plus 2. So again, it's not exactly like our first equation, 5 to the power of 2x and 5 to the x. When we can substitute 5 to the x as y square, but close. And here's what we can do. We will still do this substitution. And what it means now, 2 to the power 2x plus 6, since these are additions, this is 2 to the power of 6 and then 2 to the power of 2x, right? Because if you multiply these expressions, the exponent parts are summarized, which is 64 times 2 to the power of 2x, which is 64y square, right? Because 2 to the power of 2x is 2x square, which is y square. Now, 2 to the power of x plus 2, this one, is 2 to the power of 2 times 2 to the power of x, which is 4 times 2x, which is 4y, right? So now we can rewrite this equation as 64y square minus 4y equals 14. And we can solve this equation for y. This is just a quadratic equation. Now, I probably would like to skip the solution of this particular equation. However, I can warn you that y is supposed to be greater than 0. So any solution which is greater than 0 in this particular case is good. Anyone which is not, obviously, should be rejected. I'm just thinking if I can guess the solution to this equation. If it's solvable in some, OK, I just divided by 2. Well, let's just check maybe something like. Now, I'm not really sure. Well, whatever the solution is, if there is a positive solution to this equation, then this particular 2 to the power of x equals to y can be resolved. And x would be equal to log base 2 of y. It's strange. I was thinking that I was kind of putting together problems with easy solutions like this. But this seems to be a little bit too much. But well, whatever. By the way, if there is no positive solution to this equation, maybe there are only negatives or maybe there are no solutions at all. Well, let me try. Now, I'm curious myself. So we will transfer it to this form. And we will have the solution to this would be, now I'm curious myself. So it's 64, 2 plus or minus square root of 4 times plus, since this is minus. So it's 128, 135. So it's 139, 139. This is not a square of anything. Now, it doesn't seem to be an integer solution. But in any case, whenever there is a plus here, then that's definitely a good number because the result is positive and you can always find this. If it's minus, then obviously the result will be negative and it should be rejected. Only the plus should be here, which is 2 plus 139 divided by 64. This is y. And now you have to do the log base 2 of this thing and that's the solution. If I didn't make any arithmetic mistake, if I did, I'm sorry. But in any case, the whole way how you solve this equation is obvious in this case. Let's go next. Log 3 of x square plus 1 equals to log base 2 of x cube. All right, well, the first and obvious simplification of this is since you have the log of something to some power, as you know from the properties of logarithms, this is the exponent gets basically multiplied by logarithms. Plus 1 is equal to, and here we can do exactly the same thing. Since this is x to the power of 3, then I can use this property of logarithms. Now, everything would be great and dangerous if I had exactly the same basis. But the bases are different. What should I do in this case? Well, let's just remember there is a great formula. Log base a of b times log base b of c is equal to log base a of c. It's like if you basically reduce by b in the middle. Now, what does it mean for us? Well, it means that I can always change the base of the logarithm, let's say it's b, to the base a with a proper multiplier. In this case, log c base b is equal to log c a log b a. Provided log b base a is not equal to 0, this formula is correct. Now, in this particular case, applicable to us, I don't like log base 3. I would like to have log base 2. So how can I use this formula? In our case, x is c, 3 is b. So instead of this, I can put log x base 2 divided by log 3 base 2. Instead of log x base 3, I can write this. And this is a constant. So that's just a coefficient of this. Plus 1 is equal to 3 log x 2. Now, as you see, this is a simple linear equation where log x base 2 can be substituted as y, let's say. It's linear equation, which means I can always write something like this. 2 divided by log 3 base 2 y plus 1 is equal to 3 y. Now, linear equation has, obviously, one and only one solution in this particular case. And whatever that solution is, is y. Not going to solve it, but we have to remember that y is always log of x by the base 2. So the real answer, the x would be equal to 2 to the power of y, whatever y we get from this equation. By the way, I'm not really finishing up every particular calculations here, but I do recommend you do it yourself completely. After you are comfortable with the way how this particular equation gets solved. After that, I do recommend you to really, accurately, complete it to some real numbers. All right, now next, line to the power of 1 over x is equal to square root of 3 to the power of x. All right, let's do some manipulation, obviously. It looks like 3 is the base for our future transformation. So 9 is 3 square to the power of 1x, which is actually 3 to the power of we are multiplying. So it's 2 over x. That's what this is. Now this is 3 to the power of x to the power of 1 half, square root, which means this is 3 to x over 2. And that's basically an equation which we have. Now, let's just think from the very beginning about what values x can take. Well, obviously, x should not be equal to 0, because this is the main to this. Now for this, we know that exponential function can take any argument, and the result is always positive. So there is no extra restriction. So this is the only restrictions we have. Now, judging what we have here, we can say that since exponential function establish one-to-one correspondence between domain and the range, if functions are equal, the arguments are equal, which means 2 over x is equal x over 2. Or x square is equal to 4, x is equal to 2, and x is equal to minus 2. These are two solutions, and both actually work fine. Both are solutions to this equation, because both satisfy this particular inequality. Now, another thing which I don't really do, I didn't do the check. When you go through these problems yourself, I do recommend you not only complete all the solutions, but also do the checking. It's a very good practice. Next problem has an interesting financial sense. Let's say you have 100 units of some currency. Well, let's say dollars. You have $100. You would like to invest it in some kind of a bank, and you would expect that in 10 years, it would double, using whatever the interest bank pays you. So let's assume that bank pays you interest once a year on every anniversary. So if the bank pays you an interest, let's say x. For instance, x can be, well, if x is equal to 5%, then x is actually 0.05. So this is a fraction of the capital, which will be added after the first year to whatever it was in the beginning of the first year. It will be added at the end of the second year, relative to whatever it was in the beginning of the second year, et cetera. So after each year, if x is the interest, it's not in percentage, just as a fraction. If x is an interest, then your original sum, whatever the sum you have in the beginning of that year, would actually be multiplied by 1 plus x. And that would be, so if s is the beginning, then s would be transferred after a year into s multiplied by 1 plus x. So let's say bank pays you 10% interest. For instance, 10% interest, which is 0.1. And if you had $100 in the beginning of the year, then at the end, it should add 10%, which is 1.10 of this, which is another $10. So it would be 1.10, which is exactly $100 times 1.1. That's what it is. So every year, your capital is multiplied by 1 plus x. So the problem is, let's say I would like to double my money in 10 years. Which banks should they choose? Which, no, different banks pay different interest. So which bank should they choose to achieve that goal? Well, let's think about it. After the first year, from some amount of money, I will get this amount of money multiplied by 1 plus x. Then the second year, I will get s times 1 plus x. That was in the beginning. And then I should multiply it again by 1 plus x. So it will be 1 plus x squared, et cetera. So if I would like in 10 years to double my capital, then the equation which we are talking about should be 2s, which is double original amount. So from amount s, I am going to 1s times 1 plus x after the first year. s times 1 plus x squared after the second year, et cetera. Up to s times 1 plus x in the 10th degree to get to double the amount which I have. So my equation is basically this. So if my original amount is $100, then in 10 years it would be. And it should be equal to $200. And I have to choose the bank which would give me the interest, which is a solution to this equation. OK. How can I solve this equation? All right. Using logarithms is easy. Obviously, we should reduce it by 100. So I will have 1 plus x to the 10th degree is equal to 2. And what I can do now, I can take logarithm from both sides. Let's take logarithm by base 2, because then this part will be easier. So we took the logarithm from both sides. If my arguments are equal, their logarithms, base 2, should also be equal to each other. If arguments are equal, the functions are equal. Now, I know the property of the logarithm. If you have an exponent here under the logarithm, it can be actually extracted as a multiplier. Now, log 2 base 2 is 1, obviously, as we all know. This is exponent. I have to use for the base 2 to get 2. So from here, we go to log 2, 1 plus x is equal to 1 tenths, which means what? If I raise my base into this power, I will get whatever is under the logarithm, 2 to the power of 1 tenths. x is equal to 2 to the power of 1 tenths minus 1. So that's the solution. So I have to calculate this using whatever the methodology I can. Well, nowadays, we have calculators, obviously. Now, 2 to the power of 1 tenths will obviously be greater than 1, because if you remember, the graph for base 2 would look like this. This is 1. This is 1. And here is 1 tenths. So it's somewhere here. And they have to subtract 1. So it will be this value. So I have to calculate this value using whatever ways and means I have. Minus 1, that will be an x. So I know the x, and I can go to the bank, and I will ask, can you pay me this particular interest? If they can, well, maybe, or greater, that's good too, then you invest money into this bank. If not, go to another bank. By the way, in this particular case, the original amount, 100, is called present value of your investment. And the future amount, whatever you want to get in 10 years, $200. It's a future value. So basically knowing present value and future value and knowing the term, which is 10 years, you can determine an interest. Similarly, if you know the present value and term and interest, you can find out the future value. I mean, it's all interrelated, obviously. And the formula is s present value times 1 plus whatever interest is, let's put it a, whatever term is 10 years, will be equal to s future value. So this is basically equation which combines together all these elements. So knowing, we have one, two, three, four. We have four different variables here. Knowing any three is sufficient to find out the force. And that's what all the bankers are doing. They're just calculating present value, future value, interest, et cetera, et cetera. That's their main involvement. All right, and another problem is related more to physics. There is an element called radium. It's a radioactive element. And radioactive elements have so-called half-life, half-life. Now, because of radioactivity, the mass of whatever original mass we have of this radioactive element is decreasing. Let's say every year, it's decreasing by certain fraction. We will use the same methodology as with the banks. But in case of the bank, your interest actually increasing pure amount, in case of radioactive decay, your amount of radioactive material is decreasing by certain percentage, whatever the percentage is. Now, half-life is basically the number of years. Usually, in case like this, it's a year. It's calculated in years. Number of years it takes to diminish the original mass by half. So if you have some original mass, then certain number of years, it's diminished by some percentage. Every year, some fraction is reduced by some fraction. And the number of years is this. This is your future mass. This is present mass. This is future mass. So in radioactive case, it's exactly the same as interest in the bank, except this is minus, because we are decreasing the amount rather than increasing. Or if you wish, it's a negative interest, so to speak. So here is the problem. For radium, obviously this is the clue into whatever I wrote before. It was present here and future there. It's just divided by present value. So in case of radium, in 1,601 years, amount will decrease by half, which means this ratio is equal to 1 half. So the question is, what's the fraction of losing mass of radioactive radium in one year? Well, this is a very similar problem. You have the term, you have this ratio, so you have to find 1 minus x. I mean, you have to find x. And here it is. Let's just again apply logarithm in both cases. And I will use also logarithm base 2 of 1 minus x, 1601 is equal to log 2 of 1 half. Well, obviously this is equal to minus 1, right? 2 to the power of minus 1 will give you 1 half. Now this is 1601 log of 1 minus x base 2. So from here, having this equation, I know I can calculate what's my 161 times log 2 1 minus x equals to minus 1. So obviously, log 2 1 minus x is equal to minus 1 over 1601. OK, I hope you're not discouraged by this minus sign because we are subtracting from 1. If you remember, log logarithm of 1 with any base is equal to 0. But if you go less than 1, this is y is equal to log base 2 of x, right? So this is 1 and log is equal to 0. Less than 1 argument will give you negative log. So everything is fine. So all you have to do right now is basically to say the following. 1 minus x is equal to 2 to the power minus 1, 1601. So x is equal to 1 minus 2 to the power 1601. So that's the solution. This is how much what's the fraction actually of decreasing the amount of radioactive radium happens in one year, considering that 1601 years is the half-life of radium. Well, that concludes actually all the problems I wanted to present to you today. As you saw, some of them are really kind of practical. The banking problem is practical, very practical. And the physical problem is also quite practical. So you have to know what exactly is the reduction of mass in radioactive case. That's it. Thank you very much. I do remind you to go through these problems again. Go to the very end of every problem, get the solution, and check this solution. And as usually, I do encourage everybody to check in and register on unizord.com as a student, have somebody, as a supervisor or a parent, enroll you in whatever the course. They decide and check your marks on exams. It's very important. And well, until the next set of problems, thank you very much.